derivation of return type based on max range of input possible in C++
I was recently asked this question in an interview of C++ where I
was asked to improve the below piece of code which fails when
adding two int
's results in the result being long
and return
type needs accordingly to be derived.
Here the below code fails because the decltype()
based derivation is not intelligent enough to identify based on the actual range of values of input but the type and derives return type as same. Hence we need perhaps some metaprogramming template technique to derive the return type as long
if T
is int
.
How can this be generalized any hints or clues?
I feel that decltype()
won't be helpful here.
#include<iostream>
#include<string>
#include<climits>
using namespace std;
template<typename T> auto adder(const T& i1, const T& i2) -> decltype(i1+i2)
{
return(i1+i2);
}
int main(int argc, char* argv)
{
cout << adder(INT_MAX-10, INT_MAX-3) << endl; // wrong.
cout << adder<long>(INT_MAX-10, INT_MAX-3) << endl; // correct!!.
return(0);
}
c++11 templates template-meta-programming integer-overflow template-deduction
add a comment |
I was recently asked this question in an interview of C++ where I
was asked to improve the below piece of code which fails when
adding two int
's results in the result being long
and return
type needs accordingly to be derived.
Here the below code fails because the decltype()
based derivation is not intelligent enough to identify based on the actual range of values of input but the type and derives return type as same. Hence we need perhaps some metaprogramming template technique to derive the return type as long
if T
is int
.
How can this be generalized any hints or clues?
I feel that decltype()
won't be helpful here.
#include<iostream>
#include<string>
#include<climits>
using namespace std;
template<typename T> auto adder(const T& i1, const T& i2) -> decltype(i1+i2)
{
return(i1+i2);
}
int main(int argc, char* argv)
{
cout << adder(INT_MAX-10, INT_MAX-3) << endl; // wrong.
cout << adder<long>(INT_MAX-10, INT_MAX-3) << endl; // correct!!.
return(0);
}
c++11 templates template-meta-programming integer-overflow template-deduction
For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04
add a comment |
I was recently asked this question in an interview of C++ where I
was asked to improve the below piece of code which fails when
adding two int
's results in the result being long
and return
type needs accordingly to be derived.
Here the below code fails because the decltype()
based derivation is not intelligent enough to identify based on the actual range of values of input but the type and derives return type as same. Hence we need perhaps some metaprogramming template technique to derive the return type as long
if T
is int
.
How can this be generalized any hints or clues?
I feel that decltype()
won't be helpful here.
#include<iostream>
#include<string>
#include<climits>
using namespace std;
template<typename T> auto adder(const T& i1, const T& i2) -> decltype(i1+i2)
{
return(i1+i2);
}
int main(int argc, char* argv)
{
cout << adder(INT_MAX-10, INT_MAX-3) << endl; // wrong.
cout << adder<long>(INT_MAX-10, INT_MAX-3) << endl; // correct!!.
return(0);
}
c++11 templates template-meta-programming integer-overflow template-deduction
I was recently asked this question in an interview of C++ where I
was asked to improve the below piece of code which fails when
adding two int
's results in the result being long
and return
type needs accordingly to be derived.
Here the below code fails because the decltype()
based derivation is not intelligent enough to identify based on the actual range of values of input but the type and derives return type as same. Hence we need perhaps some metaprogramming template technique to derive the return type as long
if T
is int
.
How can this be generalized any hints or clues?
I feel that decltype()
won't be helpful here.
#include<iostream>
#include<string>
#include<climits>
using namespace std;
template<typename T> auto adder(const T& i1, const T& i2) -> decltype(i1+i2)
{
return(i1+i2);
}
int main(int argc, char* argv)
{
cout << adder(INT_MAX-10, INT_MAX-3) << endl; // wrong.
cout << adder<long>(INT_MAX-10, INT_MAX-3) << endl; // correct!!.
return(0);
}
c++11 templates template-meta-programming integer-overflow template-deduction
c++11 templates template-meta-programming integer-overflow template-deduction
edited Nov 18 '18 at 3:51
max66
36.3k74165
36.3k74165
asked Nov 18 '18 at 1:46
Anand KulkarniAnand Kulkarni
748
748
For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04
add a comment |
For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04
For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04
For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04
add a comment |
1 Answer
1
active
oldest
votes
Hence we need perhaps some metaprogramming template technique to derive the return type as long if T is int.
Not so simple.
If T
is int
, you're non sure that long
is enough.
The standard say only that
1) the number of bits for int
(sizeof(int) * CHAR_BIT
) is at least 16
2) the number of bits for long
(sizeof(long) * CHAR_BIT
) is at least 32
3) sizeof(int) <= sizeof(long)
So if a compiler manage a int
with sizeof(int) == sizeof(long)
, this is perfectly legal and
adder<long>(INT_MAX-10, INT_MAX-3);
doesn't works because long
can be not enough to contain (without overflow) the sum between two int
's.
I don't see a simple and elegant solution.
The best that come in my mind is based on the fact that C++11 introduced the following types
1) std::int_least8_t
, smallest integer type with at least 8 bits
2) std::int_least16_t
, smallest integer type with at least 16 bits
3) std::int_least32_t
, smallest integer type with at least 32 bits
4) std::int_least64_t
, smallest integer type with at least 64 bits
C++11 also introduce std::intmax_t
as the maximum width integer type.
So I propose the following template type selector
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
that, given a number of bits, define the corresponding smallest "at least" integer type.
I propose also the following using
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
that, given a type T
, detect the smallest integer type that surely contain a sum between two T
values (a integer with a number of bits that is at least the number of bits of T
plus one).
So your adder()
simply become
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
Observe that th returned value isn't simply
return i1 + i2;
otherwise you return the correct type but with the wrong value: i1 + i2
is calculated as a T
value so you can have overflow, then the sum is assigned to a typeNext<T>
variable.
To avoid this problem, you have to initialize a typeNext<T>
temporary variable with one of two values (typeNext<T>{i1}
), then add the other (typeNext<T>{i1} + i2
) obtaining a typeNext<T>
value, finally return the computed value. This way the sum in calculated as a typeNext<T>
sum and you doesn't have overflow.
The following is a full compiling example
#include <cstdint>
#include <climits>
#include <iostream>
#include <type_traits>
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
int main()
{
auto x = adder(INT_MAX-10, INT_MAX-3);
std::cout << "int: " << sizeof(int)*CHAR_BIT << std::endl;
std::cout << "long: " << sizeof(long)*CHAR_BIT << std::endl;
std::cout << "x: " << sizeof(x)*CHAR_BIT << std::endl;
std::cout << std::is_same<long, decltype(x)>::value << std::endl;
}
In my Linux 64bit platform, i get 32bit for int
, 64bit for long
and for x
and also that long
and decltype(x)
are the same type.
But this is true for my platform; nothing guaranties that long
and decltype(x)
are ever the same.
Observe also that trying to get a type for the sum of two std::intmax_t
's
std::intmax_t y {};
auto z = adder(y, y);
gives an error and doesn't compile because isn't defined a typeFor
for a N
bigger that sizeof(std::intmax_t)*CHAR_BIT
.
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see theadder()
description.
– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
add a comment |
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Hence we need perhaps some metaprogramming template technique to derive the return type as long if T is int.
Not so simple.
If T
is int
, you're non sure that long
is enough.
The standard say only that
1) the number of bits for int
(sizeof(int) * CHAR_BIT
) is at least 16
2) the number of bits for long
(sizeof(long) * CHAR_BIT
) is at least 32
3) sizeof(int) <= sizeof(long)
So if a compiler manage a int
with sizeof(int) == sizeof(long)
, this is perfectly legal and
adder<long>(INT_MAX-10, INT_MAX-3);
doesn't works because long
can be not enough to contain (without overflow) the sum between two int
's.
I don't see a simple and elegant solution.
The best that come in my mind is based on the fact that C++11 introduced the following types
1) std::int_least8_t
, smallest integer type with at least 8 bits
2) std::int_least16_t
, smallest integer type with at least 16 bits
3) std::int_least32_t
, smallest integer type with at least 32 bits
4) std::int_least64_t
, smallest integer type with at least 64 bits
C++11 also introduce std::intmax_t
as the maximum width integer type.
So I propose the following template type selector
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
that, given a number of bits, define the corresponding smallest "at least" integer type.
I propose also the following using
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
that, given a type T
, detect the smallest integer type that surely contain a sum between two T
values (a integer with a number of bits that is at least the number of bits of T
plus one).
So your adder()
simply become
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
Observe that th returned value isn't simply
return i1 + i2;
otherwise you return the correct type but with the wrong value: i1 + i2
is calculated as a T
value so you can have overflow, then the sum is assigned to a typeNext<T>
variable.
To avoid this problem, you have to initialize a typeNext<T>
temporary variable with one of two values (typeNext<T>{i1}
), then add the other (typeNext<T>{i1} + i2
) obtaining a typeNext<T>
value, finally return the computed value. This way the sum in calculated as a typeNext<T>
sum and you doesn't have overflow.
The following is a full compiling example
#include <cstdint>
#include <climits>
#include <iostream>
#include <type_traits>
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
int main()
{
auto x = adder(INT_MAX-10, INT_MAX-3);
std::cout << "int: " << sizeof(int)*CHAR_BIT << std::endl;
std::cout << "long: " << sizeof(long)*CHAR_BIT << std::endl;
std::cout << "x: " << sizeof(x)*CHAR_BIT << std::endl;
std::cout << std::is_same<long, decltype(x)>::value << std::endl;
}
In my Linux 64bit platform, i get 32bit for int
, 64bit for long
and for x
and also that long
and decltype(x)
are the same type.
But this is true for my platform; nothing guaranties that long
and decltype(x)
are ever the same.
Observe also that trying to get a type for the sum of two std::intmax_t
's
std::intmax_t y {};
auto z = adder(y, y);
gives an error and doesn't compile because isn't defined a typeFor
for a N
bigger that sizeof(std::intmax_t)*CHAR_BIT
.
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see theadder()
description.
– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
add a comment |
Hence we need perhaps some metaprogramming template technique to derive the return type as long if T is int.
Not so simple.
If T
is int
, you're non sure that long
is enough.
The standard say only that
1) the number of bits for int
(sizeof(int) * CHAR_BIT
) is at least 16
2) the number of bits for long
(sizeof(long) * CHAR_BIT
) is at least 32
3) sizeof(int) <= sizeof(long)
So if a compiler manage a int
with sizeof(int) == sizeof(long)
, this is perfectly legal and
adder<long>(INT_MAX-10, INT_MAX-3);
doesn't works because long
can be not enough to contain (without overflow) the sum between two int
's.
I don't see a simple and elegant solution.
The best that come in my mind is based on the fact that C++11 introduced the following types
1) std::int_least8_t
, smallest integer type with at least 8 bits
2) std::int_least16_t
, smallest integer type with at least 16 bits
3) std::int_least32_t
, smallest integer type with at least 32 bits
4) std::int_least64_t
, smallest integer type with at least 64 bits
C++11 also introduce std::intmax_t
as the maximum width integer type.
So I propose the following template type selector
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
that, given a number of bits, define the corresponding smallest "at least" integer type.
I propose also the following using
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
that, given a type T
, detect the smallest integer type that surely contain a sum between two T
values (a integer with a number of bits that is at least the number of bits of T
plus one).
So your adder()
simply become
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
Observe that th returned value isn't simply
return i1 + i2;
otherwise you return the correct type but with the wrong value: i1 + i2
is calculated as a T
value so you can have overflow, then the sum is assigned to a typeNext<T>
variable.
To avoid this problem, you have to initialize a typeNext<T>
temporary variable with one of two values (typeNext<T>{i1}
), then add the other (typeNext<T>{i1} + i2
) obtaining a typeNext<T>
value, finally return the computed value. This way the sum in calculated as a typeNext<T>
sum and you doesn't have overflow.
The following is a full compiling example
#include <cstdint>
#include <climits>
#include <iostream>
#include <type_traits>
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
int main()
{
auto x = adder(INT_MAX-10, INT_MAX-3);
std::cout << "int: " << sizeof(int)*CHAR_BIT << std::endl;
std::cout << "long: " << sizeof(long)*CHAR_BIT << std::endl;
std::cout << "x: " << sizeof(x)*CHAR_BIT << std::endl;
std::cout << std::is_same<long, decltype(x)>::value << std::endl;
}
In my Linux 64bit platform, i get 32bit for int
, 64bit for long
and for x
and also that long
and decltype(x)
are the same type.
But this is true for my platform; nothing guaranties that long
and decltype(x)
are ever the same.
Observe also that trying to get a type for the sum of two std::intmax_t
's
std::intmax_t y {};
auto z = adder(y, y);
gives an error and doesn't compile because isn't defined a typeFor
for a N
bigger that sizeof(std::intmax_t)*CHAR_BIT
.
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see theadder()
description.
– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
add a comment |
Hence we need perhaps some metaprogramming template technique to derive the return type as long if T is int.
Not so simple.
If T
is int
, you're non sure that long
is enough.
The standard say only that
1) the number of bits for int
(sizeof(int) * CHAR_BIT
) is at least 16
2) the number of bits for long
(sizeof(long) * CHAR_BIT
) is at least 32
3) sizeof(int) <= sizeof(long)
So if a compiler manage a int
with sizeof(int) == sizeof(long)
, this is perfectly legal and
adder<long>(INT_MAX-10, INT_MAX-3);
doesn't works because long
can be not enough to contain (without overflow) the sum between two int
's.
I don't see a simple and elegant solution.
The best that come in my mind is based on the fact that C++11 introduced the following types
1) std::int_least8_t
, smallest integer type with at least 8 bits
2) std::int_least16_t
, smallest integer type with at least 16 bits
3) std::int_least32_t
, smallest integer type with at least 32 bits
4) std::int_least64_t
, smallest integer type with at least 64 bits
C++11 also introduce std::intmax_t
as the maximum width integer type.
So I propose the following template type selector
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
that, given a number of bits, define the corresponding smallest "at least" integer type.
I propose also the following using
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
that, given a type T
, detect the smallest integer type that surely contain a sum between two T
values (a integer with a number of bits that is at least the number of bits of T
plus one).
So your adder()
simply become
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
Observe that th returned value isn't simply
return i1 + i2;
otherwise you return the correct type but with the wrong value: i1 + i2
is calculated as a T
value so you can have overflow, then the sum is assigned to a typeNext<T>
variable.
To avoid this problem, you have to initialize a typeNext<T>
temporary variable with one of two values (typeNext<T>{i1}
), then add the other (typeNext<T>{i1} + i2
) obtaining a typeNext<T>
value, finally return the computed value. This way the sum in calculated as a typeNext<T>
sum and you doesn't have overflow.
The following is a full compiling example
#include <cstdint>
#include <climits>
#include <iostream>
#include <type_traits>
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
int main()
{
auto x = adder(INT_MAX-10, INT_MAX-3);
std::cout << "int: " << sizeof(int)*CHAR_BIT << std::endl;
std::cout << "long: " << sizeof(long)*CHAR_BIT << std::endl;
std::cout << "x: " << sizeof(x)*CHAR_BIT << std::endl;
std::cout << std::is_same<long, decltype(x)>::value << std::endl;
}
In my Linux 64bit platform, i get 32bit for int
, 64bit for long
and for x
and also that long
and decltype(x)
are the same type.
But this is true for my platform; nothing guaranties that long
and decltype(x)
are ever the same.
Observe also that trying to get a type for the sum of two std::intmax_t
's
std::intmax_t y {};
auto z = adder(y, y);
gives an error and doesn't compile because isn't defined a typeFor
for a N
bigger that sizeof(std::intmax_t)*CHAR_BIT
.
Hence we need perhaps some metaprogramming template technique to derive the return type as long if T is int.
Not so simple.
If T
is int
, you're non sure that long
is enough.
The standard say only that
1) the number of bits for int
(sizeof(int) * CHAR_BIT
) is at least 16
2) the number of bits for long
(sizeof(long) * CHAR_BIT
) is at least 32
3) sizeof(int) <= sizeof(long)
So if a compiler manage a int
with sizeof(int) == sizeof(long)
, this is perfectly legal and
adder<long>(INT_MAX-10, INT_MAX-3);
doesn't works because long
can be not enough to contain (without overflow) the sum between two int
's.
I don't see a simple and elegant solution.
The best that come in my mind is based on the fact that C++11 introduced the following types
1) std::int_least8_t
, smallest integer type with at least 8 bits
2) std::int_least16_t
, smallest integer type with at least 16 bits
3) std::int_least32_t
, smallest integer type with at least 32 bits
4) std::int_least64_t
, smallest integer type with at least 64 bits
C++11 also introduce std::intmax_t
as the maximum width integer type.
So I propose the following template type selector
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
that, given a number of bits, define the corresponding smallest "at least" integer type.
I propose also the following using
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
that, given a type T
, detect the smallest integer type that surely contain a sum between two T
values (a integer with a number of bits that is at least the number of bits of T
plus one).
So your adder()
simply become
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
Observe that th returned value isn't simply
return i1 + i2;
otherwise you return the correct type but with the wrong value: i1 + i2
is calculated as a T
value so you can have overflow, then the sum is assigned to a typeNext<T>
variable.
To avoid this problem, you have to initialize a typeNext<T>
temporary variable with one of two values (typeNext<T>{i1}
), then add the other (typeNext<T>{i1} + i2
) obtaining a typeNext<T>
value, finally return the computed value. This way the sum in calculated as a typeNext<T>
sum and you doesn't have overflow.
The following is a full compiling example
#include <cstdint>
#include <climits>
#include <iostream>
#include <type_traits>
template <std::size_t N, typename = std::true_type>
struct typeFor;
/* in case std::intmax_t is bigger than 64 bits */
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool,
(N > 64u) && (N <= sizeof(std::intmax_t)*CHAR_BIT)>>
{ using type = std::intmax_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 32u) && (N <= 64u)>>
{ using type = std::int_least64_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 16u) && (N <= 32u)>>
{ using type = std::int_least32_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N > 8u) && (N <= 16u)>>
{ using type = std::int_least16_t; };
template <std::size_t N>
struct typeFor<N, std::integral_constant<bool, (N <= 8u)>>
{ using type = std::int_least8_t; };
template <typename T>
using typeNext = typename typeFor<1u+sizeof(T)*CHAR_BIT>::type;
template<typename T>
typeNext<T> adder (T const & i1, T const & i2)
{ return {typeNext<T>{i1} + i2}; }
int main()
{
auto x = adder(INT_MAX-10, INT_MAX-3);
std::cout << "int: " << sizeof(int)*CHAR_BIT << std::endl;
std::cout << "long: " << sizeof(long)*CHAR_BIT << std::endl;
std::cout << "x: " << sizeof(x)*CHAR_BIT << std::endl;
std::cout << std::is_same<long, decltype(x)>::value << std::endl;
}
In my Linux 64bit platform, i get 32bit for int
, 64bit for long
and for x
and also that long
and decltype(x)
are the same type.
But this is true for my platform; nothing guaranties that long
and decltype(x)
are ever the same.
Observe also that trying to get a type for the sum of two std::intmax_t
's
std::intmax_t y {};
auto z = adder(y, y);
gives an error and doesn't compile because isn't defined a typeFor
for a N
bigger that sizeof(std::intmax_t)*CHAR_BIT
.
edited Nov 18 '18 at 10:36
answered Nov 18 '18 at 3:38
max66max66
36.3k74165
36.3k74165
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see theadder()
description.
– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
add a comment |
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see theadder()
description.
– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
Thank you max66, I will take a look and try to understand the approach. As you say it perhaps its black and white and there are more intercacies to it than simply int->long
– Anand Kulkarni
Nov 18 '18 at 4:23
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see the
adder()
description.– max66
Nov 18 '18 at 10:37
@DeeptiKulkarni - Sorry: I've forgotten an important point. Answer modified: see the
adder()
description.– max66
Nov 18 '18 at 10:37
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
I checked with interviewer later on a lighter note of chat and he explained that the above is good enough solution. All they needed was to cover int to long, double to long double. :)
– Anand Kulkarni
Nov 20 '18 at 11:50
add a comment |
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For class, you can use User-defined deduction guides;but for function, only explicit instantiation can specify the type you need, otherwise the compiler will deduct the template argument.
– rsy56640
Nov 18 '18 at 2:04