%p Format specifier in c
up vote
13
down vote
favorite
How are the specifiers %p and %Fp working in the following code?
void main()
{
int i=85;
printf("%p %Fp",i,i);
getch();
}
I am getting the o/p as 0000000000000055 0000000000000055
c format-specifiers
edited Sep 28 '12 at 3:14
jonsca
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
add a comment |
up vote
13
down vote
favorite
How are the specifiers %p and %Fp working in the following code?
void main()
{
int i=85;
printf("%p %Fp",i,i);
getch();
}
I am getting the o/p as 0000000000000055 0000000000000055
c format-specifiers
edited Sep 28 '12 at 3:14
jonsca
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
How are the specifiers %p and %Fp working in the following code?
void main()
{
int i=85;
printf("%p %Fp",i,i);
getch();
}
I am getting the o/p as 0000000000000055 0000000000000055
c format-specifiers
edited Sep 28 '12 at 3:14
jonsca
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
How are the specifiers %p and %Fp working in the following code?
void main()
{
int i=85;
printf("%p %Fp",i,i);
getch();
}
I am getting the o/p as 0000000000000055 0000000000000055
c format-specifiers
c format-specifiers
edited Sep 28 '12 at 3:14
jonsca
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
edited Sep 28 '12 at 3:14
jonsca
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
edited Sep 28 '12 at 3:14
jonsca
8,598114757
edited Sep 28 '12 at 3:14
jonsca
8,598114757
edited Sep 28 '12 at 3:14
jonsca
8,598114757
8,598114757
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
asked Sep 28 '12 at 3:05
poorvankBhatia
4,000104890
4,000104890
working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14
add a comment |
working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14
working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14
working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14
add a comment |
5 Answers
5
active
oldest
votes
up vote
18
down vote
accepted
If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.
In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.
And of course, 55 in hex is 85 in decimal.
answered Sep 28 '12 at 3:17
David Christo
691311
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The followingpwill be treated as a normal char to print.
– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
add a comment |
up vote
2
down vote
%p is for printing a pointer address.
85 in decimal is 55 in hexadecimal.
On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
add a comment |
up vote
2
down vote
Here is the compilation output from my machine:
format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]
format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]
so there are warnings but it does compile and the output is:
0x55 0.000000p
I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.
btw: the typical usage of %p would be to print an address i.e. &i as opposed an int
answered Sep 28 '12 at 4:19
user657862
261210
add a comment |
up vote
2
down vote
It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.
And %p is used to printing the address of a pointer the addresses are depending by our system bit.
answered Sep 28 '12 at 5:18
reegan vijay
13410
add a comment |
up vote
1
down vote
Addition to what @Myforwik said
%p is for printing a pointer address.
%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)
and 85 in decimal is 55 in hexadecimal.
I hope its okay now.
References :
http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html
http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.
In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.
And of course, 55 in hex is 85 in decimal.
answered Sep 28 '12 at 3:17
David Christo
691311
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The followingpwill be treated as a normal char to print.
– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
add a comment |
up vote
18
down vote
accepted
If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.
In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.
And of course, 55 in hex is 85 in decimal.
answered Sep 28 '12 at 3:17
David Christo
691311
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The followingpwill be treated as a normal char to print.
– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
add a comment |
up vote
18
down vote
accepted
up vote
18
down vote
accepted
If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.
In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.
And of course, 55 in hex is 85 in decimal.
answered Sep 28 '12 at 3:17
David Christo
691311
If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.
In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.
And of course, 55 in hex is 85 in decimal.
answered Sep 28 '12 at 3:17
David Christo
691311
answered Sep 28 '12 at 3:17
David Christo
691311
answered Sep 28 '12 at 3:17
David Christo
691311
answered Sep 28 '12 at 3:17
David Christo
691311
691311
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The followingpwill be treated as a normal char to print.
– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
add a comment |
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The followingpwill be treated as a normal char to print.
– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
1
1
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The following
p will be treated as a normal char to print.– halfelf
Sep 28 '12 at 3:23
%Fp is not for pointer, depending on compilers. For gcc/clang, %F is for float, similar to %f with a little diff. The following
p will be treated as a normal char to print.– halfelf
Sep 28 '12 at 3:23
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
hmm, so why is he getting 000.... 55 twice and no p? (i agree with you, as i just ran it myself)
– David Christo
Sep 28 '12 at 3:28
add a comment |
up vote
2
down vote
%p is for printing a pointer address.
85 in decimal is 55 in hexadecimal.
On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
add a comment |
up vote
2
down vote
%p is for printing a pointer address.
85 in decimal is 55 in hexadecimal.
On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
add a comment |
up vote
2
down vote
up vote
2
down vote
%p is for printing a pointer address.
85 in decimal is 55 in hexadecimal.
On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
%p is for printing a pointer address.
85 in decimal is 55 in hexadecimal.
On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
answered Sep 28 '12 at 3:14
Myforwik
1,86142537
1,86142537
add a comment |
add a comment |
up vote
2
down vote
Here is the compilation output from my machine:
format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]
format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]
so there are warnings but it does compile and the output is:
0x55 0.000000p
I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.
btw: the typical usage of %p would be to print an address i.e. &i as opposed an int
answered Sep 28 '12 at 4:19
user657862
261210
add a comment |
up vote
2
down vote
Here is the compilation output from my machine:
format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]
format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]
so there are warnings but it does compile and the output is:
0x55 0.000000p
I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.
btw: the typical usage of %p would be to print an address i.e. &i as opposed an int
answered Sep 28 '12 at 4:19
user657862
261210
add a comment |
up vote
2
down vote
up vote
2
down vote
Here is the compilation output from my machine:
format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]
format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]
so there are warnings but it does compile and the output is:
0x55 0.000000p
I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.
btw: the typical usage of %p would be to print an address i.e. &i as opposed an int
answered Sep 28 '12 at 4:19
user657862
261210
Here is the compilation output from my machine:
format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]
format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]
so there are warnings but it does compile and the output is:
0x55 0.000000p
I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.
btw: the typical usage of %p would be to print an address i.e. &i as opposed an int
answered Sep 28 '12 at 4:19
user657862
261210
answered Sep 28 '12 at 4:19
user657862
261210
answered Sep 28 '12 at 4:19
user657862
261210
answered Sep 28 '12 at 4:19
user657862
261210
261210
add a comment |
add a comment |
up vote
2
down vote
It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.
And %p is used to printing the address of a pointer the addresses are depending by our system bit.
answered Sep 28 '12 at 5:18
reegan vijay
13410
add a comment |
up vote
2
down vote
It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.
And %p is used to printing the address of a pointer the addresses are depending by our system bit.
answered Sep 28 '12 at 5:18
reegan vijay
13410
add a comment |
up vote
2
down vote
up vote
2
down vote
It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.
And %p is used to printing the address of a pointer the addresses are depending by our system bit.
answered Sep 28 '12 at 5:18
reegan vijay
13410
It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.
And %p is used to printing the address of a pointer the addresses are depending by our system bit.
answered Sep 28 '12 at 5:18
reegan vijay
13410
edited Sep 28 '12 at 8:04
user647772
answered Sep 28 '12 at 5:18
reegan vijay
13410
answered Sep 28 '12 at 5:18
reegan vijay
13410
answered Sep 28 '12 at 5:18
reegan vijay
13410
13410
add a comment |
add a comment |
up vote
1
down vote
Addition to what @Myforwik said
%p is for printing a pointer address.
%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)
and 85 in decimal is 55 in hexadecimal.
I hope its okay now.
References :
http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html
http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
add a comment |
up vote
1
down vote
Addition to what @Myforwik said
%p is for printing a pointer address.
%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)
and 85 in decimal is 55 in hexadecimal.
I hope its okay now.
References :
http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html
http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
add a comment |
up vote
1
down vote
up vote
1
down vote
Addition to what @Myforwik said
%p is for printing a pointer address.
%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)
and 85 in decimal is 55 in hexadecimal.
I hope its okay now.
References :
http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html
http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
Addition to what @Myforwik said
%p is for printing a pointer address.
%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)
and 85 in decimal is 55 in hexadecimal.
I hope its okay now.
References :
http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html
http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
answered Aug 1 '13 at 15:06
Saurabh Rana
1,60611218
1,60611218
add a comment |
add a comment |
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working as it should be, pointer and float, what's the problem?
– halfelf
Sep 28 '12 at 3:14