Size of formatted string
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I am struggling to understand what happens during snprintf.
Let's say I have two numbers:
int i =11; int k = 3;
I want to format them like this "[%02d] %03dt" and use snprintf.
Afterwards I use the resulting string with write().
snprintf needs the length/bytes n.
I do not understand what is the length I need to provide...
I have 2 theories:
a) It is
sizeof(int)*2
b) I check how many chars the formatted string will contain by counting the digits of the two integers and adding the other chars that the output will have:
2*sizeof(char) + 1*sizeof(char) + 2*sizeof(char) + 3*sizeof(char)+ 1*sizeof(char)
-> digits of i + digits of k + zeros added to first int + zeros added to second int + tab
I am struggling to understand what is the "n" I have to give to snprintf
c printf sizeof
edited Nov 23 '18 at 15:52
Stoogy
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
add a comment |
I am struggling to understand what happens during snprintf.
Let's say I have two numbers:
int i =11; int k = 3;
I want to format them like this "[%02d] %03dt" and use snprintf.
Afterwards I use the resulting string with write().
snprintf needs the length/bytes n.
I do not understand what is the length I need to provide...
I have 2 theories:
a) It is
sizeof(int)*2
b) I check how many chars the formatted string will contain by counting the digits of the two integers and adding the other chars that the output will have:
2*sizeof(char) + 1*sizeof(char) + 2*sizeof(char) + 3*sizeof(char)+ 1*sizeof(char)
-> digits of i + digits of k + zeros added to first int + zeros added to second int + tab
I am struggling to understand what is the "n" I have to give to snprintf
c printf sizeof
edited Nov 23 '18 at 15:52
Stoogy
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
The size you provide as the second argument is the size of the buffer thatsnprintfwill write the text into. And there's a "trick" to get to know the amount of characters needed by usingsnprintfitself (hint: what does it return?). Please see e.g. thisprintf(and family) reference for details.
– Some programmer dude
Nov 23 '18 at 13:46
3
By the way,sizeof(int)gives you the size of anint, not the number of digits it can hold.
– Some programmer dude
Nov 23 '18 at 13:47
add a comment |
I am struggling to understand what happens during snprintf.
Let's say I have two numbers:
int i =11; int k = 3;
I want to format them like this "[%02d] %03dt" and use snprintf.
Afterwards I use the resulting string with write().
snprintf needs the length/bytes n.
I do not understand what is the length I need to provide...
I have 2 theories:
a) It is
sizeof(int)*2
b) I check how many chars the formatted string will contain by counting the digits of the two integers and adding the other chars that the output will have:
2*sizeof(char) + 1*sizeof(char) + 2*sizeof(char) + 3*sizeof(char)+ 1*sizeof(char)
-> digits of i + digits of k + zeros added to first int + zeros added to second int + tab
I am struggling to understand what is the "n" I have to give to snprintf
c printf sizeof
edited Nov 23 '18 at 15:52
Stoogy
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
I am struggling to understand what happens during snprintf.
Let's say I have two numbers:
int i =11; int k = 3;
I want to format them like this "[%02d] %03dt" and use snprintf.
Afterwards I use the resulting string with write().
snprintf needs the length/bytes n.
I do not understand what is the length I need to provide...
I have 2 theories:
a) It is
sizeof(int)*2
b) I check how many chars the formatted string will contain by counting the digits of the two integers and adding the other chars that the output will have:
2*sizeof(char) + 1*sizeof(char) + 2*sizeof(char) + 3*sizeof(char)+ 1*sizeof(char)
-> digits of i + digits of k + zeros added to first int + zeros added to second int + tab
I am struggling to understand what is the "n" I have to give to snprintf
c printf sizeof
c printf sizeof
edited Nov 23 '18 at 15:52
Stoogy
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
edited Nov 23 '18 at 15:52
Stoogy
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
edited Nov 23 '18 at 15:52
Stoogy
761825
edited Nov 23 '18 at 15:52
Stoogy
761825
edited Nov 23 '18 at 15:52
Stoogy
761825
761825
asked Nov 23 '18 at 13:42
wolverinewolverine
345
asked Nov 23 '18 at 13:42
wolverinewolverine
345
asked Nov 23 '18 at 13:42
wolverinewolverine
345
345
The size you provide as the second argument is the size of the buffer thatsnprintfwill write the text into. And there's a "trick" to get to know the amount of characters needed by usingsnprintfitself (hint: what does it return?). Please see e.g. thisprintf(and family) reference for details.
– Some programmer dude
Nov 23 '18 at 13:46
3
By the way,sizeof(int)gives you the size of anint, not the number of digits it can hold.
– Some programmer dude
Nov 23 '18 at 13:47
add a comment |
The size you provide as the second argument is the size of the buffer thatsnprintfwill write the text into. And there's a "trick" to get to know the amount of characters needed by usingsnprintfitself (hint: what does it return?). Please see e.g. thisprintf(and family) reference for details.
– Some programmer dude
Nov 23 '18 at 13:46
3
By the way,sizeof(int)gives you the size of anint, not the number of digits it can hold.
– Some programmer dude
Nov 23 '18 at 13:47
The size you provide as the second argument is the size of the buffer that
snprintf will write the text into. And there's a "trick" to get to know the amount of characters needed by using snprintf itself (hint: what does it return?). Please see e.g. this printf (and family) reference for details.– Some programmer dude
Nov 23 '18 at 13:46
The size you provide as the second argument is the size of the buffer that
snprintf will write the text into. And there's a "trick" to get to know the amount of characters needed by using snprintf itself (hint: what does it return?). Please see e.g. this printf (and family) reference for details.– Some programmer dude
Nov 23 '18 at 13:46
3
3
By the way,
sizeof(int) gives you the size of an int, not the number of digits it can hold.– Some programmer dude
Nov 23 '18 at 13:47
By the way,
sizeof(int) gives you the size of an int, not the number of digits it can hold.– Some programmer dude
Nov 23 '18 at 13:47
add a comment |
3 Answers
3
active
oldest
votes
It is the buffer size
According to a documentation:
Maximum number of bytes to be used in the buffer. The generated string
has a length of at most n-1, leaving space for the additional
terminating null character. size_t is an unsigned integral type.
Suppose you write to an array such as this:
char buf[32];
The buffer can hold 32 chars (including the null terminator). Therefore we call the function like this:
snprintf (buf, 32, "[%02d] %03dt", i, k);
You can also check the return value to see how many chars have been written (or would have been written). In this case, if it's bigger than 32, then that would mean that some characters had to be discarded because they didn't fit.
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
add a comment |
Pass 0 and NULL first to obtain an exact amount
int n = snprintf(NULL, 0, "[%02d] %03dt", i, k);
Then you know you need n + 1
char *buf = malloc(n + 1);
snprintf(buf, n + 1, "[%02d] %03dt", i, k);
free(buf);
See it on ideone: https://ideone.com/pt0cOQ
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
add a comment |
n is the size of the string you're passing into snprintf, so it knows when to stop writing to the buffer. This is to prevent a category of errors knows as buffer overflows. snprintf will write n - 1 characters into the passed-in buffer and then terminate it with the null character.
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is the buffer size
According to a documentation:
Maximum number of bytes to be used in the buffer. The generated string
has a length of at most n-1, leaving space for the additional
terminating null character. size_t is an unsigned integral type.
Suppose you write to an array such as this:
char buf[32];
The buffer can hold 32 chars (including the null terminator). Therefore we call the function like this:
snprintf (buf, 32, "[%02d] %03dt", i, k);
You can also check the return value to see how many chars have been written (or would have been written). In this case, if it's bigger than 32, then that would mean that some characters had to be discarded because they didn't fit.
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
add a comment |
It is the buffer size
According to a documentation:
Maximum number of bytes to be used in the buffer. The generated string
has a length of at most n-1, leaving space for the additional
terminating null character. size_t is an unsigned integral type.
Suppose you write to an array such as this:
char buf[32];
The buffer can hold 32 chars (including the null terminator). Therefore we call the function like this:
snprintf (buf, 32, "[%02d] %03dt", i, k);
You can also check the return value to see how many chars have been written (or would have been written). In this case, if it's bigger than 32, then that would mean that some characters had to be discarded because they didn't fit.
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
add a comment |
It is the buffer size
According to a documentation:
Maximum number of bytes to be used in the buffer. The generated string
has a length of at most n-1, leaving space for the additional
terminating null character. size_t is an unsigned integral type.
Suppose you write to an array such as this:
char buf[32];
The buffer can hold 32 chars (including the null terminator). Therefore we call the function like this:
snprintf (buf, 32, "[%02d] %03dt", i, k);
You can also check the return value to see how many chars have been written (or would have been written). In this case, if it's bigger than 32, then that would mean that some characters had to be discarded because they didn't fit.
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
It is the buffer size
According to a documentation:
Maximum number of bytes to be used in the buffer. The generated string
has a length of at most n-1, leaving space for the additional
terminating null character. size_t is an unsigned integral type.
Suppose you write to an array such as this:
char buf[32];
The buffer can hold 32 chars (including the null terminator). Therefore we call the function like this:
snprintf (buf, 32, "[%02d] %03dt", i, k);
You can also check the return value to see how many chars have been written (or would have been written). In this case, if it's bigger than 32, then that would mean that some characters had to be discarded because they didn't fit.
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
edited Nov 23 '18 at 13:54
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
answered Nov 23 '18 at 13:48
BlazeBlaze
7,5241832
7,5241832
add a comment |
add a comment |
Pass 0 and NULL first to obtain an exact amount
int n = snprintf(NULL, 0, "[%02d] %03dt", i, k);
Then you know you need n + 1
char *buf = malloc(n + 1);
snprintf(buf, n + 1, "[%02d] %03dt", i, k);
free(buf);
See it on ideone: https://ideone.com/pt0cOQ
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
add a comment |
Pass 0 and NULL first to obtain an exact amount
int n = snprintf(NULL, 0, "[%02d] %03dt", i, k);
Then you know you need n + 1
char *buf = malloc(n + 1);
snprintf(buf, n + 1, "[%02d] %03dt", i, k);
free(buf);
See it on ideone: https://ideone.com/pt0cOQ
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
add a comment |
Pass 0 and NULL first to obtain an exact amount
int n = snprintf(NULL, 0, "[%02d] %03dt", i, k);
Then you know you need n + 1
char *buf = malloc(n + 1);
snprintf(buf, n + 1, "[%02d] %03dt", i, k);
free(buf);
See it on ideone: https://ideone.com/pt0cOQ
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
Pass 0 and NULL first to obtain an exact amount
int n = snprintf(NULL, 0, "[%02d] %03dt", i, k);
Then you know you need n + 1
char *buf = malloc(n + 1);
snprintf(buf, n + 1, "[%02d] %03dt", i, k);
free(buf);
See it on ideone: https://ideone.com/pt0cOQ
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
edited Nov 23 '18 at 13:51
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
answered Nov 23 '18 at 13:46
pmgpmg
84.8k999170
84.8k999170
add a comment |
add a comment |
n is the size of the string you're passing into snprintf, so it knows when to stop writing to the buffer. This is to prevent a category of errors knows as buffer overflows. snprintf will write n - 1 characters into the passed-in buffer and then terminate it with the null character.
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
add a comment |
n is the size of the string you're passing into snprintf, so it knows when to stop writing to the buffer. This is to prevent a category of errors knows as buffer overflows. snprintf will write n - 1 characters into the passed-in buffer and then terminate it with the null character.
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
add a comment |
n is the size of the string you're passing into snprintf, so it knows when to stop writing to the buffer. This is to prevent a category of errors knows as buffer overflows. snprintf will write n - 1 characters into the passed-in buffer and then terminate it with the null character.
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
n is the size of the string you're passing into snprintf, so it knows when to stop writing to the buffer. This is to prevent a category of errors knows as buffer overflows. snprintf will write n - 1 characters into the passed-in buffer and then terminate it with the null character.
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
answered Nov 23 '18 at 13:51
mnisticmnistic
7,21611025
7,21611025
add a comment |
add a comment |
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The size you provide as the second argument is the size of the buffer that
snprintfwill write the text into. And there's a "trick" to get to know the amount of characters needed by usingsnprintfitself (hint: what does it return?). Please see e.g. thisprintf(and family) reference for details.– Some programmer dude
Nov 23 '18 at 13:46
3
By the way,
sizeof(int)gives you the size of anint, not the number of digits it can hold.– Some programmer dude
Nov 23 '18 at 13:47