size_t Format specifiers in c?

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I want to print out a variable of type size_t in C but it appears that size_t is aliased to different variable types on different architectures. For example, on one machine (64-bit) the following code does not throw any warnings: size_t size = 1; printf("the size is %ld", size); but on my other machine (32-bit) the above code produces the following warning message: warning: format '%ld' expects type 'long int *', but argument 3 has type 'size_t *' I suspect this is due to the difference in pointer size, so that on my 64-bit machine size_t is aliased to a long int ("%ld"), whereas on my 32-bit machine size_t is aliased to another type. Is there a format specifier specifically for size_t? c  platform-independent  size-t  format-string  format-specifiers share improve this question edited Jun 3 at 17:41 PJTraill 878 2 20 asked  Jan 24 '10 at 3:47 Ethan Heilman 5,725 7 48 81

Your warning message does not match the code. The warning mentions pointers, your code doesn't have any. Did you remove some & somewhere? –  Jens  Apr 17 '12 at 14:38 Pointers? No I don't get any warnings about pointers, in fact depending on what machine I run that code on sometimes I get no warnings at all. Try the following test code: #include <stdio.h> int main(){ size_t size = 1; printf("the size is %ld", size); return 0; } –  Ethan Heilman  Apr 17 '12 at 16:39 1   See also Cross platform format string for variables of type size_t?. –  legoscia  Sep 21 '12 at 10:10 @EthanHeilman He's referring to the fact that your warnings say warning: format '%ld' expects type 'long int *', but argument 3 has type 'size_t *' when it probably should be saying warning: format '%ld' expects type 'long int', but argument 3 has type 'size_t'. Were you maybe using scanf() instead when you got these warnings? –  RastaJedi  Aug 20 at 19:04 add a comment

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Yes: use the z length modifier: size_t size = sizeof(char); printf("the size is %zd\n", size); // decimal size_t printf("the size is %zx\n", size); // hex size_t The other length modifiers that are available are hh (for char), h (for short), l (for long), ll(for long long), j (for intmax_t), t (for ptrdiff_t), and L (for long double). See §7.19.6.1 (7) of the C99 standard. share improve this answer answered  Jan 24 '10 at 3:49 Adam Rosenfield 239k 64 365 491

whats the difference between zd and zu? I get that zd is decimal, but is it signed, if so how does zd being signed effect things. –  Ethan Heilman  Jan 24 '10 at 3:51 1   It's the difference between a size_t and an ssize_t; the latter is seldomly used. –  Adam Rosenfield  Jan 24 '10 at 3:53 16   Right, so in this case, you should be using %zu, because the argument is unsigned. –  caf  Jan 24 '10 at 23:03 The other options available are explained in the printf manual page: linux.die.net/man/3/printf –  INS  Jan 7 '14 at 22:00 2   @detly: No, the z length modifier is not part of C89/C90. If you're aiming for C89-compliant code, the best you can do is cast to unsigned long and use the l length modifier instead, e.g. printf("the size is %lu\n", (unsigned long)size);; supporting both C89 and systems with size_t larger than long is trickier and would require using a number of preprocessor macros. –  Adam Rosenfield  Mar 25 '14 at 6:01 show 1 more comment

up vote 19 down vote	Yes, there is. It is %zu (as specified in ANSI C99). size_t size = 1; printf("the size is %zu", size); Note that size_t is unsigned, thus %ld is double wrong: wrong length modifier and wrong format conversion specifier. In case you wonder, %zd is for ssize_t (which is signed). share improve this answer answered  Mar 1 '14 at 13:24 maxschlepzig 7,180 3 40 67
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up vote 0 down vote	MSDN, says that Visual Studio supports the "I" prefix for code portable on 32 and 64 bit platforms. size_t size = 10; printf("size is %Iu", size); share improve this answer answered  Oct 14 '15 at 19:03 Arkantos 11 3
	1   it's MS specific, which is not standard conforming, so it's not platform independent –  Lưu Vĩnh Phúc  Jun 24 at 7:24 add a comment

45 down vote favorite 14

I want to print out a variable of type size_t in C but it appears that size_t is aliased to different variable types on different architectures. For example, on one machine (64-bit) the following code does not throw any warnings: size_t size = 1; printf("the size is %ld", size); but on my other machine (32-bit) the above code produces the following warning message: warning: format '%ld' expects type 'long int *', but argument 3 has type 'size_t *' I suspect this is due to the difference in pointer size, so that on my 64-bit machine size_t is aliased to a long int ("%ld"), whereas on my 32-bit machine size_t is aliased to another type. Is there a format specifier specifically for size_t? c  platform-independent  size-t  format-string  format-specifiers share improve this question edited Jun 3 at 17:41 PJTraill 878 2 20 asked  Jan 24 '10 at 3:47 Ethan Heilman 5,725 7 48 81

Your warning message does not match the code. The warning mentions pointers, your code doesn't have any. Did you remove some & somewhere? –  Jens  Apr 17 '12 at 14:38 Pointers? No I don't get any warnings about pointers, in fact depending on what machine I run that code on sometimes I get no warnings at all. Try the following test code: #include <stdio.h> int main(){ size_t size = 1; printf("the size is %ld", size); return 0; } –  Ethan Heilman  Apr 17 '12 at 16:39 1   See also Cross platform format string for variables of type size_t?. –  legoscia  Sep 21 '12 at 10:10 @EthanHeilman He's referring to the fact that your warnings say warning: format '%ld' expects type 'long int *', but argument 3 has type 'size_t *' when it probably should be saying warning: format '%ld' expects type 'long int', but argument 3 has type 'size_t'. Were you maybe using scanf() instead when you got these warnings? –  RastaJedi  Aug 20 at 19:04 add a comment

3 Answers

active oldest votes

up vote 70 down vote accepted

Yes: use the z length modifier: size_t size = sizeof(char); printf("the size is %zd\n", size); // decimal size_t printf("the size is %zx\n", size); // hex size_t The other length modifiers that are available are hh (for char), h (for short), l (for long), ll(for long long), j (for intmax_t), t (for ptrdiff_t), and L (for long double). See §7.19.6.1 (7) of the C99 standard. share improve this answer answered  Jan 24 '10 at 3:49 Adam Rosenfield 239k 64 365 491

whats the difference between zd and zu? I get that zd is decimal, but is it signed, if so how does zd being signed effect things. –  Ethan Heilman  Jan 24 '10 at 3:51 1   It's the difference between a size_t and an ssize_t; the latter is seldomly used. –  Adam Rosenfield  Jan 24 '10 at 3:53 16   Right, so in this case, you should be using %zu, because the argument is unsigned. –  caf  Jan 24 '10 at 23:03 The other options available are explained in the printf manual page: linux.die.net/man/3/printf –  INS  Jan 7 '14 at 22:00 2   @detly: No, the z length modifier is not part of C89/C90. If you're aiming for C89-compliant code, the best you can do is cast to unsigned long and use the l length modifier instead, e.g. printf("the size is %lu\n", (unsigned long)size);; supporting both C89 and systems with size_t larger than long is trickier and would require using a number of preprocessor macros. –  Adam Rosenfield  Mar 25 '14 at 6:01 show 1 more comment

up vote 19 down vote	Yes, there is. It is %zu (as specified in ANSI C99). size_t size = 1; printf("the size is %zu", size); Note that size_t is unsigned, thus %ld is double wrong: wrong length modifier and wrong format conversion specifier. In case you wonder, %zd is for ssize_t (which is signed). share improve this answer answered  Mar 1 '14 at 13:24 maxschlepzig 7,180 3 40 67
	add a comment

up vote 0 down vote	MSDN, says that Visual Studio supports the "I" prefix for code portable on 32 and 64 bit platforms. size_t size = 10; printf("size is %Iu", size); share improve this answer answered  Oct 14 '15 at 19:03 Arkantos 11 3
	1   it's MS specific, which is not standard conforming, so it's not platform independent –  Lưu Vĩnh Phúc  Jun 24 at 7:24 add a comment