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Question

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?

Your warning message does not match the code. The warning mentions pointers, your code doesn't have any. Did you remove some & somewhere? — 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 int main(){ size_t size = 1; printf("the size is %ld", size); return 0; } — Apr 17 '12 at 16:39
See also Cross platform format string for variables of type size_t?. — 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? — Aug 20 at 19:04

Adam Rosenfield 239k 64 365 491 · Accepted Answer · Jan 24 '10 at 3:49

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.

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

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maxschlepzig 7,180 3 40 67 · Answer 2 · Mar 1 '14 at 13:24

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).

Arkantos 11 3 · Answer 3 · Oct 14 '15 at 19:03

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);

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

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Adam Rosenfield 239k 64 365 491 · Accepted Answer · Jan 24 '10 at 3:49