How to one print a size_t variable portably using the printf family

cprintf

I have a variable of type size_t, and I want to print it using printf(). What format specifier do I use to print it portably?

In 32-bit machine, %u seems right. I compiled with g++ -g -W -Wall -Werror -ansi -pedantic, and there was no warning. But when I compile that code in 64-bit machine, it produces warning.

size_t x = <something>;
printf("size = %u\n", x);

warning: format '%u' expects type 'unsigned int', 
    but argument 2 has type 'long unsigned int'

The warning goes away, as expected, if I change that to %lu.

The question is, how can I write the code, so that it compiles warning free on both 32- and 64- bit machines?

Edit: As a workaround, I guess one answer might be to "cast" the variable into an integer that is big enough, say unsigned long, and print using %lu. That would work in both cases. I am looking if there is any other idea.

Best Answer

Use the z modifier:

size_t x = ...;
ssize_t y = ...;
printf("%zu\n", x);  // prints as unsigned decimal
printf("%zx\n", x);  // prints as hex
printf("%zd\n", y);  // prints as signed decimal

Related Solutions

Platform independent size_t Format specifiers in c

Yes: use the z length modifier:

size_t size = sizeof(char);
printf("the size is %zu\n", size);  // decimal size_t ("u" for unsigned)
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.

How to print the size of int in C

The sizeof function returns a size_t type. Try using %zu as the conversion specifier instead of %d.

printf("The size of integer is %zu\n", sizeof(n));

To clarify, use %zu if your compiler supports C99; otherwise, or if you want maximum portability, the best way to print a size_t value is to convert it to unsigned long and use %lu.

printf("The size of integer is %lu\n", (unsigned long)sizeof(n));

The reason for this is that the size_t is guaranteed by the standard to be an unsigned type; however the standard does not specify that it must be of any particular size, (just large enough to represent the size of any object). In fact, if unsigned long cannot represent the largest object for your environment, you might even need to use an unsigned long long cast and %llu specifier.

In C99 the z length modifier was added to provide a way to specify that the value being printed is the size of a size_t type. By using %zu you are indicating the value being printed is an unsigned value of size_t size.

This is one of those things where it seems like you shouldn't have to think about it, but you do.

Best Answer

Related Solutions

Platform independent size_t Format specifiers in c

How to print the size of int in C

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