t Format specifiers in c?
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 ?
Yes: use the z length modifier:
size_t size = sizeof(char);
printf("the size is %zdn", size); // decimal size_t
printf("the size is %zxn", 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.
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).
MSDN表示，Visual Studio支持在32位和64位平台上可移植代码的“I”前缀。
size_t size = 10;
printf("size is %Iu", size);
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