I just found this code on Wikipedia.
Link: http://en.wikipedia.org/wiki/Sizeof#Use
The code:
/* the following code illustrates the use of sizeof
* with variables and expressions (no parentheses needed),
* and with type names (parentheses needed)
*/
char c;
printf("%zu,%zu", sizeof c, sizeof(int));
It states that: "The z prefix should be used to print it, because the actual size can differ on each architecture."
I tried it on my compiler, but it gives the following result:
zu,zu
The syntax of printf() function is given below: printf("format string",argument_list);
The C++ Printf Prototype When you use the printf() function, it prints the string pointed out by the format to the standard output stdout. The format can also contain some specifiers that start with a % and replace values of variables with the printf() function.
"printf" is the name of one of the main C output functions, and stands for "print formatted". printf format strings are complementary to scanf format strings, which provide formatted input (lexing aka. parsing).
The first argument to printf is a string of identifiers. %s refers to a string %d refers to an integer %c refers to a character. Therefore: %s%d%s%c\n prints the string "The first character in sting ", %d prints i, %s prints " is ", and %c prints str[0]. Follow this answer to receive notifications.
Yes that syntax is correct (at least for C99). Looks like your compiler isn't set up to handle it though. Just take out the z
and you'll probably be fine. To be correct, make sure your printf
format specifiers match the size of the types. Turning on all the warnings your compiler will give you probably helps out in that respect.
Your quotation:
The z prefix should be used to print it, because the actual size can differ on each architecture
is referring to the fact that size_t
(which is the type returned by the sizeof
operator) can vary from architecture to architecture. The z
is intended to make your code more portable. However, if your compiler doesn't support it, that's not going to work out. Just fiddle with combinations of %u
, %lu
, etc. until you get the output making sense.
The z
length modifier was added to C in the C99 standard; you might have a compiler that doesn't support C99.
If your C compiler doesn't support that, you can probably treat the sizes as unsigned long:
printf("%lu,%lu", (unsigned long)sizeof c, (unsigned long)sizeof(int));
Yes, but it only works on C99-compliant compilers. From wikipedia:
z: For integer types, causes printf to expect a size_t sized integer argument.
Did you tell your compiler that you want it thinking with a C99 brain? There is probably a switch to do that. For instance, -std=c99
for gcc.
If your compiler does not support it, but you know others will, you can do a PRId64
style work around (disclaimer - PSEUDO CODE AHEAD ..):
#ifdef __SOME_KNOWN_C99_COMPILER
#define PORTUNSIGNED "zu"
#else
#define PORTUNSIGNED "u"
#endif
printf("%-11" PORTUNSIGNED " ways to skin a cat\n");
Its probably better to get a compiler that has functional support for c99, however.
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