An interesting code of "printf"

Code:

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#include <stdint.h>                                                           
#include <string>                                                             
#include <stdio.h>                                                             
                                                                               
int main(){                                                                   
    uint64_t a=3;                                                             
    std::size_t b = 91;                                                       
    printf("%llu,%llu\n",a, b);                                             
    printf("%llu,%llu\n",b, a);                                             
    return 0;                                                                 
}

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As you can see, the first line

Code:

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    printf("%llu,%llu\n",a, b);

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prints out weird output for b, because std::size_t is not %llu
But when you switch the position of a and b, as in the second line:

Code:

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    printf("%llu,%llu\n",b, a);

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The output of a is also wrong. What happened?

Right, depending on the architecture you compile for (e.g. 32 or 64-bit x86), size_t may be a 32 or a 64-bit value. Since the printf implementation reads 64 bits to make the %llu output, it will not read the correct data if one of those two are a 32-bit value.

If you printf("%z", sizeof(size_t)), it should show you the size of size_t.

--
Mats

. . . with C99, of course. %z doesn't exist in C89, nor C++ for that matter.

Something like this would work just about anywhere:

Code:

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printf("%lu\n", sizeof(size_t));

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Of course, in C++, you could let cout handle the type.

Note that %llu is also C99.

@OP: Except for <string>, your code looks very much like C. If most of your code is C, perhaps you should consider removing this header file and any other C++-specific stuff and compiling as C . . . .

Quote:

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Originally Posted by dwks

. . . with C99, of course. %z doesn't exist in C89, nor C++ for that matter.

Something like this would work just about anywhere:

Code:

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printf("%lu\n", sizeof(size_t));

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.

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It would be safer to do:

Code:

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printf("%lu\n", (unsigned long)sizeof(size_t));

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Lying to printf is bad.

Instead it is better to cast size_t to a type of your choosing (like unsigned long).

Quote:

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Originally Posted by matsp

If you printf("&#37;z", sizeof(size_t)), it should show you the size of size_t.

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The length modifier (the z) is optional, but AFAIK, the type specifier (u) is not. So it should be like this: printf("%zu", sizeof(size_t))

edit:

printf("%llu,%llu\n",a, b);

llu is the conversion specification for a long long unsigned int, not a uint64_t.

Use PRIu64 in inttypes.h (or cinttypes) instead:

printf(PRIu64 ",%zu\n",a, b);

edit2:

Quote:

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Originally Posted by dwks

. . . with C99, of course. %z doesn't exist in C89, nor C++ for that matter.

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Quote:

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Originally Posted by C++ Chapter 17

9 The descriptions of many library functions rely on the Standard C99 Library for the signatures and semantics of those
functions. In all such cases, any use of the restrict qualifier shall be omitted.

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So I think the z length modifier is supported in C++.

Quote:

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Originally Posted by dwks

. . . with C99, of course. %z doesn't exist in C89
Note that %llu is also C99.

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If you have %llu, then %zu should also work - sorry about missing the u.

--
Mats

Quote:

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Originally Posted by robwhit

The length modifier (the z) is optional, but AFAIK, the type specifier (u) is not. So it should be like this: printf("%zu", sizeof(size_t))

edit:

printf("%llu,%llu\n",a, b);

llu is the conversion specification for a long long unsigned int, not a uint64_t.

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Yes, in a 32-bit architecture, uint64_t is probably unsigned long long, whilst in a 64-bit architecture, it may be unsigned long.

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mats

Quote:

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The length modifier (the z) is optional, but AFAIK, the type specifier (u) is not. So it should be like this: printf("&#37;zu", sizeof(size_t))

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Quote:

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If you have %llu, then %zu should also work - sorry about missing the u.

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Thanks for that -- I was wondering why my compiler was complaining about the non-existent "%z" . . . :)

Quote:

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So I think the z length modifier is supported in C++.

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I don't think it is. My C++ compiler complains about it, and C++ is a superset of C89, not of C99. C99 was then more or less based on some ideas from C++ . . . IIRC, of course.

Quote:

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It would be safer to do:

Code:

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printf("%lu\n", (unsigned long)sizeof(size_t));

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Lying to printf is bad.

Instead it is better to cast size_t to a type of your choosing (like unsigned long).

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Erg, that's exactly what I meant to put. Oops. :)

The C++ standard was published in 1998, so it can't rely on C99. I think the C++09 draft does, though.

Quote:

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Something like this would work just about anywhere:

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Except on 64-bit MS compilers, where long is still 32 bits.

Quote:

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Originally Posted by CornedBee

The C++ standard was published in 1998, so it can't rely on C99. I think the C++09 draft does, though.


Except on 64-bit MS compilers, where long is still 32 bits.

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But size_t is 64-bit.

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Mats

Yes. That's why dwks's snippet would break.

I already mentioned that I had intended to cast the value of sizeof() to an unsigned long, which would let sizeof() work on all values up to the limits of that data type. So unless you had a very, very large data type, this would work:

Code:

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printf("%lu\n", (unsigned long)sizeof(size_t));

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Of course, if sizeof() evaluated to a value larger than ULONG_MAX, the value would wrap.

[edit] And of course, if you didn't cast the value, strange things might happen. :) [/edit]