Formatting 16 bytes in hex

I have this program and it appears the %016X formatting spec is not working right. Am I doing something wrong?

Code:

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#include <stdio.h>
#include <string.h>

int main (void) {
        char a[4] ;
        char b[4] ;
        char c[9] ;
        unsigned int int1 = 1 ;
        unsigned int int2 = 2 ;
       
        memcpy(a, &int1, 4) ; 
        memcpy(b, &int2, 4) ; 
       
        memcpy(c,a,4) ;
        memcpy(c+4,b,4) ;
       
        c[8] = 0 ;
       
        printf("a = %08X\n", *a) ; 
        printf("b = %08X\n", *b );
        printf("c = %016X\n", *c );  // <-- why only 8 bytes?
        printf("c+0(4) = %08X\n", *c ) ;
        printf("c+4(4) = %08X\n", *(c+4) ) ;
               
    return 0;
}

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This produces:

Code:

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a = 00000001
b = 00000002
c = 0000000000000001
c+0(4) = 00000001
c+4(4) = 00000002

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Is 8 bytes the max %x can handle? I'm expecting 0000000100000002

Todd

Oops. My bad. X will only take a single character, as proven by this:

Code:

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#include <stdio.h>
#include <string.h>

int main (void) {
        char a[4] ;
        char b[4] ;
        char c[9] ;
        unsigned int int1 = 257 ; // <-- prints as 00000001!
        unsigned int int2 = 258 ; // <-- prints as 00000002!
       
        memcpy(a, &int1, 4) ; 
        memcpy(b, &int2, 4) ; 
       
        memcpy(c,a,4) ;
        memcpy(c+4,b,4) ;
       
        c[8] = 0 ;
       
        printf("a = &#37;08X\n", *a) ; 
        printf("b = %08X\n", *b );
        printf("c = %016X\n", *c ); 
        printf("c+0(4) = %08X\n", *c ) ;
        printf("c+4(4) = %08X\n", *(c+4) ) ;
               
    return 0;
}

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Nevermind.

Quote:

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X will only take a single character, as proven by this:

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I believe this doesn't prove anything. In fact, using %X is like using %d, except the output is in hexadecimal instead of being in decimal. Why is it looking like it's taking only one character is because you have a char array and you are doing

Code:

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printf("%X", *a);

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where *a is of type char. So you are asking to output the value of that char as hexadecimal. And it's doing what it's suppose to. If you would like to output a 64-bit value, i think this is the correct syntax (let's say 'int3' is of type unsigned long long)

Code:

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printf("c = %016llX\n", int3);

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Quote:

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Originally Posted by foxman

(let's say 'int3' is of type unsigned long long)

Code:

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printf("c = %016llX\n", int3);

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for gcc

Quote:

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for gcc

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It seems to be correct according to C99.

Why not just do it portably:

Code:

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for(i = 0; i < 16; i++)
{
    printf("%02X", c[i]);
}

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%016llX might be "standard" but I seriously doubt it would work everywhere.

What I learned from this is that %x formatting works on the datatype being referenced.

So, for chars, it works on 1 byte. Any number between % and X is how many bytes to pad or fill or reserve to the left of the byte. Simple simple.

Quote:

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Originally Posted by brewbuck

%016llX might be "standard" but I seriously doubt it would work everywhere.

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Works on gcc & MS VC++ 6.0! But you also don't need the "ll" if the var is already defined as _int64.

Todd

Quote:

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Originally Posted by Todd Burch

But you also don't need the "ll" if the var is already defined as _int64.

Todd

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I doubt it. printf does not know how the var is defined, so the only way to know it is _int64 - looking on format
if you use %X format - it will put 8 bytes on stack, but printf will only read 4 bytes, interprete it as int and print in the hex format, so it will work ok for numbers <= UINT_MAX and while the number put to printf will be last argument

But it won't work fine if you're trying to pass other info to the function after the 64-bit int, since it will read the second dword of the 64-bit int instead. At least this is on a 32-bit compiler. It might be different on a 64-bit system. I don't know.

You're right. I did a quick test, and neither &#37;016X nor %016llx work. After 2**32, the hex prints zeros. printf() looks like it's only working on the first word.

on MSVC you use I64 prefix for _int64

Quote:

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Originally Posted by vart

on MSVC you use I64 prefix for _int64

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Got an example?

Code:

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UINT64 n = 55;
printf("%I64u", n);

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Quote:

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on MSVC you use I64 prefix for _int64

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Ah, am I right to say that this is because even later versions of MSVC do not support the part of C99 concerning long long, providing _int64 instead?

OK, &#37;016I64x works! Cool. Thanks.