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Old September 28th, 2004, 02:47 AM
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Default Use of long long int

I'm pretty new to C coding, and I'm trying to use long long (64-bit) integers. I ran the following test program on z/OS (under USS), and on Windows 2000, but I don't understand the results:
Does anyone have an idea why the HEX print only prints the content of the first 4 bytes? Or why on Windows, the DECIMAL print produces a negative?
I think I understand the result of the BYTE print, with Windows being a little Endian, and so the entire value is stored in reverse byte order..

Test program:

#include <stdio.h>
#include <stdlib.h>
void main(void)
   unsigned long long juul = 123456789012345678LL;
   unsigned char *pjuul;
   int i;

   printf("juuld: %lld\n",juul);
   printf("juulx: %016X\n",juul);
   pjuul = (char *) &juul;
   printf("juulb: ");
   for (i = 0;i < 8; i++) printf("%02X",*(pjuul + i));
   exit (0);
/* PRODUCES on z/OS: on Windows:
    juuld: 123456789012345678 -1506741426
    juulx: 0000000001B69B4B 00000000A630F34E
    juulb: 01B69B4BA630F34E 4EF330A64B9BB601

Thanks beforehand,

Old November 30th, 2005, 04:16 PM
tnd tnd is offline
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I believe its because microsoft C doesn't understand your format codes.
try using
printf("%.I64", juul);
printf("%16.I64X", juul);

Althought I haven't checked but I think signed 32 bit BCD representation of A630F34E is -1506741426

Old September 20th, 2006, 06:07 PM
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You stored an int type and read it as a decimal: the highest order bit in any decimal number is either 0 for positive numbers, or 1 for negative numbers. Assuming your int value was large enough to change the value of the highest order bit to a 1, reading as it as a decimal number interpreted it to be a negative number.


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