how to put an unsigned int into a char array and extract it back

Question

can someone explain why this isn't working? i am trying to put an unsigned int into a char buffer and then fetch it back into another unsigned int.

  1 #include<stdio.h>
  2 #include<string.h>
  3 int main(){
  4   unsigned int tester = 320;
  5   char buffer[512];
  6   memset(buffer,0,512);
  7   memcpy(buffer,&tester,sizeof(unsigned int));
  8   /*buffer[0]|=tester;
  9   buffer[1]|=(tester>>8);
 10   buffer[2]|=(tester>>16);
 11   buffer[3]|=(tester>>24);*/
 12   unsigned int tested;
 13   memcpy(&tested,buffer,sizeof(unsigned int));
 14 
 15   /*tested|=(buffer[3]<<24);
 16   tested|=(buffer[2]<<16);
 17   tested|=(buffer[1]<<8);
 18   tested|=(buffer[0]);*/
 19   printf("\n %d\n",tested);
 20 }

when i do memcpy it works. but when i take the bitwise approach it doesn't work. funny thing is when i put buffer size as 20 it works. but when i use large buffers, or even 50, it always prints 368 in the bitwise appraoch. again the memcpy works fine.

octopusgrabbus · Accepted Answer · 2012-04-29 14:24:39Z

10

When you declare test on line 12 unsigned int tested; it should be initialized to zero unsigned int tested =0; Also, compiling with -Wuninitialized or -Wall would have warned you tested was uninitialized.

edited Apr 29, 2012 at 14:24

octopusgrabbus

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answered Apr 28, 2012 at 21:04

Richard Pennington

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5 Comments

Christoph Over a year ago

gcc warns about this if you provide -Wuninitialized, which is included in -Wall

Richard Pennington Over a year ago

I warned the OP without any options. ;-)

user949110 Over a year ago

yup you are right!! thanks!! is it because tested gets initialized with garbage and all i'm doing is overwriting on that?

Richard Pennington Over a year ago

Yes. A word of warning however. You probably want to make the array an array of unsigned char or you'll be bit by a sign bit if your compiler treats plain char as signed. (Pun intended)

user949110 Over a year ago

i know. but in my main application i am not really in control of the buffer, it's going to be passed to me. thanks for the warning :)

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