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Error in bitwise OR operation?

Posted on 2004-07-30
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Last Modified: 2010-04-15
I am getting an error for the following code...

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

#define hex 0x00020000;
void main()
{
      unsigned r;
       r=hex | 100;
      printf("%u",r);
      getchar();
}

error C2143: syntax error : missing ';' before '|'

What is the problem?

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Question by:gopikrish
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4 Comments
 
LVL 55

Accepted Solution

by:
Jaime Olivares earned 120 total points
ID: 11681739
Hi gopikrish,
> #define hex 0x00020000;
the last comma is not valid. Would be:
#define hex 0x00020000

Good luck,
Jaime.
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LVL 55

Expert Comment

by:Jaime Olivares
ID: 11681749
To be clearer: #defines don't need an ending colon (;) like regular C code lines.
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LVL 5

Expert Comment

by:ddunlea
ID: 11682033
Whilst it's not causing the problem you're seeing, are you sure you want to be ORing decimal 100 i.e. 0x64? You probably want to write 0x100, which will give you a single bit set high,  as opposed to just plain 100 which will give 01100100b
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LVL 23

Expert Comment

by:Mysidia
ID: 11682715
Since you have
 #define hex 0x00020000;

The line:
       r=hex | 100;

is actually expanding to
        r=0x00020000; | 100;

Which is not what you want: "| 100;"  is not a valid C statement
Removing the trailing semicolon in the #define hex   line should remedy this
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