HI,
Here are some fact related to macros..
its a preprocessor directory so before comping the code will replaced by
appropriate macros... these are not comping or executing ..so macros
replaced one by one and other fact macro will not replaced the content of
other macros...
so code of 1st code will be like that after preprocessing
1st all A will be replaced by B except #define B A,
2nd replaced in code where ever is B is replaced by A so like..
sorry for my 1st post giving wrong explanation...
@saurabh in ur 2nd post
after preprocessing of 1st macro code be like that
include<stdio.h>
int main(){
int A =4;
#define B A
printf("%d",B);
return 0;
}
after 2nd preprocessing
include<stdio.h>
int main(){
int A =4;
printf("%d",A);
return 0;
}
so out put is 4
if need more explanation of above logic you can see by running those example
#include<stdio.h>
int main()
{
int A =4;
#define A B
#define B A*3
printf("%d",A);
return 0;
}
#include<stdio.h>
int main()
{
int A =4;
#define A B*B
#define B A
printf("%d",A);
return 0;
}
the 1st statement will not affect by preprocessor becoz those macros
definition are coming after that otherwise u can see
#include<stdio.h>
int main()
{
int A =4;
printf("%d",B)
#define A B
return 0;
}
for some more
#include<stdio.h>
int main()
{
int A =4;
printf("%d",A);
printf("%d",B);
#define A B
printf("%d",B);
#define B A
printf("%d",B);
printf("%d",A);
return 0;
}
by looking compilation error u will able to see the scope of macros
and their preprocessing ..
thanks for reading long post.. :)
On Fri, Sep 10, 2010 at 3:28 PM, praveen <praveen200...@gmail.com> wrote:
> After preprocessing
>
> Where ever there is 'A' in your program, it is replaced by 'B'
> Where ever there is 'B' in your program, it is replaced by 'B'
>
> this will be the program look like
> #include<stdio.h>
> int main(){
> int B =4;
>
> printf("%d",B);
> return 0;
>
> }
> therefore, the program will compile and run :)
>
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Thanks & Regards
Umesh kewat
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