c program output

Question

Hello, I am trying some program and confused with the output of the program

#include<stdio.h>

#define a(x) (x*x) 

int main() 
{
 int i=3,j;

 j=a(i+1);
 printf("%d",j);
 return 0;
}

I want to know why the program is not giving the output 16(as instead to that i an getting the output 7 for the above program )

i understood the point very much but if the program like this

#include<stdio.h>

#define a(x) (x*x)

int main()
{
 int i=3 ,j,k;

 j=a(i++);
 k=a(++i);
 printf("%d\n%d",j,k);
 return 0;

} 

then why the above program would give output 9 and 49....

plz help

Answer 1

Read up on C operator precedence and think about what the macro a expands to in this case.

Answer 2

Because you made a bad macro:

a(i + 1)

expands to

i + 1 * i + 1

which is equivalent to

i + i + 1

or

2 * i + 1

Use parenthesis:

#define a(x) ((x) * (x))

And then you'll get it to expand to

(i + 1) * (i + 1)

which does what you want.

Answer 3

Another write-up for the explanation is at http://en.wikipedia.org/wiki/C_preprocessor#Precedence

Answer 4

After preprocessing the line

j=a(i+1);

will be:

j=(i+1*i+1);

which when evaluated for i=3 will give j=7:

j=(3+1*3+1);

To get the desired result you need to define the macro as:

#define a(x) ((x)*(x)) 

which results in:

j=((i+1)*(i+1));

and gives the result 16 when i=3

Answer 5

Because a(i+1) gets preprocessed into (i+1*i+1).

And 3+3+1 = 7.

You might want to use parenthesis around x.

edit: Wow, is this redundant or what. :/

Answer 6

Because your macro is wrong. Apparently it works, but the error is more subtle (not quite, but still), as the expanded code has some issues, by not following the expected order of operations.

j = a(i+1) will expand to j = i + 1 * i + 1 which is 7.

If you want to resolve your problem redefine your macro as:

#define a(x) ((x)*(x))

It's good that you've encountered this problem now, and not later. Those type of errors, are sometimes very hard to debug, but now you'll know how to write "professional" macros :).

Answer 7

Because a(i+1) gets preprocessed into (i+1*i+1).

And 3+3+1 = 7.

You might want to use parenthesis around x.

edit: Wow, is this redundant or what. :/

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