The following program is printing 49
#include <stdio.h>
#define SQR(x) (x*x)
void main(void) {
int x, y; x = 5;
y = SQR(++x);
printf("y is %d\n", y);
}
I don't understand what I am missing.
Here is what I am understanding how it might be working
In SQR(++x), ++x will be evaluated to be 6 and after x*x the 36 should be returned.
Therefore y = 36
But the answer is 49
Read the wikipage about the C preprocessor and understand that the C preprocessor operate at a purely textual level (as the first phase of the compiler). Read also GNU cpp documentation. Your y = SQR(++x);is expanded as
y = ++x * ++x;
Which is not what you should want. Read about undefined behavior (and this answer).
Think about the horror happening (using your original SQR) with SQR(y+3).
Look at the preprocessed form of your source code. With GCC, use
gcc -C -E foo.c > foo.i
to obtain into foo.i (which you should examine with an editor or a pager) the preprocessed form of foo.c
So, at the very least, you should define your macro as
#define SQR(x) ((x)*(x))
Otherwise SQR(y+3) won't be expanded like you want. In general when using a macro you should know (and document, e.g. in a comment, if you are writing the macro) what is happening.
In fact, in your case, you should not use a macro at all, but define an inline function
static inline int square(int x) { return x*x; }
this would be as fast as your macro, but much more safe, since square(++x) would increment (rightly) x only once, and square(y+3) is doing what you want!
Remember that macros are replaced pretty much verbatim, that means your
SQR(++x);
will be replaced by
++x * ++x;
And, as noted by dyp, it's undefined behavior.
