C preprocessor??Plz Help!!

man s · 12-20-2009, 11:27 AM

Hello folks!
I have recently encountered a snippet of C preprocessor code which has been troubling me since the past few days.
Here it is:

#define SQR(x) (x*x)
main(){

int a,b=3;
a=SQR(b+2);
printf("\n%d",a);
}

This gives answer 11 which completely baffles me.

Any suggestions on it will be highly appreciated.

Thanx in advance.

MTK358 · 12-20-2009, 11:50 AM

I don't understand it either. It's like it's doing "a = sqr(b) + 2" instead of "a = sqr(b + 2)".

JohnGraham · 12-20-2009, 12:09 PM

You have to be REALLLY REEEEEEEEEEEEEALLY careful with macros. To see why, we can get the c preprocessor to show us how it's expanded the macro:

Code:

$ cat > a.c << 'EOF'
> #define SQR(x) (x*x)
> main(){
> 
> int a,b=3;
> a=SQR(b+2);
> printf("\n%d",a);
> }
> EOF


$ gcc a.c -E
# 1 "a.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "a.c"

main(){

int a,b=3;
a=(b+2*b+2);
printf("\n%d",a);
}

As you can see, SQR(b+2) gets expanded (correctly - i.e. this is not a bug in the preprocessor!) to (b+2*b+2), which will be evaluated as (b+(2*b)+2) since multiplication has higher precedence than addition.

This can be solved by defining SQR like:

Code:

#define SQR(x) ((x)*(x))

More generally, always put your arguments in parentheses each time they're used.

Note that there's still a flaw in this macro - a call to SQR(b++) would be expanded to ((b++)*(b++)), which is clearly not what you want. When you use a macro, you should try and only use each argument once, as it might have side effects.

You can cure this with:

Code:

#include <math.h>
#define SQR(x) pow((x),2)

John G

paulsm4 · 12-20-2009, 12:09 PM

Hi -

The issue is "precedence":

Code:

/* BAD */
#define SQR(x) (x*x)
int
main()
{
  int a=2,b=3;
  a=SQR(b+2); 
  printf("\n%d",a);
  return 0;
}

Code:

/* BETTER */
#define SQR(x) ((x) * (x))
int
main()
{
  int a=2,b=3;
  a=SQR((b+2)); 
  printf("\n%d",a);
  return 0;
}

And here's why:
http://www.difranco.net/cop2220/op-prec.htm

Because the first example was actually doing this:

Code:

a = SQR(b+2*b+2)
        = 3 + (2*3) + 2
        = 11

btmiller · 12-20-2009, 12:14 PM

If you run cpp yourself on the source file, you'll see exactly why this is happening. Remember a #define is just a macro expansion for the preprocessor -- it just replaces one set of text with another, so your line becomes:

a=(b+2*b+2)

since b = 3, by order-of-operation (* takes operator precendence over +), the answer is clearly 11. To avoid this, change

a=SQR(b+2)

to

a=SQR((b+2))

or better yet, don't use macros for something that should be a function :-).

smeezekitty · 12-20-2009, 01:13 PM

The multiplication is being done first.
Change it to:

Code:

#define SQR(x) ((x)*(x))

man s · 12-20-2009, 09:20 PM

Thanx a lot guys!It was really helpful.
From what I got,the expansion is something like:

3+2*3+2
where order of precedence gives it 11.And since () has higher precedence than *,it gave the right answer.

Thanx again for the help.