macro definitions problem

**acpower** · 06-14-2012

Code:

#include <stdio.h>
#define next(x) x+1
#define max(x,y) x >= y ? x : y


int a = ..;
int b = ..;
void main(void) {


printf("%d ", max(next(a)*2,++b));
printf("%d %d\n", a, b);


}

Lets say we have inputs a=6 and b=2:

max(next(a)*2,3) ----> max(6+1*2,3) ----> max(8,3) ---> 8

Output: 8 6 3 (which is correct)

However, if we have inputs a=4 and b=7:

max(next(4)*2,8) ----> max(4+1*2,8) -----> max(6,8) ----8

Output: 8 4 8

But when I run this code for this input I get "9 4 8". Where did I go wrong?

Thanks

**anduril462** · 06-14-2012

Expand the macro by hand, on paper. Take a careful look at the results, then take a read through this: Question 3.8.

**acpower** · 06-14-2012

I have one more of these:

Code:

#include <stdio.h>


#define min(a,b)  ((a) > (b) ? (b) : (a))
#define inc(a)    a++
#define mult(a,b) (a * b)


int main(void) {

    int x = ..;
    int y = ..;
    
    printf("min(%d,inc(%d))",x,y);
    printf("=%d\n",min(x,inc(y)));
    printf("min(mult(%d,%d+2),11)",x,y);
    printf("=%d\n",min(mult(x,y+2),11));
    return 0;


}

For inputs x=2 and y=3:

min(2,inc(3)) ---> min(2,3++) ---> min(2,4) ---> 2

(note y is now 4)

min(mult(2,4+2),11) ---> min(mult(2,6),11) ---> min(12,11) ---> 10

Output:

min(2,inc(3))=2
min(mult(2,4+2),11)=10

However for negative numbers say, x=-1 and y=-2 I get a strange result:

min(-1,inc(-2)) ---> min(-1,-2++) ---> min(-1,-1) ---> -1

y is now -1

min(mult(-1,-1+2),11) ---> min(mult(-1,1),11) ---> min(-1,11) ---> -1

Output:
min(2,inc(3))=-1
min(mult(2,4+2),11)=-1

But this is incorrect, the output is actually supposed to be -1 and 2.

If you run the code for the above inputs it shows "min(mult(-1,0+2),11)". Some how y has been incremented twice to get zero, not sure how this has happened.

**acpower** · 06-14-2012

@anduril462: I've already tried this by hand, first thing I did. And I've made many attempts, ran the code etc

**christop** · 06-14-2012

If you actually expanded the macros by hand, you would have seen the problem:

Code:

min(x,inc(y))
// expands to
min(x,y++)
// expands to
((x) > (y++) ? (y++) : (x))

y is incremented twice because y++ appears twice. Do the same expansion by hand with your mult() macro and you'll see the problem when you try to do something like "mult(x,y+2)".

**acpower** · 06-14-2012

oops thanks!

**Salem** · 06-14-2012

Or you can get the compiler to show you what the pre-processor has done.

Code:

$ cat foo.c
#include <stdio.h>

#define min(a,b)  ((a) > (b) ? (b) : (a))
#define inc(a)    a++
#define mult(a,b) (a * b)

int main(void) {
    int x = 0;
    int y = 0;
    printf("min(%d,inc(%d))",x,y);
    printf("=%d\n",min(x,inc(y)));
    printf("min(mult(%d,%d+2),11)",x,y);
    printf("=%d\n",min(mult(x,y+2),11));
    return 0;
}
$ gcc -E foo.c | tail 

int main(void) {
    int x = 0;
    int y = 0;
    printf("min(%d,inc(%d))",x,y);
    printf("=%d\n",((x) > (y++) ? (y++) : (x)));
    printf("min(mult(%d,%d+2),11)",x,y);
    printf("=%d\n",(((x * y+2)) > (11) ? (11) : ((x * y+2))));
    return 0;
}

**acpower** · 06-15-2012

Hang on, if y was incremented twice how come for inputs x=2,y=3, y is 4 and not 5? Why was y incremented twice for y=-2 (to get zero) but not for y=3?

**laserlight** · 06-15-2012

Originally Posted by acpower

if y was incremented twice how come for inputs x=2,y=3, y is 4 and not 5? Why was y incremented twice for y=-2 (to get zero) but not for y=3?

Because (x) is then evaluated, not (y++), since 2 > 3 is false.

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