pointer

[jason42009]

hello guys, I have this code:

#include <stdlib.h>
#include <stdio.h>

int *integer(int *a, int *p);

int main(){
	int k = 5;
	int *p = &k;
	int n = 10;
	int m = 30;
	p = (&m, &n);
	printf("p is: %d, k is : %d \n", *p, k);
	
}

int *integer(int *a, int *p){
	if(*a  > *p ){
		return a;
	}
	else {
		return p;
	}

}

My question is: isn't that supposed to print out 30 ? (because 30 > 10) , but when I run it , it prints out 10.

Thanks guys

[laserlight]

Look at this line:

p = (&m, &n);

You forgot to call the function, i.e., to write:

p = integer(&m, &n);

As such, what happened is that &m was evaluated, then &n was evaluated, and the result of (&m, &n) was &n, hence *p is 10, since n is 10.

[KBriggs]

	p = (&m, &n);

Is that supposed to be a function call? If it is, you forgot to name the function you are calling.

EDIT: too slow

[jason42009]

ya that is supposed to be a function call . wow, you guys are too fast in responding ..

Thanks