Function pointers uasage
Posted
by chaitanyavarma
on Stack Overflow
See other posts from Stack Overflow
or by chaitanyavarma
Published on 2010-06-14T15:16:26Z
Indexed on
2010/06/14
15:22 UTC
Read the original article
Hit count: 148
c++
|function-pointers
Hi All, Why these two codes give the same output, Case - 1:
#include <stdio.h>
typedef void (*mycall) (int a ,int b);
void addme(int a,int b);
void mulme(int a,int b);
void subme(int a,int b);
main()
{
mycall x[10];
x[0] = &addme;
x[1] = &subme;
x[2] = &mulme;
(x[0])(5,2);
(x[1])(5,2);
(x[2])(5,2);
}
void addme(int a, int b) {
printf("the value is %d\n",(a+b));
}
void mulme(int a, int b) {
printf("the value is %d\n",(a*b));
}
void subme(int a, int b) {
printf("the value is %d\n",(a-b));
}
Output:
the value is 7
the value is 3
the value is 10
Case -2 :
#include <stdio.h>
typedef void (*mycall) (int a ,int b);
void addme(int a,int b);
void mulme(int a,int b);
void subme(int a,int b);
main()
{
mycall x[10];
x[0] = &addme;
x[1] = &subme;
x[2] = &mulme;
(*x[0])(5,2);
(*x[1])(5,2);
(*x[2])(5,2);
}
void addme(int a, int b) {
printf("the value is %d\n",(a+b));
}
void mulme(int a, int b) {
printf("the value is %d\n",(a*b));
}
void subme(int a, int b) {
printf("the value is %d\n",(a-b));
}
Output:
the value is 7
the value is 3
the value is 10
© Stack Overflow or respective owner