Home » Community » Newbie corner » array of function pointer
| array of function pointer [message #36149] |
Sat, 05 May 2012 13:08  |
varu
Messages: 21
Registered: March 2012
Location: Bangalore,India
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Promising Member |
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hi
i am doing something similar as show below in c . can u please help me to do the same in u++
#include <stdio.h>
int sum(int a, int b);
int subtract(int a, int b);
int main(void)
{
p[0] = sum; // address of sum()
p[1] = subtract; // address of subtract()
int result;
int i = 2, j = 2, op;
printf("0: Add, 1: Subtract\n");
do {
printf("Enter number of operation: ");
scanf("%d", &op);
} while(op<0 || op>2);
result = (*p[op]) (i, j);
printf("%d", result);
return 0;
}
int sum(int a, int b)
{
return a + b;
}
int subtract(int a, int b)
{
return a - b;
}
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| Re: array of function pointer [message #36164 is a reply to message #36149] |
Tue, 08 May 2012 11:05  |
navi
Messages: 107
Registered: February 2012
Location: Sydney, Australia
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Experienced Member |
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U++ is C/C++; U++ is not a new language its a new cross platform GUI Framework using the C++. See the U++ Overview page to learn what is U++
In your code you did not declare any 'p'. also 'p's should be assigned the '&' of the functions. attached binary is complied using MinGW compiler with U++
int (*p[2])(int,int);
p[0] = ∑ // address of sum()
p[1] = &subtract; // address of subtract()
#include "stdio.h"
int sum(int a, int b);
int subtract(int a, int b);
int main(int argc, const char *argv[]){
int (*p[2])(int,int);
int result;
int i = 2, j = 3, op;
p[0] = ∑ // address of sum()
p[1] = &subtract; // address of subtract()
printf("0: Add, 1: Subtract\n");
do {
printf("Enter number of operation: ");
scanf("%d", &op);
} while(op<0 || op>2);
result = (*p[op]) (i, j);
printf("%d", result);
return 0;
}
int sum(int a, int b){
return a + b;
}
int subtract(int a, int b){
return a - b;
}
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