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Function that returns a function pointer

Started by ICanC Mar 11, 2017 at 5:28 PM 4 replies 24.5k views

Original Post

Mar 11, 2017 at 5:28 PM

hello, I'm trying to make a calculator and I'm struggling to make a function invoke the right function based on a function pointer (for practice)

I'm struggling to understand how to define a function pointer, and the syntax on how it should be implemented. Please see below the attempt I've made so far :


#include <stdio.h>

int add(int, int);
int sub(int, int);
int mul(int, int);
int (*getfunc())(char);

int main(void)
{
	puts("\t\tCalculator\n\n");
	
	return 0;
}

int add(int a, int b)
{
	return a + b;
}

int sub(int a, int b)
{
	return a - b;
}

int mul(int a, int b)
{
	return a * b;
}

int (*getfunc())(char op)
{
	switch(op)
	{
		case '+':  return getfunc = add;
		case '-':  return getfunc = sub;
	}
}

however the compiler never likes the syntax for the function implementation, can someone please point me in the right direction? Thanks

apatriarca

Mar 11, 2017 at 5:49 PM

The easier way to define the type correctly is to first create a typedef for your function type and then define your function returning that type. getfunc can't set his own value in its implementation. And it can't be a function pointer if you are trying to implement it as a function getting a char and returning a function pointer. getfunc would also have a different type than add or sub..


typedef int (*op_func)(int, int);

// .. define add and sub

op_func getfunc(char op)
{
    switch(op) {
        case '+': return add;
        case '-': return sub;
    }
}

int main()
{
    fprintf("5 + 7 = %d", getfunc('+')(5, 7));
    fprintf("5 - 7 = %d", getfunc('-')(5, 7));
}

ICanC

Mar 11, 2017 at 5:55 PM

Thanks a lot.

Just for understanding, can you advise what the code would look like without using typedef?

Alberth

Mar 11, 2017 at 5:57 PM


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

int add(int a, int b) { return a + b; }
int sub(int a, int b) { return a - b; }

typedef int (*FNP)(int, int);

FNP get_func(int op)
{
    if (op == '+') return &add
    return ⊂
}

int main(void)
{
/* 2 equivalent ways to write 'fnc', a point to an "int f(int, int)" function */
/*    FNP fnc; */
    int (*fnc)(int, int);

    int a = 3;
    int b = 5;
    int result;

    fnc = get_func('+');

    result = (*fnc)(a, b);
    printf("%d + %d = %d\n", a, b, result);
    return 0;
}

Thanks a lot.

Just for understanding, can you advise what the code would look like without using typedef?

http://stackoverflow.com/questions/997821/how-to-make-a-function-return-a-pointer-to-a-function-c

but I couldn't make it work.

Anyway, it looks too weird to even try without typedef

alvaro

Mar 11, 2017 at 6:19 PM

Thanks a lot.

Just for understanding, can you advise what the code would look like without using typedef?

Something like this:

#include <stdio.h>

int add(int x, int y) {
  return x + y;
}

int subtract(int x, int y) {
  return x - y;
}

int (*get_function(char c))(int,int) {
  return (c == '+') ? add : subtract;
}

int main() {
  printf("%d\n", get_function('+')(5,7));
}

Here's a useful link: http://stackoverflow.com/questions/10758811/c-syntax-for-functions-returning-function-pointers

ICanC

Mar 11, 2017 at 6:36 PM

Thanks Alvaro, that code works and the link is good too

appreciate all the repsonses

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