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Naman Tamrakar
Naman Tamrakar

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Can we pass a function to a function in C as a parameter?

Whenever we work in high-level languages like python and JavaScript we can pass functions as parameters to a function and call it there, But is this also possible in low-level languages like C/C++?

Yes! This is also possible in C/C++ but in a little different way. We know that In C/C++ we have access to memory addresses by pointers. So using a pointer we can also point to a function the same as we do for array, int, char, or any other data type.

So let's start with a very simple program

Code

#include<stdio.h>

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

//                  function as param
//                ___________^_____________
int do_operation(int (* operation)(int, int), int a, int b) {
    return operation(a, b);
}

int main(int argc, char const *argv[])
{
    printf("%d\n", do_operation(add, 10, 20));
    return 0;
}
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Output

30
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What's going on in function do_operation in the above code?

The parameter int (* operation)(int, int) is the function signature with the name operation. Let me explain more about it.

  int  (* operation) (int, int)
   ^          ^         ^
  left    identifier   right
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  • identifier: function name (operation)
  • left: function return type (int)
  • right: params type function accept (int, int)

So we can declare a function like this:

return_type (* function_name) (param1, param2);
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But what if we have to use this function in multiple functions? We will have to declare it like this everywhere?

No. Here comes our life saver typedef. We can typedef the function signature and use it anywhere. For example, we can declare our above function signature as:

typedef int (* operation)(int, int);

int do_operation(operation op, int a, int b) {
    return op(a, b);
}
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Isn't this very simple and readable?

YES


Source: freepik.com


Now Let's take a complex example of how passing functions as parameters can be useful.

Code

#include <stdio.h>


/**
 * Defining internal functions
*/
int _multiply(int a, int b) {
    return a*b;
}

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

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


// Functions name
typedef enum {
    ADD, SUBTRACT, MULTIPLY
} Operations;

// declaring function type
typedef int (*function_t)(int, int);

// struct to hold multiple functions
typedef struct {
    Operations op;
    char name[31];
    function_t fun;
} Operation_t;

// Initialize functions array
Operation_t operations[] = {
    {ADD, "Add", _add}, 
    {SUBTRACT, "Subtract", _subtract}, 
    {MULTIPLY, "Multiply", _multiply}
};

/**
 * Function to do multiple operating by just function name
*/
void do_operation(Operations op, int a, int b) {
    for (int i=0; i<sizeof(operations)/sizeof(Operation_t); i++) {
        if (op == operations[i].op) {
            printf("Operation: %s\n", operations[i].name);
            printf("Result: %d\n\n", operations[i].fun(a, b));
            return;
        }
    }

    printf("Invalid Operation\n");
}


int main() {
    // calling the with function name to perform operation
    do_operation(ADD, 10, 20);
    do_operation(SUBTRACT, 10, 20);
    do_operation(MULTIPLY, 10, 20);
    return 0;
}
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Output

Operation: Add
Result: 30

Operation: Subtract
Result: -10

Operation: Multiply
Result: 200
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From the above code, We can observe that just by passing the function name we can execute it.


This article is highly inspired by the content provided by Jacob Sorber on his youtube channel.

❀️Thank you so much for reading this article. I'm a passionate engineering student learning new things so If you find any mistakes or have any suggestions please let me know in the comments.

Also, consider sharing and giving a thumbs up If this post helps you in any way.

Top comments (2)

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pauljlucas profile image
Paul J. Lucas

Your code doesn't even compile. Specifically, you can't create a typedef named operation then declare a function having the same name.

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namantam1 profile image
Naman Tamrakar

My bad I mistyped function do_operation as operation.

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