Return function pointer Dry Run in C

Return function pointer is an interactive C dry run visualizer from the Pointersfunctions programs section. Study the source code, then use the execution controls to follow each step, variable update, highlighted line, and console output.

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Return function pointer Program Code

#include <stdio.h>

int add(int a, int b) { return a + b; }
int subtract(int a, int b) { return a - b; }
int multiply(int a, int b) { return a * b; }
int divide(int a, int b) { return b != 0 ? a / b : 0; }

// Function that returns a function pointer based on operator character
int (*getOperation(char op))(int, int) {
    switch(op) {
        case '+': return add;
        case '-': return subtract;
        case '*': return multiply;
        case '/': return divide;
        default: return NULL;
    }
}

// Function that returns one of several math operations
typedef int (*MathOp)(int, int);

MathOp getMathOperation(int code) {
    MathOp operations[] = {add, subtract, multiply, divide};
    if(code >= 0 && code < 4) {
        return operations[code];
    }
    return add;
}

// Demonstration of returning different function pointers
int main() {
    int i;
    int code;
    int x = 20, y = 5;
    char ops[] = {'+', '-', '*', '/'};
    int (*op)(int, int);
    MathOp mathOp;
    MathOp dynamicOp;
    
    printf("=== Returning function pointer by operator ===\n");
    
    for(i = 0; i < 4; i++) {
        op = getOperation(ops[i]);
        if(op != NULL) {
            printf("%d %c %d = %d\n", x, ops[i], y, op(x, y));
        }
    }
    
    printf("\n=== Returning function pointer by code ===\n");
    for(code = 0; code < 4; code++) {
        mathOp = getMathOperation(code);
        printf("Operation %d: %d and %d = %d\n", code, x, y, mathOp(x, y));
    }
    
    // Chaining function pointer returns
    printf("\n=== Chaining ===\n");
    dynamicOp = getMathOperation(2);  // multiply
    printf("multiply(20,5) = %d\n", dynamicOp(20, 5));
    
    return 0;
}

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