Calculator using function pointers Dry Run in C

Calculator using function pointers 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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Calculator using function pointers Program Code

#include <stdio.h>
#include <math.h>

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

// Advanced operations
double power(double a, double b) { return pow(a, b); }
double modulus(double a, double b) { return fmod(a, b); }
double max(double a, double b) { return a > b ? a : b; }
double min(double a, double b) { return a < b ? a : b; }

typedef double (*MathFunction)(double, double);

typedef struct {
    char *name;
    char symbol;
    MathFunction func;
} Operation;

int main() {
    int i;
    // Table of operations (easy to extend!)
    Operation operations[] = {
        {"Add", '+', add},
        {"Subtract", '-', subtract},
        {"Multiply", '*', multiply},
        {"Divide", '/', divide},
        {"Power", '^', power},
        {"Modulus", '%', modulus},
        {"Max", 'M', max},
        {"Min", 'm', min}
    };
    int opCount = sizeof(operations)/sizeof(operations[0]);
    
    double a, b;
    double result;
    double x = 5, y = 3;
    int choice;
    MathFunction ops[] = {add, multiply, power};
    
    printf("=== Advanced Calculator ===\n\n");
    
    do {
        // Display menu
        printf("\nAvailable Operations:\n");
        for(i = 0; i < opCount; i++) {
            printf("%d. %s (%c)\n", i+1, operations[i].name, operations[i].symbol);
        }
        printf("0. Exit\n");
        printf("Enter choice: ");
        scanf("%d", &choice);
        
        if(choice == 0) {
            printf("Goodbye!\n");
            break;
        }
        
        if(choice >= 1 && choice <= opCount) {
            printf("Enter two numbers: ");
            scanf("%lf %lf", &a, &b);
            
            result = operations[choice-1].func(a, b);
            printf("%.2f %c %.2f = %.2f\n", 
                   a, operations[choice-1].symbol, b, result);
        } else {
            printf("Invalid choice!\n");
        }
    } while(1);
    
    // Demonstrate function pointer flexibility
    printf("\n=== Function Pointer Flexibility ===\n");

    for(i = 0; i < 3; i++) {
        printf("Operation %d: %.0f op %.0f = %.0f\n", i+1, x, y, ops[i](x, y));
    }
    
    return 0;
}

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