c pow with enumeration-CodePudding

I created 3 void functions that will return the result of the given pow (either int, float or double) to the origin.

#include <assert.h>
#include <float.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

enum base_type { BASE_TYPE_INT, BASE_TYPE_FLOAT, BASE_TYPE_DOUBLE };

void my_pow_int(int *b, int n) {
    int a = *b;
    *b = pow(a,n);
}

void my_pow_float(float *b, int n) {
    float a = *b;
    *b = powf(a,n);
}

void my_pow_double(double *b, int n) {
    double a = *b;
    *b = pow(a,n);
}

Next I created a function that takes 3 parameters and then decides what function from above should be executed.

void my_pow(void *b, int n, enum base_type type) {
    if (type == BASE_TYPE_INT){
        *b = my_pow_int(b, n);
    }
    if (type == BASE_TYPE_FLOAT){
        *b = my_pow_float(b, n);
    }
    else{
        *b = my_pow_double(b, n);
    }
}

I now do have the problem that when I give he function a void pointer, It will not compile because of the casting. I don't get what I am doing wrong here.

CodePudding user response：

The functions that you're calling dereference b with the proper pointer type, and those functions don't return anything, so no need to set anything in my_pow:

void my_pow(void *b, int n, enum base_type type) {
    if (type == BASE_TYPE_INT){
        my_pow_int(b, n);
    }
    else if (type == BASE_TYPE_FLOAT){
        my_pow_float(b, n);
    }
    else{
        my_pow_double(b, n);
    }
}

CodePudding user response：

First problem: You my_pow_* functions return void, so *b = my_pow... doesn't work. It doesn't need to, since you're doing an in-place assignment in your functions.

If you fix that, you can just call your function as

double val = 3.0;
my_pow(&val, 2, BASE_TYPE_DOUBLE);