Issue with making a swap function with pointers

Question

I am trying to make a swap function on my own with float pointers and it just does not work. For some reason I think I don't pass the float pointers to the function in the correct way.

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

void swap(float *a, float *b);

int main() {
    float num1 = 0.0;
    float num2 = 0.0;

    float *px = NULL;
    float *py = NULL;

    printf("Please enter a decimal number: ");
    scanf("%f", &num1);
    getchar();

    printf("Please enter another decimal number: ");
    scanf("%f", &num2);

    px = &num1;
    py = &num2;

    printf("The numbers before swapping - \nNum1 = %f\nNum2 = %f\n", num1, num2);

    swap(&px, &py);

    printf("\nThe numbers after swapping - \nNum1 = %f\nNum2 = %f\n\n", num1, num2);

    system("PAUSE");
    return 0;
}

void swap(float *a, float *b) {
    float temp = *a;
    *a = *b;
    *b = temp;
}

Delete all occurrences of px and py in this code, passing &num1 and &num2 to your swap instead. Or lose the & in front of px and py in your swap call. If you didn't get at least a warning about float** not being compatible with float* in this, I'm shocked. — WhozCraig
– WhozCraig, Commented Mar 18, 2016 at 23:00
when compiling, always enable all the warnings, then fix those warnings. Using gcc -c -Wall -Wextra -Wconversion -std=gnu99 file.c -o file.o the compiler output: 1) line:4: passing argument 1 of 'swap' from incompatible pointer type 2) line 25: passing argument 2 of 'swap' from incompatible pointer type 3) line4 expected 'float*' but argument is of type 'float**'. Suggest fixing those problems before trying anything else. Note: do not ignore warnings. — user3629249
– user3629249, Commented Mar 19, 2016 at 18:23

chqrlie · Accepted Answer · 2016-03-18 23:02:03Z

2

You pass addresses of pointers to float instead of just addresses of float values. Remove the px and py variables and simplify your call to:

swap(&num1, &num2);

The compiler should have issued a warning about this type mismatch. Do not ignore these warnings, they indicate programming errors. Better even, enable more warnings: gcc -Wall -Wextra -Werror or clang -Weverything will produce more warnings about potential errors.

answered Mar 18, 2016 at 23:02

chqrlie

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Comments

ram · Accepted Answer · 2016-03-18 23:15:42Z

0

Modify the prototype of swap to void swap(float** a, float** b );. Then modify the function's definition to:

void swap(float** a, float** b) {
    float temp = **a;
    **a = **b;
    **b = temp;
}

answered Mar 18, 2016 at 23:15

ram

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2 Comments

user3185968 Over a year ago

More indirection is obviously better. Make it three stars please.

ram Over a year ago

I mistakenly thought about this as the problem of passing pointers as parameters in C, hence the more convoluted example. Thanks for your very lovely comment though.

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