CODEWITHSUNDEEP
c
saegis
saegis

Reputation: 3

Why does it says my variables are doubles when i set them as floats?

#include <stdio.h>

int main() {
    float a, b, c, d, e, f, x, y;

    printf("Insert the 4 coefficients\n");
    scanf("%f%f%f%f\n", a, b, d, e);
    printf("Insert the 2 constants\n");
    scanf("%f%f\n", c, f);
    x=(c*e-b*f)/(a*e-b*d);
    y=(a*f-c*d)/(a*e-b*d);
    if (a*e-b*d!=0)
        printf("The system is possible and determined. The results are x=%f and y=%f\n", x, y);
    else if (a*e-b*d==0 && a*f-c*d==0 && c*e-b*f==0)
        printf("The system is possible and undetermined\n");
    else
        printf("The system is impossible");
    return 0;

}

I have this code in C and whenever I go to compile it i get this:

 warning: format ‘%f’ expects argument of type ‘float *’, but argument 2 has type ‘double’ [-Wformat=]
    7 |     scanf("%f%f%f%f\n", a, b, d, e);
      |            ~^           ~
      |             |           |
      |             float *     double

This error repeats for all the variables on both the scans. I don't understand what's wrong, please help!

Upvotes: 0

Views: 66

Answers (1)

Sayed Hussainullah Sadat
Sayed Hussainullah Sadat

Reputation: 338

scanf function expects a pointer to your variable, not the variable itself. You can easily fix it by passing the address of your variable using the unary & (address-of) operator:

scanf("%f%f%f%f\n", &a, &b, &d, &e);

Note: only string (char str[n]) does not need this address(& sign) in the scanf function.

Upvotes: 1

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