C function not calculating properly [closed] - c

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In my function when I set *area = 2 or to any other int, my program works as intended. For some reason I cannot calculate the length by the width and have it out put properly, all I get are 0's. What am I missing? Is it something in the call? Also is %f an appropriate conversion for double or is there something better to use? My code is below:
#include <stdio.h>
void area_perimeter(double width, double length, double *area, double *perimeter);
int main(){
double width, length, are, peri;
printf("Enter the width of the rectangle: ");
scanf("%f",&width);
printf("Enter the length of the rectangle: ");
scanf("%f",&length);
area_perimeter(width, length, &are, &peri);
printf("The area of the rectangle is: %f\n",are);
printf("The perimeter of the rectangle is: %f\n",peri);
return 0;
}
void area_perimeter(double width, double length, double *area, double *perimeter){
*area = length * width;
*perimeter = (*area * 2);
}

Answer 1

You invoked undefined behavior by passing a pointer to object having wrong type: %f in scanf()callls for float*, but you passed double*. Use %lf to read double.
Using %f in printf() to print double is good. C99 compiler will also accept %lf.
Try this:
#include <stdio.h>
void area_perimeter(double width, double length, double *area, double *perimeter);
int main(void){
double width, length, are, peri;
printf("Enter the width of the rectangle: ");
if (scanf("%lf",&width) != 1) {
fputs("read error width\n", stderr);
return 1;
}
printf("Enter the length of the rectangle: ");
if (scanf("%lf",&length) != 1) {
fputs("read error length\n", stderr);
return 1;
}
area_perimeter(width, length, &are, &peri);
printf("The area of the rectangle is: %f\n",are);
printf("The perimeter of the rectangle is: %f\n",peri);
return 0;
}
void area_perimeter(double width, double length, double *area, double *perimeter){
*area = length * width;
*perimeter = (*area * 2);
}