In this code I seem to get zero, I'm not too familiar with as to why I can't change the variable length with the function I created. Any help could be useful.
#include <stdio.h> double get_length(double a); int main(int argc, char* argv[]) { double length = 0; get_length(length); printf("%lf", length); return 0; } double get_length(double a) { printf("What is the rectangle's length?\n"); scanf("%lf", &a); return a; } When it prints it returns 0.0000
You're not storing the return value. Change:
get_length(length); to:
length = get_length(length); There's no need to pass length when you do this.
The other way to do it is to pass an address:
#include <stdio.h> void get_length(double * a); int main(int argc, char* argv[]) { double length = 0; get_length(&length); printf("%f", length); return 0; } void get_length(double * a) { printf("What is the rectangle's length?\n"); scanf("%lf", a); } Note that %f, not %lf, is the correct format specifier for a double in printf(). Using %lf is correct for scanf(), however.
C is "pass-by-value", which means the value is actually copied in. So changing the value by using the reference doesn't actually change the original reference.
There are two ways around this:
1) store to a local value and then capture the return value:
length = get_length() ... double get_length() { double a; printf("What is the rectangle's length?\n"); scanf("%lf", &a); return a; } 2) pass a pointer:
get_length(&length) ... double get_length(double *length) { printf("What is the rectangle's length?\n"); scanf("%lf", length); } 
get_length(x);has no effect on value ofx.