Taking input using function in C is not working - c

Question

I have this silly program with my silly problem. I am a very beginner. Let me tell you what I want to do and what is happening here:
What I want to do: I am basically trying to make a function for input which is named here as inp(). This function will ask for input using these two lines:
printf("Enter the Number: ");
scanf("%d", &dip);
When my program will get the number from the user, it will store that inside a variable, let's say dip and will use this number in our another two functions named squarefn and cubefn. I don't know what's going wrong here. But, I can't use the inp() properly to get the number from user.
Why I want to use the inp() function?: Basically, I just want to keep everything inside each function so whenever I need, I will just call my functions. I created inp() so that, I don't need to ask twice or type twice for input.
**What is the output?: ** It's showing some random value or trash value.
Need more information? Feel free to ask!
#include<stdio.h>
int squarefn(int x);
int cubefn(int cube);
int coic();
int printkor(int printkortesi);
int inp();
int main()
{
coic();
}
int squarefn(int input)
{
input = input * input;
return input;
}
int cubefn(int input)
{
input = input * input * input;
return input;
}
int coic()
{
int coi;
int x;
printf("Which one you want?\n");
printf("1. Square\n");
printf("2. Cube\n");
printf("Enter here: ");
scanf("%d", &coi);
int cubed = cubefn(x);
switch (coi)
{
case 1:
// printf("Enter the Number: ");
// scanf("%d", &x);
inp();
int dear = inp(x);
squarefn(dear);
int squared = squarefn(x);
printkor(squared);
break;
case 2:
printf("Enter the Number: ");
scanf("%d", &x);
cubefn(x);
int cubed = cubefn(x);
printkor(cubed);
break;
default:
printf("Error.");
break;
}
}
int inp()
{
int dip;
printf("Enter the Number: ");
scanf("%d", &dip);
return dip;
}
int printkor(int printkortesi)
{
printf("Printed: %d", printkortesi);
}

Answer 1

There is a lot of redundant code in your program, many of the statements don't do anything except taking CPU time . Read about Function Calls in C. Most of your doubts will be solved after a good read
return_type function_name( parameter list ) {
body of the function
}
You have return type of int, to receive value of function foo
int foo(int x)
{return x*x;}
int main()
{
int y =foo(5); //integer y takes value returned by function foo, that is 5
printf("%d\n",y); // This will print value 25
return 0;
}
Your code altered as below, should work now:
#include<stdio.h>
int squarefn(int x);
int cubefn(int cube);
int coic();
int printkor(int printkortesi);
int inp();
int main()
{
coic();
}
int squarefn(int input)
{
input = input * input;
return input;
}
int cubefn(int input)
{
input = input * input * input;
return input;
}
int coic()
{
int coi;
int x;
printf("Which one you want?\n");
printf("1. Square\n");
printf("2. Cube\n");
printf("Enter here: ");
scanf("%d", &coi);
int cubed = cubefn(x);
int dear = inp();
switch (coi)
{
case 1:
{int squared = squarefn(dear);
printkor(squared);
break;}
case 2:
{int cubed = cubefn(dear);
printkor(cubed);
break;}
default:
printf("Error.");
break;
}
}
int inp()
{
int dip;
printf("Enter the Number: ");
scanf("%d", &dip);
return dip;
}
int printkor(int printkortesi)
{
printf("Printed: %d\n", printkortesi);
}
Edit:
Fixed it, I just noticed there was variable declaration in the switch statement. Case statements are only labels. This means the compiler will interpret this as a jump directly to the label. So I added { } in switch and it should work now.