From int to int

[Jjoraisnt32]

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

int getInput(int userInput);

int main()
{
    int userInput;
    
    getInput(&userInput);
    printf("%d", userInput);
    
    system ("pause");
    return 0;
}

int getInput(int userInput)
{
     char whichShape;

     printf("P: Display a parallelogram\n");
     printf("R: Display a rectangle\n");
     printf("S: Display a square\n");
     printf("T: Display a right-trangle\n");
     printf("X: Exit program\n");
     printf("Enter your choice\n");
     whichShape = getchar ();
     if (whichShape == 'P' || whichShape == 'p')
 {
     userInput = 1;
 }
     else if (whichShape == 'R' || whichShape == 'r')
 {
     userInput = 2;
 }
     else if (whichShape == 'S' || whichShape == 's')
 {
     userInput = 3;
 }
     else if (whichShape == 'T' || whichShape == 't')
 {
     userInput = 4;
 }
     else if (whichShape == 'X' || whichShape == 'x')
 {
     userInput = 5;
 }
}

I get invalid conversnio from int to int at line 10. I do not understand why

[itsme86]

Read the error message again. It probably says invalid conversion from int * to int.

The problem is, getInput() is expecting an int, but you're passing &userInput (the address of an int, or an int *) to it.

Also, your getInput() is declared to return an int, but you're not returning anything. You need to figure out if you want to design your getInput() function to modify userInput via a pointer, or if you want to modify it in the caller by using getInput()'s return value. Once you've figured out the design, work on the implementation.

[Jjoraisnt32]

Oh yea, I meant to include the return statement. I just put

 return userInput;

after the if statements. and the error was int* to int. I still do not understand why. i call the function in main and i want it to give me the value of userInput i thought that is why i need the &userInput.

[itsme86]

The reason you're getting the error is because the types don't match. Here's a quick primer on the design choices for you...

Method 1 - Pointer:

void caller()
{
  int num = 6;
  getNum(&num);
  printf("%d\n", num);
}

void getNum(int *num)
{
  *num /= 2;
}

Method 2 - Return Value:

void caller()
{
  int num = 6;
  num = getNum(num);
  printf("%d\n", num);
}

int getNum(int num)
{
  return num / 2;
}

Both versions should print "3", but you can see the difference between the methods of getting that value to the caller.

[hk_mp5kpdw]

int getInput(int userInput)
{
     char whichShape;

     printf("P: Display a parallelogram\n");
     printf("R: Display a rectangle\n");
     printf("S: Display a square\n");
     printf("T: Display a right-trangle\n");
     printf("X: Exit program\n");
     printf("Enter your choice\n");
     whichShape = getchar ();

Bit of a nit in this particular case but... getchar returns an int, not a char.