Passing value from one function to another function in C - c

Question

I'm attempting to pass a value determined by one function to another function that comes after it. Essentially: Function 1 determines a value, lets say x. Then X is passed to another function that specifically displays X as text. It's in a separate function to be used by other functions who can also determine the value of X. However, i'm only getting 0 instead of the variable x. What is wrong here?
float fahToCel(int userInput)
{
float result = 0;
do
{
printf("\nPlease input your number in Fahrenheit units:");
scanf("%d", &userInput);
} while (result = 0);
result = (float)(((float)5.0f / (float)9.0f) * (float)(userInput - 32));
return result;
}
float celToFah(int userInput){
float result = 0;
do
{
printf("\nPlease input your number in Celcius units:");
scanf("%d", &userInput);
} while (result = 0);
result = (float)(((float)9.0f / (float)5.0f) * (float)(userInput + 32));
return result;
}
float displayResult(float result) {
printf("\nThe equivalent tempature is: %f\n", result);
return result;
}
void inputInformation() {
printf("1. Convert temperature input from the user in degrees Fahrenheit to
degrees Celsius.\n");
printf("2. Convert temperature input from the user in degrees Celsius to
degrees Fahrenheit.\n");
printf("3. Quit.\n ");
}
int main()
{
int menuChoice = 0;
int userInput = 0;
int result = 0;
while (1)
{
inputInformation();
printf("Please make a selection now:");
scanf("%d", &menuChoice);
switch (menuChoice)
{
case 1:
fahToCel(userInput);
displayResult(result);
break;
case 2:
celToFah(userInput);
displayResult(result);
break;
case 3:
return 0;
default:
printf("\nThat is not a choice!\n");
break;
}
}
}

Answer 1

You never assign the result to result. Change like this:
result = fahToCel(userInput);
displayResult(result);