error: called object '0' is not a function [duplicate] - c

Question

This question already has answers here:
Why am I getting a 'called object 0 is not a function error'?
(4 answers)
Closed 9 years ago.
I'm running into the following error in both functions in the following code:
In function 'Celsius_Converter':
Line 64: error: called object '0' is not a function
In function 'Fahrenheit_Converter':
Line 90: error: called object '1' is not a function
Here is my code:
/*
*Program Name: COP 2220-10018 Project 3
*
* Author: Nathan Gamble
*
* Description: Convert either Celsius to Fahrenheit, or Fahrenheit to Celsius.
*
* Input: Celsius or Fahrenheit Temperature
*
* Output: Fahrenheit or Celsius, depending on which one the User requested.
*/
#include <stdio.h>
#include <math.h>
int main (void)
{
//Local Declarations
char choice;
int fahrenheit;
int celsius;
//Statements
printf("This program converts Celsius temperature to Fahrenheit degree and Fahrenheit temperature to Celsius degree. \n");
printf("If you want to convert Celsius to Fahrenheit, please enter C.");
printf("If you want to convert Fahrenheit to Celsius, please enter F.");
scanf("%s", &choice);
//Process
if (choice == 'C') {
fahrenheit = Celsius_Converter();
printf("The temperature in Fahrenheit degree is %d", &fahrenheit);
}
else {
celsius = Fahreinheit_Converter();
printf("The temperature in Celsius degree is %d", &celsius);
}
return 0;
} // End main
/*
*Program Name: COP 2220-10018 Project 3
*
* Author: Nathan Gamble
*
* Description: Convert Celsius to Fahrenheit
*
* Input: Celsius Temperature
*
* Output: Fahrenheit
*/
int Celsius_Converter ()
{
//Local Declarations
double fahrenheit;
double celsius;
printf("Enter a temperature in Celsius.");
scanf("%d", &fahrenheit);
celsius = (5/9) (&fahrenheit - 32);
return celsius;
} //end Celsius_Converter
/*
*Program Name: COP 2220-10018 Project 3
*
* Author: Nathan Gamble
*
* Description: Convert Fahrenheit to Celsius
*
* Input: Fahrenheit Temperature
*
* Output: Celsius
*/
int Fahrenheit_Converter ()
{
//Local Declarations
double fahrenheit;
double celsius;
printf("Enter a temperature in Fahrenheit.");
scanf("%d", &celsius);
fahrenheit = (9/5) (&celsius + 32);
return fahrenheit;
} //end Fahrenheit_Converter

Answer 1

You have a few problems here:
You're using the address of some values instead of the values theselves. You use the address of celsius when you call scanf (which is right, because scanf needs to know where to store the value):
double celsius;
scanf("%d", &celsius);
but you need to celsius itself when you do arithmetic, not its address:
fahrenheit = (9/5) (celsius + 32); // instead of &celsius
There's a similar error in the other conversion routine, where you use &fahrenheit rather than fahrenheit. This is also an issue in your output. When you're printing, you need to use the value of the variable, not its address. So
printf("The temperature in Celsius degree is %d", &celsius);
should also use celsius, not &celsius.
Once you've got those worked out, though, you can address your main issue.
something (...)
is function call syntax in C. This means that when you do, e.g., (9/5)(celsius + 32), you're trying to call a function (9/5), which is, by integer arithmetic, 0. So, you need to make the multiplication explicit, using *, so you'd have
(9/5)*(celsius+32)
but then you'll still run into the problem that 9/5 is 1, because it's integer arithmetic. You can fix this by making at least one of those numbers a non-integer, e.g., by using (9/5.0):
fahrenheit = (9/5.0) * (celsius + 32);
A similar explanation holds for the other case, except that 5/9 is 0.
The return types of some of your methods aren't right. E.g., since you're trying to return a double, the function needs to be declared to return a double, not an int.
int Celsius_Converter () // needs to be double
{
double celsius;
...
celsius = (5/9) (&fahrenheit - 32);
return celsius;
} //end Celsius_Converter

Answer 2

First, the answer to your question:
(9/5) (&celsius + 32) is a function invocation, more precisely, you are trying to invoke the function 9/5, which is 1, passing &celsius + 32 (which is an error) as its parameter.
What you want to do is probably (9/5) * (&celsius + 32). There is no such thing as implicit operators in C. Actually, it should be (9/5) * (celsius + 32): the & is wrong here.
The same goes for the other function.
Then, some other considerations on your code:
scanf("%s", &choice); is a hazard, because you are asking for buffer overruns. You'd better do scanf("%c", &choice); or
char choice[2];
...
scanf("%1s", choice);
printf("The temperature in Fahrenheit degree is %d", &fahrenheit); is wrong. It should be printf("The temperature in Fahrenheit degree is %d", fahrenheit);. The same goes for Celsius. The & is needed for scanfs, when you're reading into a primitive type (you'll learn the reason behind this as you go further with your studies, or you can post another question, or check a online C guide).
Moreover, I have some doubts on the rationale behind your decomposition in functions, but, again, this is something you don't learn now.