C programming function calling to return values - c

Question

I'm a college student in my first year of software engineering. I am in the fourth week of my semester and am having trouble in my programming class. Currently, I was given this assignment in which I was given a function called "getNum()" and I had to use it in another function where the program user, would input a number and the function that I program (must be named isOdd()) would determine if the number is odd or even. Then the main function would print whether the number is odd or even. This is the way that my professor worded it:
" Write a program that uses the getNum() function provided to you in Assignment 2 to get anumber from the user (prompting them first, as always). Create and use a function called isOddwith parameters (the number) and return values (1 ifthe number is odd, 0 if the number is evenOR use a bool or boolean data type, your choice) to determine if thenumber is odd. In main(), tell the user (by displaying using printf()or cout) whether the number is evenor odd."
Now, the problem I am having is understanding programming as I am fairly new to it and some words confuse me, such as parameter and return value. To give you and idea of what I have written so far,
#include <stdio.h>
int isOdd(int numInput);
int getNum(void);
int main(void)
{
int number = 0;
while (number > -1)
{
if (isOdd(0))
{
printf("The number is even.\n");
}
else if (isOdd(1))
{
printf("The number is odd.\n");
}
}
return 0;
}
int isOdd(int numInput)
{
int myNumber = 0;
printf("Please enter a number: ", numInput);
myNumber = getNum();
if (myNumber % 2 == 0)
{
myNumber == 0;
}
else
{
myNumber == 1;
}
return myNumber;
}
#pragma warning(disable: 4996)
int getNum(void)
{
/* the array is 121 bytes in size; we'll see in a later lecture how we can improve this code */
char record[121] = { 0 }; /* record stores the string */
int number = 0;
/* NOTE to student: indent and brace this function consistent with your others */
/* use fgets() to get a string from the keyboard */
fgets(record, 121, stdin);
/* extract the number from the string; sscanf() returns a number
* corresponding with the number of items it found in the string */
if (sscanf(record, "%d", &number) != 1)
{
/* if the user did not enter a number recognizable by
* the system, set number to -1 */
number = -1;
}
return number;
}
This is what I have written, trying to do things accordingly to my professor's instructions, as I do not yet know how to properly use booleans. As you can see, at the bottom is the getNum() function that my professor has said is mandatory for this assignment. As of now, everything I input, I am told is even. I am not asking for you guys to solve and do everything for me but I want to understand what I am doing wrong, what my thinking is doing wrong and to overall better my understanding for future programming. Thank you

Answer 1

It's hard to help you without knowing why you did what you did. A lot of the code is just baffling:
if (isOdd(0))
Why are you passing a zero to isOdd?
printf("Please enter a number: ", numInput);
myNumber = getNum();
Is numInput supposed to be the number they enter or is myNumber supposed to be?
if (myNumber % 2 == 0)
{
myNumber == 0;
}
The statement myNumber == 0 compares myNumber to zero to see if they're equal. It does nothing useful here since you ignore the result of the comparison.

Answer 2

The function:
int getNum(void)
Takes no parameters (void), and returns an integer (int) value. The function itself accepts input from the standard input (stdin) stream - this is normally from the keyboard.
To complete your assignment you should write a function:
int isOdd( int value ) ;
Where value is an integer parameter and the return value is 1 if value is odd and 0 if it is even. Alternatively you are allowed to use the Boolean type:
#include "bool.h"
bool isOdd( int value ) ;
In which case you would return true for odd, and false for even.
Your isOdd() includes the getNumber() call and user prompt code. Not only is that not specified in the assignment, it is poor design making isOdd() difficult to use in more general situations. The assignment explicitly requires you to pass the value to be tested as a parameter.
The assignment does not require you to iterate the test (the while loop is not needed). The user input prompt and acceptance should be in main as follows:
int main(void)
{
printf( "Please enter a number: " ) ;
fflush( stdin ) ; // you'll may this on some platforms
int myNumber = getNum();
if( isOdd( myNumber ) )
{
printf("The number is odd.\n");
}
else
{
printf("The number is even.\n");
}
}
return 0;
}
Note that there are only two possible outcomes, so you do not need an else if( !isOdd( myNumber ) ... - if it is not odd it is implicitly even.
Your isOdd() function will work, but is over complicated. You are required to return a Boolean (a type with two possible values true/false - or an integer 1/0 which can be implicitly converted to a Boolean), and (myNumber % 2 == 0) is a Boolean expression (an expression with two possible results true or false). Anywhere you see the pattern:
if( <boolean expression> )
{
b = true ;
}
else
{
b = false ;
}
you can simply write:
b = <boolean expression> ;
In your case the Boolean determination of odd-ness is simply value % 2 != 0. You can return that directly:
bool isOdd( int value )
{
return value % 2 != 0 ;
}