Another problem with a program

When you have a function that returns nothing, both the function prototype and function definition need to specify a return type of "void".

Todd

When I type return void, it says:

(52) : warning C4033: 'outputToScreen' must return a value

One last comment on your (now) "if / else" ladder... :)

Code:

---

int calcLetterGrade (numgrade)
{
        int lettergrade;
        if (numgrade < 60)
                lettergrade = 'F';
        else if (numgrade <70)
                lettergrade = 'D';
        else if(numgrade < 80)
                lettergrade = 'C';
        else if (numgrade <90)
                lettergrade = 'B';
        else if (numgrade <100)
                lettergrade = 'A';
        return lettergrade;
}

---

You don't need the last "if", just let the 'A' grade be an all-inclusive "anything 90 or above" situation. If someone gets a 101 for a score, your current program will return an uninitialized value.

Todd

Quote:

---

Originally Posted by tristangreer

When I type return void, it says:

(52) : warning C4033: 'outputToScreen' must return a value

---

That's because you haven't changed both the prototype and the definition.

That's where I'm totally confused. How do I correct the problems with the outputToScreen function? At least give me a hint! haha

Ok, I edited the program and completely took out the outputToScreen function, and placed what was needed in the main function. Now, when I enter the numerical value for the grade, the lettergrade value always outputs as 70. Why is this?

Code:

---


#include<stdio.h>
#pragma warning(disable:4996)

int getNumGrade ();
int calcLetterGrade();

int main (void)
{
        int numgrade = 0;
        int lettergrade = 0;
        char cgoAgain = 'y';
        printf ("Enter numerical grade value now:");
        numgrade = getNumGrade ();
        lettergrade = calcLetterGrade();
        printf ("\n When numerical grade is %d,",numgrade);
        printf ("letter grade is %d.", lettergrade);

while (cgoAgain =='y' || cgoAgain == 'Y')
{
        printf("\nDo you want to enter another grade?");
        scanf("%c%*c",&cgoAgain);
}
return 0;
}

int getNumGrade()
{
int numgrade = 0;
scanf("%d%*c",&numgrade);
return numgrade;
}

int calcLetterGrade (numgrade)
{
        int lettergrade;
        if (numgrade < 60)
                lettergrade = 'F';
        else if (numgrade <70)
                lettergrade = 'D';
        else if(numgrade < 80)
                lettergrade = 'C';
        else if (numgrade <90)
                lettergrade = 'B';
        else if (numgrade <100)
                lettergrade = 'A';
        return lettergrade;
}

---

Two problems now. Besides the fact that any number produces a lettergrade of 70, when I type "y" to enter another letter grade, it doesn't ask for a new numerical value, it just asked if I want to type in another letter grade. There's obviously a problem with the loop, but i'm unsure what it is.

You are calling calcLetterGrade without a parm:

Code:

---

lettergrade = calcLetterGrade();

---

Yet your function prototype, and function definition state that you should be passing a parm.

You should be getting compiler warnings about this stuff.

Todd

You seem to be confused about prototypes.
A prototype is information that tells the compiler how the real function, which is defined later, looks like. Therefore, they should be indentical.
A pretty good rule is to copy the first line of the definition and paste it as your declaration and append a ; afterwards.
Remember: The function prototype and definition MUST match or you WILL get errors or undefined behavior.