Another problem with a program

I'm trying to make a program that will take a number grade and turn it into a letter grade. As I said in another post, this is my first day programming. Here is my program:

Code:

---

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

int getNumGrade ();
int calcLetterGrade();
int outputToScreen (int, int);

int main (void)
{
int numgrade = 0;
int lettergrade = 0;
char cgoAgain = 'y';
printf ("Enter numerical grade value now:");
numgrade = getNumGrade ();
lettergrade = calcLetterGrade();
int ouputToScreen (numgrade, lettergrade);

while (cgoAgain =='y' || cgoAgain == 'Y');
{
printf("\nEnter numerical grade value now:");
numgrade = getNumGrade(numgrade);

return 0;
}

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

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

int outputToScreen (int numgrade, int lettergrade)
{
        printf ("\n When Numerical grade is %d, letter grade is %d", numgrade, lettergrade);
}

---

I get the following error:


(16) : error C2143: syntax error : missing ';' before 'type'
(26) : error C2143: syntax error : missing ';' before 'type'

How can I fix those?

Here is one of the problems:

Code:

---

while (cgoAgain =='y' || cgoAgain == 'Y');

---

Thank you, I corrected that one. Where does the line 16 error come from, though?

Also, since you are eagerly learning, check out this piece of your code:

Code:

---

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

---

If the numgrade is 50, lettergrade will be set to 'F'.

However, your logic will continue to check for numgrade being equal to other values. What can you do to your "if" ladder to make it more efficient?

Todd

You've defined a function...

Code:

---

int ouputToScreen (numgrade, lettergrade);

---

in the middle of main()!

Could I make else if statements?

I tried that earlier, but it said they were illegal else statements.

I thought I defined outputToScreen earlier in the prototype? Should it be

outputToScreen (numgrade, lettergrade);

in the prototype?

Quote:

---

Originally Posted by tristangreer

Could I make else if statements?

I tried that earlier, but it said they were illegal else statements.

---

Yes, that is what you should do. Try again and post if you get errors.

Is this any better or am I taking steps backward? I'll bold the changes.

Code:

---

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

int getNumGrade ();
int calcLetterGrade();
int outputToScreen (int, int);

int main (void)
{
int numgrade = 0;
int lettergrade = 0;
char cgoAgain = 'y';
printf ("Enter numerical grade value now:");
numgrade = getNumGrade ();
lettergrade = calcLetterGrade();
ouputToScreen (numgrade, lettergrade);
printf ("\nWould you like to enter another grade? ");

while (cgoAgain =='y' || cgoAgain == 'Y')
{
printf("\nEnter numerical grade value now:");
numgrade = getNumGrade(numgrade);
}
return 0;
}

int getNumGrade()
{
int numgrade = 0;
scanf("&#37;d%*c",&numgrade);
return numgrade;
}

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

int outputToScreen (int numgrade, int lettergrade)
{
        printf ("\n When Numerical grade is %d, letter grade is %d", numgrade, lettergrade);
        return 0;
}

---

That is to attempt to fix the line 16 error, this is the new error code I get.

1>(16) : warning C4013: 'ouputToScreen' undefined; assuming extern returning int
1>Linking...
1>Lab2a.obj : error LNK2019: unresolved external symbol _ouputToScreen referenced in function _main

Quote:

---

Originally Posted by tristangreer

I thought I defined outputToScreen earlier in the prototype? Should it be

outputToScreen (numgrade, lettergrade);

in the prototype?

---

You sure did. Get rid of the errant one in the middle of main().

1>(16) : warning C4013: 'ouputToScreen' undefined; assuming extern returning int
1>Linking...
1>Lab2a.obj : error LNK2019: unresolved external symbol _ouputToScreen referenced in function _main

How do I fix this one? I'll make the else statements and see if they work.

Here's the code with the else statements, no errors on those:

Code:

---

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

int getNumGrade ();
int calcLetterGrade();
int outputToScreen (int, int);

int main (void)
{
int numgrade = 0;
int lettergrade = 0;
char cgoAgain = 'y';
printf ("Enter numerical grade value now:");
numgrade = getNumGrade ();
lettergrade = calcLetterGrade();
ouputToScreen (numgrade, lettergrade);
printf ("Would you like to enter another grade?");

while (cgoAgain =='y' || cgoAgain == 'Y')
{
printf("\nEnter numerical grade value now:");
numgrade = getNumGrade(numgrade);
}
return 0;
}

int getNumGrade()
{
int numgrade = 0;
scanf("&#37;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;
}

int outputToScreen (int numgrade, int lettergrade)
{
        printf ("\n When Numerical grade is %d, letter grade is %d", numgrade, lettergrade);
        return 0;
}

---

Here's the error again:

(16) : warning C4013: 'ouputToScreen' undefined; assuming extern returning int
1>Linking...
1>Lab2a.obj : error LNK2019: unresolved external symbol _ouputToScreen referenced in function _main

You do seem to be doing well for a newbie :)

Code:

---

int ouputToScreen (numgrade, lettergrade);

---

This is a function prototype.
I'm not sure if you are trying to call the function or declare a prototype here?

Code:

---

while (cgoAgain =='y' || cgoAgain == 'Y');

---

As others have pointed out, you must remove the ; because otherwise it's interpreted as a single-line while statement, which isn't what you want.

Code:

---

{
printf("\nEnter numerical grade value now:");
numgrade = getNumGrade(numgrade);

---

You have a {, but no }, thus causing all sorts of loads of problems. Essentially, you're putting everything inside that while block. And you can't define a function inside a function.
Indent properly and you'll quickly and easily find mis-matched brackets!

http://cpwiki.sourceforge.net/Indentation
http://cpwiki.sourceforge.net/User:Elysia/Indentation

Code:

---

int outputToScreen (int numgrade, int lettergrade)
{
        printf ("\n When Numerical grade is %d, letter grade is %d", numgrade, lettergrade);
}

---

Unfortunately, you're invoking undefined behavior here. You say the function returns an int, but you never return anything.
Since you are returning nothing, you might as well make it return void - aka, nothing.

And I believe I told you before, but... don't strip argument names from your function prototypes. If anything, they destroy a lot of things.

In the outputToScreen function, I'm wanting the program to write that information on screen. I'm not sure I understand where the information should be returned, if it should be returned at all.

As far as the prototype goes, I'm wanting to declare the prototype. Is that the wrong way to do it?

It is saying outputToScreen must return a value, but where should it be returned, I'm totally confused on that.

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.