Noob Programmer having trouble with if and switch statements in program

Heresy · #1 17-Feb-2006, 14:52

Currently, I am working on a program for a beginning C Programming course. I understand the majority of the content and have debugged as much as I can but am now at a loss. We are assigned a project that is supposed to get a number from -99 to 99 then output its name. I compiled the program without any errors but it is doing what I want it to do. The coding is below and I do realize that I completely forgot to add the negative element, but I can do that after I post. Right now, we are limited to using if, else, and switch statements. So no loops.

CPP / C++ / C Code:

#include <stdio.h>

int
main(void)
{
	int number;
	int ones, tens;

	printf("Input an integer number between -99 and 99: ");
		scanf("%d", &number);

	tens = number / 10;
	ones = (number - tens * 10);

	if (number > 99 || number < -99)
		printf("Sorry, number not between -99 and 99.\n");
	if (number == 0)
		printf("Zero");
	if (number == 10)
		printf("Ten");
	if (number == 11)
		printf("Eleven");
	if (number == 12)
		printf("Tweslve");
	if (number == 13)
		printf("Thirteen");
	if (number == 14)
		printf("Fourteen");
	if (number == 15)
		printf("Fifteen");
	if (number == 16)
		printf("Sixteen");
	if (number == 17)
		printf("Seventeen");
	if (number == 18)
		printf("Eighteen");
	if (number == 19)
		printf("Nineteen");
	else

switch (tens)
{
            case '0':
	        	printf("");
				break;
            case '2':
                printf("Twenty");
                break;
            case '3':
                printf("Thirty");
                break;
            case '4':
                printf("Forty");
                break;
            case '5':
                printf("Fifty");
                break;
            case '6':
                printf("Sixty");
                break;
            case '7':
                printf("Seventy");
                break;
            case '8':
                printf("Eighty");
                break;
            case '9':
                printf("Ninety");
                break;
}
switch (ones)
{
            case '0':
                printf("");
                break;
            case '1':
                printf("One");
                break;
            case '2':
                printf("Two");
                break;
            case '3':
                printf("Three");
                break;
            case '4':
                printf("Four");
                break;
            case '5':
                printf("Five");
                break;
            case '6':
                printf("Six");
                break;
            case '7':
                printf("Seven");
                break;
            case '8':
                printf("Eight");
                break;
            case '9':
                printf("Nine");
                break;
}

davekw7x · #2 17-Feb-2006, 15:56

Quote:

Originally Posted by Heresy

am now at a loss.

Your variables "tens" and "ones" are ints.

SO: the case labels should be integet constants, not character literals

CPP / C++ / C Code:

case 1: ....   /* not case '1': */

case 2: ....  /* not case '2': */

...

When you get the case stuff straightned out you might find other things to fix, but maybe this will get you unstuck at least temporarily.

Regards,

Dave

Heresy · #3 17-Feb-2006, 16:29

Quote:

Originally Posted by davekw7x

Your variables "tens" and "ones" are ints.

SO: the case labels should be integet constants, not character literals

When you get the case stuff straightned out you might find other things to fix, but maybe this will get you unstuck at least temporarily.

I did that except that now I am getting an error now. Also, how does the compiler know to go from the else statement to the two switch statements? I think I am just missing something simple or not connect everything together properly.

Guidelines Plz · #4 17-Feb-2006, 16:51

Please read the Guidelines so you can post better questions. And proofread your question. This post:

Quote:

Originally Posted by Heresy

I understand the majority of the content and have debugged as much as I can but am now at a loss... I compiled the program without any errors but it is doing what I want it to do....

says you have a problem, "but it is doing what I want it to do" meaning it's working. I realize this is a typo. But you wrote and ran the program, we didn't, so we don't know what it is doing! So what can we look for? Tell us...

Quote:

Originally Posted by Heresy

I did that except that now I am getting an error now.

And?

Concentrate on Guideline #1, #2, #3.

davekw7x · #5 17-Feb-2006, 17:55

Quote:

Originally Posted by Heresy

I did that except that now I am getting an error now.

If you would post your code and state exactly what error you are getting, maybe someone here can help. I know (believe it or not) how frustrating it is at first when nothing seems to work. Even the stuff that used to work a little bit isn't working at all, right?

The code in your original post was missing a closing brace } at the end of the program. When I added that, the program compiled OK, but most of the inputs didn't give any output (for the reason that I stated in my previous response).

I think you are pretty close except for the case labels. When I changed all of the case labels, I got somewhat better action.

Quote:

Originally Posted by Heresy

Also, how does the compiler know to go from the else statement to the two switch statements? I think I am just missing something simple or not connect everything together properly.

Here is how the if/else stuff works.

First example: if the condition is "true" (value of the expression is not zero), a single statement is executed --- the single statement immediately following the "if()". If the condition is "false" (value of the expression is zero), the single statement immediately following the "if()" is skipped. Either way, the program continues on after that.

Code:

if (condition) 
  this single statement is executed if the arithmetic value of "condition" is not zero

This is the rest of the program

Second example: if the condition is "true", the single statement following the "if()" is executed. If not, the statement immediately following the "if()" is not executed, but the single statement immediately following the "else" is executed. Either way, the program continues on after that.

Code:

if (condition)
  this single statement is executed if the arithmetic value of "condition" is not zero
else
  this single statement is executed if the arithmetic value of "condition" is zero

This is the rest of the program

Here's how you can have more than one statement after the "if" and/or "else": Just enclose a block of statements in curly braces {}, then the entire block is executed or not, depending on the value of the "condition" expression.

Third example: A block of statements is executed or skipped, depending on the "condition". Either way, program execution continues on after that.

Code:

if (condition) {
  statement
 ..
 ..
  All statements enclosed by the curly braces are executed if the value of the "condition" is not zero
 ..
 ..
}

This is the rest of the program

Fourth example: Either the block of statements after "if()", or the block after the "else" is executed, depending on the value of the "condition". Either way, program execute continues on after the blocks.

Code:

if (condition) {
  ..
  these statements are executed if the arithmetic value of "condition" is not zero
 ..
}
else {
  ..
  ..
  these statements are executed if the arithmetic value of "condition" is zero
  ..
  ..
}

The rest of the program

A lot of people (including me) usually use the curly braces even if there is only one statement after the "if" or "else". All of the statements in a block are usually indented to make the control structure easy to see by us humans. The compiler ignores indentation.

Regards,

Dave

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