Thread: Basic Help with Arrays and Output

Okay, here's a simplified version of my program:

Code:

/* INCLUDES */
#include <stdio.h>

int lookupTable(char dataCharacter1, char dataCharacter2);
char encodeB(int dataCharacter, int position);

main()
{
	int i;
	int inputB[4];
	int outputB[120];
	
	*outputB = 0;

	/* test sequence 0367 */
	inputB[0] = 0;
	inputB[1] = 3;
	inputB[2] = 6;
	inputB[3] = 7;

	for (i = 0; i <= ((int)(sizeof(inputB)) / 2); i++)
	{
		*outputB += lookupTable(inputB[i], inputB[i+1]);
		i++;
	}

	printf((char *)outputB);
}

int lookupTable(char dataCharacter1, char dataCharacter2)
{
	int encodation[20];
	int j;

	for (j = 0; j <= 8; j++)
	{
		encodation[j] = encodeB(dataCharacter1, j);
		j++;
	}
}

char encodeB(int dataCharacter, int position)
{
	char bEncodation[2];

	if (dataCharacter == 0)
	{
		if (position == 0)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
		else if (position == 2)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
		else if (position == 4)
		{
			bEncodation[0] = '0';
			bEncodation[1] = '0';
			
			return *bEncodation;
		}
		else if (position == 6)
		{
                                                bEncodation[0] = '0';
			bEncodation[1] = '0';
			
			return *bEncodation;		}
		else if (position == 8)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
	}
}

If you put a breakpoint on the line: encodation[j] = encodeB(dataCharacter1, j);
in the lookupTable function, and watch the values of encodation[j] change as you step through, you'll notice that for each even position (encodation[0], encodation[2], etc.) the value is 48 ('0'). However, the value for encodation[4] (and encodation[6]) needs to be '00'. How do I get the output to have 2 zeros rather than just 1? Do I not use an array in the function: char encodeB(int dataCharacter, int position) where I have the nested if statement:

Code:

		else if (position == 4)
		{
			bEncodation[0] = '0';
			bEncodation[1] = '0';
			
			return *bEncodation;
		}

? Any help is greatly appreciated and I can always post the full program if necessary (it shows what happens to the odd positions of encodation[j]...very similar). Thanks!

You need to return a string instead of a character.

So are you saying I need to change this function:

Code:

char encodeB(int dataCharacter, int position)
{
	char bEncodation[2];

	if (dataCharacter == 0)
	{
		if (position == 0)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
		else if (position == 2)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
		else if (position == 4)
		{
			bEncodation[0] = '0';
			bEncodation[1] = '0';
			
			return *bEncodation;
		}
		else if (position == 6)
		{
                                                bEncodation[0] = '0';
			bEncodation[1] = '0';
			
			return *bEncodation;		}
		else if (position == 8)
		{
			*bEncodation = '0';
			return *bEncodation;
		}
	}
}

to something like this???:

Code:

char encodeB(int dataCharacter, int position)
{
	char bEncodation[2];

	if (dataCharacter == 0)
	{
		if (position == 0)
		{
			*bEncodation = "0";
			return *bEncodation;
		}
		else if (position == 2)
		{
			*bEncodation = "0";
			return *bEncodation;
		}
		else if (position == 4)
		{
			bEncodation[0] = "0";
			bEncodation[1] = "0";
			
			return *bEncodation;
		}
		else if (position == 6)
		{
                                                bEncodation[0] = "0";
			bEncodation[1] = "0";
			
			return *bEncodation;		}
		else if (position == 8)
		{
			*bEncodation = "0";
			return *bEncodation;
		}
	}
}

Just a little confused. Thanks for the help

MacGyver, thank you for the explanation. I'm teaching myself how to do everything via the internet and the C Programming Language book...guess I'm not teaching myself that well

Stepping through your code helped me understand your technique; however, when I implemented it into my code, it was not producing the result I wanted.

I changed my code to the below. My difficulty stands at the point of returning from the function encodeB. The variable encodation does not hold the previous returned value (whether it be 0 or 00). And therefore, when lookupTable returns, the variable outputB does not receive the entire value of various 0's. What do I need to change? Thanks for ALL your help!

Code:

/* INCLUDES */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *lookupTable(char dataCharacter1, char dataCharacter2);
char *encodeB(int dataCharacter, int position);

main()
{
	int i;
	int inputB[4];
	char *outputB = NULL;
	
	outputB = malloc(120);

	/* test sequence 0367 */
	inputB[0] = 0;
	inputB[1] = 3;
	inputB[2] = 6;
	inputB[3] = 7;

	for (i = 0; i <= ((int)(sizeof(inputB)) / 2); i++)
	{
		outputB = lookupTable(inputB[i], inputB[i+1]);
		i++;
	}

	printf(outputB);
}

char *lookupTable(char dataCharacter1, char dataCharacter2)
{
	int j;
	char *encodation = NULL;
	encodation = malloc(100);

	for (j = 0; j <= 8; j++)
	{
		encodation = encodeB(dataCharacter1, j);
		j++;
	}

	return encodation;
}

char *encodeB(int dataCharacter, int position)
{
	char *bEncodation = NULL;
	bEncodation = calloc(3,sizeof(char));

	if (dataCharacter == 0)
	{
		switch(position)
		{
			case 4:
			case 6:
				bEncodation[1] = '0';
			case 0:                                       
			case 2:
			case 8:
				bEncodation[0] = '0';
			break;
		}
	}

	return bEncodation;
}

I plotted everything out on paper and I guess I just don't know enough about C to get my result. When I translated my steps, the compiler does not like my syntax.

For the function: char *lookupTable(char dataCharacter1, char dataCharacter2)

I was thinking this line:

Code:

encodation = encodeB(dataCharacter1, j);

should be something like one of these lines, but I guess it's the wrong C syntax.

Code:

encodation += encodeB(dataCharacter1, j);

or

Code:

encodation[j] = encodeB(dataCharacter1, j);

Still looking through the C Programming book I have, but so far not seeing what I need to do to change my line of code.

slowcoder: I also learned from K&R2. Unfortunately, a few things in there weren't updated in the last edition (and that was in 1988!). You should know that

1) Today it's considered good practice (and required in C99) to explicitly declare the return type of each function, including main(), even though it defaults to int which is the right type for main(). Read

http://faq.cprogramming.com/cgi-bin/...&id=1043284376

2) Even though K&R2 explicitly casts malloc, this isn't necessary anymore (and it wasn't even necessary then - the need was removed between the 1st and 2nd editions). See

http://faq.cprogramming.com/cgi-bin/...&id=1043284351

Check your switch statement. Unless you want the next case to be run "fallen into", you need to have a break statement after each case statement.

robatino, thank you for the links. I changed my main function accordingly and bookmarked that page for future reference so I don't make that mistake again. I'm still digesting the malloc page, and will change my code accordingly with that as well. Thank you for pointing those out for me. Yes, it seems that the K&R2 book is a little outdated.

Adak, i do want the next case to be run "fallen into". There is no need in my program to have a break statement after each case statement, but thank you.

My question posed on 05-04-2007 11:45 AM (further explained in my post on 05-03-2007 03:28 PM), still remains. Any help with figuring out what syntax I need to use in order to have the variable encodation to hold the previous returned value (whether it be 0 or 00), it would be greatly appreciated.

Thanks!

Review page 103 in K&R2. Pointer additiion is not:

*ptr++; /* this increments whatever the ptr is pointing at, not what the pointer is pointing to */

it's:

ptr++; /* this is what you want for pointer addition */

Adak

Adak, I read that page over and the explanation wasn't totally clear.

In my program, I changed it to:

Code:

encodation = encodeB(dataCharacter1, j);
encodation++;

but when I did: printf(encodation);

Nothing printed, and therefore, did not work. So I wasn't sure if it meant: encodation++ = encodeB(dataCharacter1, j);

but that didn't work, because I received error C2106 left operand must be l-value. I'm going to re-read the whole chapter, but if anyone has a better explanation of this, it would be rather helpful at this point. Thanks!

I have a program that I've been using as reference (not written by me and fully works), and I noticed that there's a line of code:

Code:

*txt_ptr++ = '\\';

where \\ is added to the end when it gets printed. So that's why I thought doing:

Code:

*encodation++ = encodeB(dataCharacter1, j);

would work. I just want to add whatever encodeB() returns to the end of encodation. Still can't get it working with:

Code:

encodation = encodeB(dataCharacter1, j);
encodation++;

And when I did:

Code:

printf(*encodation);

as Adak recommended, I received two warnings: C4047: 'function' : 'const char *' differs in levels of indirection from 'char' and C4024: 'printf' : different types for formal and actual parameter 1. Am I really suppose to print *encodation? If so, what would I have to change/add to the syntax?

Thank you Adak for your explanation. Still looking through more documents online to get a better understanding. Just trying to get this working so I can really understand this stuff better....much easier when you have a working example...