Trouble with pointers, ascii values, and string manipulation

**bluej322** · 01-04-2011

Hi, i'm pretty new to programming (as you will see), and i'm having difficulty with this program that takes a string and adds a number (i) to each character's acsii value based on its index in the string (i). For example, the string "aaaaa" would become "abcde".

Here's the code:

Code:

#include <stdio.h>
#include <string.h>

main()
{
	char *origin;
	int sLimit = strlen(origin);
	int i;
	
	printf("Enter a string to encode.\n\n");
	scanf("%s",origin);

	for (i=0;i<sLimit;i++) {
		
		printf("Value %i: %i\n", i, origin[i]+i);
		
	}
}

The program successfully complies, but when i enter the string "aaaa", i get the following output:

Enter a string to encode.

aaaa
Value 0: 97
Value 1: 98
Segmentation fault

Any help would be appreciated!

**kermit** · 01-04-2011

Code:

char *origin;

Try allocating some space for your string. Then you need to move the call to strlen to a point after you have actually gotten user input into the array.

**anduril462** · 01-04-2011

Read up on the printf documentation. You're printing the value out using the %i format, which means "treat this as an integer". Try the %c format for "treat this as a character".

**anduril462** · 01-04-2011

Oh, and to follow kermit's suggestion, you can't call strlen until after your scanf. There will be garbage in the string until you place something (user input) in there.

And it's int main(void), and return 0 when you're done: http://faq.cprogramming.com/cgi-bin/...&id=1043284376

**bluej322** · 01-04-2011

Originally Posted by kermit

Code:

char *origin;

Try allocating some space for your string. Then you need to move the call to strlen to a point after you have actually gotten user input into the array.

Thanks! I moved the strlen() and allocated space for the string and it looks like it's working.
Can you explain why using "origin[10]" works and "*origin" doesn't?

**anduril462** · 01-04-2011

char origin[10] creates an array of 10 characters, which is sufficient to store a string of length 9, plus one byte for the trailing null character that tells C that the string is done. char *origin simply allocates space for a pointer to a character. This is meant to store a memory address and it's size depends on your system (typically 32 bits for a 32-bit system). Defining a pointer variable doesn't allocate space for it to point to. Think of it as an entry in your address book. Writing in somebody's address doesn't create a house, it just tells you where that house is.

**bluej322** · 01-04-2011

Originally Posted by anduril462

char origin[10] creates an array of 10 characters, which is sufficient to store a string of length 9, plus one byte for the trailing null character that tells C that the string is done. char *origin simply allocates space for a pointer to a character. This is meant to store a memory address and it's size depends on your system (typically 32 bits for a 32-bit system). Defining a pointer variable doesn't allocate space for it to point to. Think of it as an entry in your address book. Writing in somebody's address doesn't create a house, it just tells you where that house is.

Okay, that makes sense. But once i use scanf to assign data to that allocation of space am i not "building the house" at that address?

**tabstop** · 01-04-2011

Originally Posted by bluej322

Okay, that makes sense. But once i use scanf to assign data to that allocation of space am i not "building the house" at that address?

To carry this analogy a bit too far, all scanf does is put furniture in the house. If you point it somewhere where there isn't a house, then it's just going to dump it at the side of the road. (And more the point: if the address you give it is somewhere you can't legally go, then everything's going to stop.)

**bluej322** · 01-04-2011

Originally Posted by tabstop

To carry this analogy a bit too far, all scanf does is put furniture in the house. If you point it somewhere where there isn't a house, then it's just going to dump it at the side of the road. (And more the point: if the address you give it is somewhere you can't legally go, then everything's going to stop.)

I'm trying not to delve into this too much , but in what scenario would you want to use an "address", rather than a "house"?

**kermit** · 01-04-2011

Originally Posted by bluej322

I'm trying not to delve into this too much , but in what scenario would you want to use an "address", rather than a "house"?

Any time a function calls for an address, and not a house.

Bu seriously, that is not too far from the truth. One significant place is when you are using functions. As C uses a pass by value scheme, i.e., when you give arguments to a function the values are "copies", you will need to pass a pointer (the address) in to the function, should you want to manipulate the value.

For example, consider the classic case of a swap function:

Code:

void swap(int x, int y)
{
    int tmp;

    tmp = x;    /* Wrong -- We are only working on copies of the original values */
    x = y;
    y = tmp;
}

int main(void)
{
    int x = 5;
    int y = 2;

    printf("x = %d and y = %d\n", x, y);

    swap(x, y);

    printf("x = %d and y = %d\n", x, y);

    return 0;
}

The above code won't work, because the swap function is just working on copies of the variables in main. In order for it to work correctly, you need to give it access to the variables in main. The way you do this is to pass it the address of the variables (a pointer):

Code:

#include <stdio.h>

void swap(int *x, int *y)
{
    int tmp;

    tmp = *x;
    *x = *y;
    *y = tmp;
}

int main(void)
{
    int x = 5;
    int y = 2;

    printf("x = %d and y = %d\n", x, y);

    swap(&x, &y);

    printf("x = %d and y = %d\n", x, y);

    return 0;
}

**bluej322** · 01-04-2011

Thanks for the clarification!
Here's another quick question. In the swap() example you used, the swap() function has the type "void", anduril directed me to a page earlier explaining the different types of functions, there it said that void function type "takes no arguments, and returns nothing". How then, was the swap function able to take an argument from main(), and then return something? Or did i just misread the definition of void function types?

**anduril462** · 01-04-2011

The 'void' before the function name means it returns nothing. The (void) parameter list means it takes no parameters. They are distinct. The swap functions provided return nothing, but take two specific parameters.

**kermit** · 01-04-2011

Well, the void indicates that the function is not returning any value. If I wanted to return an int, for example, I would have declared it:

Code:

int swap(int *x, int *y)

If I was not passing any arguments into the function, then I would have used void inside the parentheses:

Code:

void swap(void)

That is just as an example - as a swap function, it would not do much. Just remember, when you declare a function, you have it's return type, its name, and any arguments it takes. If you aren't returning a value, declare it as void, and if you are not passing any arguments, use void in the parentheses.

Edit: Clearly, I cannot type fast enough.

At any rate, a long time ago, I wrote a series of articles on functions in the C language. They don't really touch on pointers, but they do cover some of the basic ideas of functions fairly well. You might find it helpful to browse through them. They are the 6 chapters under "Functions."

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