copy char array

[ronenk]

This code is meant to copy string from one string to another. output I get is gibberish. I suspect it is a matter of initializing (or not...).

#include<stdio.h>
void cpystring(const char *source, char *target)
{
	
	char *p=target;
	while (*p++=*source++);
	puts (target);
	puts (source);
}

int main(void)
{
	char source[6]="qwerty", target[6];
	cpystring (source, target);

	return 0;
}

[Micko]

Try this:

void cpystring(const char *source, char *target)
{
	/*
	char *p=target;
	while (*p++=*source++);
	puts (target);
	puts (source);*/
	
	
	int i=0;
	while(source[i]!='\0')
	{
		target[i]=source[i];
		++i;
	}
	target[i]='\0';
	puts(source);
	puts(target);
}

[Micko]

And main:

int main(void)
{
	char source[7]="qwerty", target[7];
	cpystring (source, target);

	return 0;
}

The reasons why your code don't work.
1. char source[6]="qwerty";
you have 6 characters + additional character calle zero character ('\0') that terminates string. Remember string is just ordinary array of characters that ends with special character '\0'. So you always must to make a room for it.
2.
You are incrementing source. That is pointer to first element in the array in the begining. After you increment it you are trying to display it with puts() but what is the address of pointer source and what is the value on which points to after while?
For example at the start of your function source has this value:
0x0012fed0, that is the address (beacuse source is a pointer) of first element (byte) in the array and that is letter 'q'. After while
value of pointer variable source is: 0x0012fed7, because there is 7 elements (including '\0') and source is pointing to a byte of memory immediately after '\0' and what is on that memory? You don't know. That's why you get rgarbage characters.

Another solution would be:

void cpystring(const char *source, char *target)
{
	
	char *p=target;
	char *m=(char*) source;
	while (*p++=*m++);
	puts (target);
	puts (source);
}

Cheers!

[ronenk]

OK, thanx!
the only thing is I'm not familiar with this syntax:

(char*) source

please explain what does it mean.

[Thantos]

(char*) is a type cast, which means to convert whatever source is into a char *. Of course it isn't needed since source is already a char *

[ronenk]

Thank's.
((-:

[Prelude]

>Of course it isn't needed since source is already a char *
It's a const char *, so the cast would silence an annoying warning. Of course, a better solution would be to make m a const char * as well and avoid the issue altogether.

[ronenk]

what does the const char mean?

[Prelude]

>what does the const char mean?
In theory you aren't allowed to change a const qualified variable. In reality you can easily subvert the type system and modify something that is const, sometimes with disasterous results.

[ronenk]

Do you recomend not to use it? How do you manage without? what is the substitute?

[Prelude]

>Do you recomend not to use it?
Of course not. If you behave when using it then const will catch bugs that would otherwise show up at run-time. I recommend liberal use of const.

>How do you manage without?
Prayer seems to work for many of my coworkers.

>what is the substitute?

/* PLEASE DON'T CHANGE THIS!!!!!!!111 */

[Thantos]

opps didn't even see the const there. Thats twice in the last 24 hours I've done something like that.

[Micko]

>Of course it isn't needed since source is already a char *
It's a const char *, so the cast would silence an annoying warning. Of course, a better solution would be to make m a const char * as well and avoid the issue altogether.

Credit to teacher Prelude!

I cast it to avoid compiler warning because of const.