Left shift a string

Gudday all
I wish to left shift a string by one char to remove a leading '\'. In concept it seems simple enough but I flounder on the strings handling still esp. initialisation, pointers etc:

Code:

#include <stdio.h>
#include <string.h>
#include <io.h>
#include <stdlib.h>
#include <cstring>

int main(int argc, char *argv[])
{
	size_t size = 14;
	char *dst[size]= {0};
	char *src[size+1] = "\0000544D.TIF";

 while (size != 0) {
	 dst[size] = src[size-1];
	 size--;
}
}

The error I get is

C:/ScanImages/leftShiftACharacter.cpp: In function `int main(int, char**)':
C:/ScanImages/leftShiftACharacter.cpp:13: error: variable-sized object `dst' may not be initialized
C:/ScanImages/leftShiftACharacter.cpp:14: error: variable-sized object `src' may not be initialized

It must be staring me in the face but I have lost it.

I really find strings and pointers to be very challenging. Is there a good on-line reference that is not too rocket science?

That means just what it says -- either explicitly put 14 in the [] when you declare your arrays, or do not initialize them when you declare them.

Also, the string "\0000544D.TIF" does not contain a '\' character -- it contains the characters '\0', '0', '0', '0', '5', '4', '4', 'D', '.', 'T', 'I', 'F', '\0' in that order.

Just add 1 to the pointer location.

Originally Posted by tabstop

That means just what it says -- either explicitly put 14 in the [] when you declare your arrays, or do not initialize them when you declare them.

Also, the string "\0000544D.TIF" does not contain a '\' character -- it contains the characters '\0', '0', '0', '0', '5', '4', '4', 'D', '.', 'T', 'I', 'F', '\0' in that order.

Your comment about the string not containing '\' puzzles me. How then is a '\' expressed if it looks like a '\0'?
How does C differentiate between a '\' and a '\0'?

The problem is that your trying to declare an array with the variable amount of size. Which is not allowed. Either you need to hard code size of the array you want or use a macro.

You really require a char array; but not a array char pointers. so take off the leading '*' char from the array declarations.

And your logic to eliminate the leading '\' and push all other characters to left won't work. Can you see why? size should really be initialised to 0.

~ssharish

Originally Posted by Tigers!

Your comment about the string not containing '\' puzzles me. How then is a '\' expressed if it looks like a '\0'?
How does C differentiate between a '\' and a '\0'?

Have you ever used \n before? Haven't you ever used \0 before? What about \t? Or \a? Or \"?

>How does C differentiate between a '\' and a '\0'?
'\0' is an escape sequence char. The parser would under stand all the escape sequnce and act accordingly. For the list of escape sequence char have a look the link.

~ssharish

Originally Posted by ssharish2005

The problem is that your trying to declare an array with the variable amount of size. Which is not allowed. Either you need to hard code size of the array you want or use a macro.

You really require a char array; but not a array char pointers. so take off the leading '*' char from the array declarations.

And your logic to eliminate the leading '\' and push all other characters to left won't work. Can you see why? size should really be initialised to 0.

~ssharish

Now you have me worried as to what I am missing or have missed.

I learnt/used C many years ago at uni. Trying to catch up again is proving to be a long process.

Originally Posted by tabstop

Have you ever used \n before? Haven't you ever used \0 before? What about \t? Or \a? Or \"?

Oh I have used those before.
Strings have been a thorn in my side these last 3 months. Trying to remember when a \n is added and when it is not. Trying to remember when \0 should be there or not.

It is very slowly coming back.

>> char *src[size+1] = "\0000544D.TIF";

That's an array of 14 char* - you probably want to drop the pointer.

>> Now you have me worried as to what I am missing or have missed.

If the string is a literal, then you'd need "\\" to represent a single "\", but that only applies to what the compiler "sees", not a string read from the user, a file, etc. At any rate, all that isn't necessary (I didn't compile this, but it should work fine):

Code:

char* remove_char( char* str, char fmt )
{
	char
		* cpy, 
		* pos;
	for( pos = cpy = str; *pos; ++pos )
		if( *pos != fmt )
			*cpy++ = *pos;
	*cpy = 0;
	return str;
}

Or using indexes, instead:

Code:

char* remove_char( char* str, char fmt )
{
	int
		cpy, 
		pos;
	for( pos = cpy = 0; str[ pos ]; ++pos )
		if( str[ pos ] != fmt )
			str[ cpy++ ] = str[ pos ];
	str[ cpy ] = 0;
	return str;
}

Example:

Code:

int main( void )
{
	char*
		buf[ 1024 ] = "\0000544D.TIF";
	puts( remove_char( buf, '\\' ) );
	return 0;
}

Of course it strips all of the chars from the text, but a similar approach could be used to remove only the first one, if desired.

Originally Posted by carrotcake1029

Just add 1 to the pointer location.

Seconded

Code:

int main(int argc, char *argv[])
{
    const char *src = "\\0000544D.TIF";
    printf("%s\n", src);  // WITH leading backslash
    printf("%s\n", src+1); // WITHOUT leading backslash
}