Switch statement and returning the correct output

I have a large application that I tried to simplify into a small program in order to ask for help on this question.

Code:

static char *the_switch(int cw);
char *txt_ptr = "";

main()
{
	int cw;
	int v[3];
	int i;
	int j;

	v[0]=8;
	v[1]=8;
	v[2]=555;
	v[3]=8;

	for (i = 0; i < 4; i++)
	{
		cw = v[i];
		txt_ptr = the_switch(cw);
		printf("%s", txt_ptr);
	}
}

static char *the_switch(int cw)
{
	switch(cw)
		{
		case 8:
			txt_ptr = "x";
			return txt_ptr;
		case 555:
			txt_ptr = "\\555\\";
			return txt_ptr;
		case 444:
			txt_ptr = "\\444\\";
			return txt_ptr;
		case 333:
			txt_ptr = "\\333\\";
			return txt_ptr;
		default:
			txt_ptr = "not a case";
			return txt_ptr;
		}
}

If I had v[2] = 555 & v[3] = 8, then I need the output to be: \555\8
which is basically just adding a backslash before cw and then adding a backslash after cw and then printing the next number in the array which is 8. If v[2] was 444 and v[3] was 8, then I basically need the output to be the same: \444\8
And finally if v[1] was 333, v[2] was 8, and v[3] was 8, then I would need the output to look like this: \333\8\8

The code that I have above only gives me \555\ instead of \555\8 and obviously the other cases only give me that type of output too. What can I add to fix this in order to get the output that I'm looking for? I'm working with someone else's code, so I'm kind of restricted on how much I can change things. Any help is greatly appreciated. Thanks

The output in may machine is xx\555\x, which is the expected output after examining the code.

Just use 8 instead of "x" in the the_switch() function.

What in the world... ? Well here's my idea:

Code:

#include <stdio.h>

char separator (int *v)
{
    int *next = v;
    if (next && *++next)
        return '\\';
    else
        return '\0';
}

int main(void)
{
    int v[] = { 8, 444, 555, 333, };
    size_t size = sizeof v / sizeof v[0];
    int i;
    for (i = 0; i < size; ++i)
    {
        printf("%d%c", v[i], separator(&v[i]));
    }
    putchar('\n');
    return 0;
}

Owner@pavilion ~
$ ./a
8\444\555\333

Will something like that work for you?

Or just something like:

Code:

printf("\\%d\\%d\\%d\n", v[1], v[2], v[3]);

fnoyan - yes, i do see that my example would produce the correct output if the x in the switch statement was an 8.

However, how would I do "And finally if v[1] was 333, v[2] was 8, and v[3] was 8, then I would need the output to look like this: \333\8\8" with the switch statement that I have in place??? The problem with changing things around as citizen and itsme86 suggested is the fact that I have pre-existing code that has MANY other cases besides mine. I only have to add these 3 cases to the switch statement, but I'm having a really hard time figuring out how to get the output look as I need it for case 333. Your help is always appreciated.

Why do your cases ever return a trailing \
It looks like you always want \#

I would think your cases should be more like...

Code:

"\\8"
"\\333"
"\\555"
...

Though I don't seen any sense in using that method to print a \ in front of a number.

This is your code, but with some modification

Code:

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

void the_switch(int cw);
char *txt_ptr;

int main()
{
	int cw;
	int v[3];
	int i;
	int j;
	txt_ptr = malloc(80);
	strcpy(txt_ptr,"");
	
	v[0]=8;
	v[1]=8;
	v[2]=555;
	v[3]=8;
	for (i = 0; i < 4; i++)
	{
		cw = v[i];
		the_switch(cw);
	  	printf("%s\n",txt_ptr);
	}
	
	getchar();
	return 0;
}

void the_switch(int cw)
{
	switch(cw)
		{
		case 8:
			strcat(txt_ptr,"\\x");
			break;
		case 555:
			strcat(txt_ptr,"\\555");
			break;
		case 444:
			strcat(txt_ptr,"\\444");
			break;
		case 333:
			strcat(txt_ptr,"\\333");
			break;
		default:
			strcat(txt_ptr, "\0");
		}
}  	

/*my output
\x
\x\x
\x\x\555
\x\x\555\x
*/

I guess this will do what u want. u need to just change X to 8 to display like \333\8\8 for 333,8,8

ssharish2005