Thread: Assigning a char value to char pointer doesn't work.

I'm trying to assign a char value to an char pointer. The problem is
when I comment

Code:

char *a = "Some Birds Can't Fly";

the printed display is

Code:

B

however without commenting above, it gives, lot of scrambled characters. Can you check the code and let me know what's happening.

Code:

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

char *assign_char(char *str, char ch)
{


    size_t len = strlen(str);
    char *strRslt = malloc( len + 2 );

    strcpy(strRslt, str);
    strRslt[len] = ch;
    strRslt[len+1] = '\0';
    
    return strRslt;

}

int main(void)
{
    char *b;
    char *a = "Some Birds Can't Fly";
    char ch = 'B';

    b = assign_char(b, ch);

    printf("%s\n", b);

    return 0;
}

compiled with

Code:

gcc main.c -o main

Originally Posted by laserlight

This is wrong:

Code:

b = assign_char(b, ch);

it should have been:

Code:

b = assign_char(a, ch);

My need is to append the ch with b, a is for demo my problem when I'm having a "a" kind of variable. Hope you get it.So isn't that correct to pass the b = assign_char(b, ch);

Originally Posted by laserlight

Incidentally, to avoid potential attempts to change a string literal, your code should be more const-correct, e.g., in main:

Code:

const char *a = "Some Birds Can't Fly";

and for the assign_char function:

Code:

char *assign_char(const char *str, char ch)

I tried this. However my result is same, see the attached image below, it shows what I'm getting.

Originally Posted by laserlight

It might be more accurate to rename assign_char to append_char.

Also, remember to free() what you malloc(). In assign_char, you should also check that malloc did not return a null pointer before you use what it returns.

Thanks a lot for the tip, I agree with you, these should definitely be implemented.

Back again to my original question, why when an unused string variable is there (a), the out put gets so wired. when I comment that "char *a = "Some Birds Can't Fly";" line. things are normal.

Originally Posted by Salem

Because if you start with a pointer in main() that is uninitialised, when you get to
strcpy(strRslt, str);
you're copying garbage.

It's really that simple.

That's really impressive, thanks for pointing me out. I never thought that way.
Can you please explain me so if I have some bunch of initialized variables after main() like "int x = 1;" will my problem solved?

Actually I tried it just now, still getting the scrabbled print.