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:

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

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the printed display is

Code:

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B

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

Code:

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#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;
}

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compiled with

Code:

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gcc main.c -o main

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This is wrong:

Code:

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b = assign_char(b, ch);

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it should have been:

Code:

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b = assign_char(a, ch);

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

Code:

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

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and for the assign_char function:

Code:

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

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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.

Quote:

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Originally Posted by laserlight

This is wrong:

Code:

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b = assign_char(b, ch);

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it should have been:

Code:

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b = assign_char(a, ch);

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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);

Quote:

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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:

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

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and for the assign_char function:

Code:

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

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I tried this. However my result is same, see the attached image below, it shows what I'm getting.
Attachment 15051

Quote:

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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.

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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.

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.

Quote:

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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.

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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.

> Can you please explain me so if I have some bunch of initialized variables after main() like "int x = 1;" will my problem solved?
No it won't.

What you're observing is undefined behaviour.
Your initial observation is down to pure dumb luck. Your 'uninitialised' pointer just happened to point to an empty string instead of a bunch of garbage.

Remember, C is a language with very few safety nets.
If you ask "I want to hang myself", C will happily give you the rope to do it.
Other languages will either ask "are you sure?" or simply say "no".