Strcpy function

I was trying to copy a string into another without with my own program
but there is no output on screen -->

Code:

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

char * copy(char*s)
{
        char *b=malloc(strlen(s));
       
       
       
        while(*s!='\0')
        {
                *b=*s;
                s++;
                b++;

        }
        *b='\0';
        return b;
}

int main(void)
{
        char *a="Hello Tarun";
        char *b;
        b=copy(a);
        puts(b);
        return 0;

}

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At the end of your loop, where is b pointing?

And, fwiw, not seeing the result isn't the only problem you have.

Quote:

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Originally Posted by CommonTater

At the end of your loop, where is b pointing?

And, fwiw, not seeing the result isn't the only problem you have.

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b is collecting the pointer returned from function
yes no output is something wrong with code ??

Quote:

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Originally Posted by tarunjain07

b is collecting the pointer returned from function

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That may be what you want it to do. But that's not what it's doing. Read what's actually there, not what you want it to be.

Quote:

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Originally Posted by tarunjain07

b is collecting the pointer returned from function
yes no output is something wrong with code ??

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There are many things wrong with the code. As with your other post you are not understanding what you are trying to do. You cannot just randomly throw code at a problem and hope a working solution comes out. I would like to point something out for you, via code, in case we have a language barrier here. Note: Even in example code I am not using single letter variables. It is an extreme bad practice outside of the world of loop iterators.

Code:

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#include <stdio.h>

int myChar(char *myString){

        int FunctionNumber=0;

        while(*myString != '\0'){
                myString++;
                FunctionNumber++;
        }

        return FunctionNumber;
}

int main(void){

        int numberReturn = 0;
        char *StartString = "Hello Tarun";

        numberReturn = myChar(StartString);

        printf("%d", numberReturn);

        getchar();
        return (0);
}

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Quote:

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Originally Posted by tarunjain07

b is collecting the pointer returned from function
yes no output is something wrong with code ??

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At the end of your function when you return b ... b is pointing at the null on the end of your string.

Not only that, but the minute you did b++ you orphaned the memory you reserved for the string, causing a "memory leak" where free() will fail to release it.

You need to preserve your pointers in their original locations or you lose your strings...

Code:

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char * copy(char*s)
  { char *b=malloc(strlen(s) + 1);  // allow for trailing null
    int i = 0;
    do
      { b[i]=s[i];
        i++; }
    while(s[1])
    return b; }

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Then in main you need to release the memory you've reserved...

Code:

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int main(void)
{
    char *a="Hello Tarun";
    char *b = NULL;
    b = copy(a);
    puts(b);
    free(b);
    return 0;
}

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