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C uses pass by value.

So mystrcpy gets a copy of str2 (string2), which you overwrite with the result of malloc. This does nothing to str2 of main, which continues to keep its old value, which is NULL.

This will works.
Allocate the memory in main()

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

void mystrcpy( char *string2 ,char *string1 )
{
while( *string1 != '\0')
{
*string2++= *string1++;

}

}
int main()
{
printf("****************program starts*************************");
char *str1 ="hello world";
char *str2 =NULL;
str2=malloc(20); //Allocate memory here
mystrcpy( str2,str1);
printf("string1 = %s string2 =%s ",str1,str2) ;
return 0;
}

So here are two different working implementations of your code

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

void mystrcpy( char **string2 ,char *string1 )
{

*string2 = malloc(20);
char *str = *string2;
while( *string1 != '\0')
{
 *str++ = *string1++;
}

}
int main()
{
printf("****************program starts*************************");
char *str1 ="hello world";
char *str2 =NULL;
mystrcpy(&str2,str1);
printf("string1 = %s string2 =%s ",str1,str2);
return 0;
}

AND

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

char* mystrcpy( char *string2 ,char *string1 )
{

string2 = malloc(20);
char *string  = string2;
while( *string1 != '\0')
{
*string2++= *string1++;
}
return string;

}
int main()
{
printf("****************program starts*************************");
char *str1 ="hello world";
char *str2 =NULL;
str2 = mystrcpy( str2,str1);
printf("string1 = %s string2 =%s ",str1,str2);
return 0;
}

It is a simple misunderstanding about Call By Value and Call by Reference...it is call by value (for variables) if you don't mention &..

1. put '\0' at the end of string2.
2. Return that string2 .

You are incrementing both ptr till it hits '\0' so, it will be priting NULL in both the case ..

To solve the problem.

Dont increment the pointer , access it via array, i.e str1[i] and str2[i] .
Return the base address of str2.

so this should solve your problem.

malloc() after successful allocation returns (void*).since we are trying to allocate memory to character strings which needs memory to be of (char*). so,typecasting is required to malloc() function by (char*). otherwise compiler can not interpret which type of variable we are allocating hence returns null.

malloc() after successful allocation returns (void*).since we are trying to allocate memory to character strings which needs memory to be of (char*). so,typecasting is required to malloc() function by (char*). otherwise compiler can not interpret which type of variable we are allocating hence returns null.

put '/0' at end you get the string2

Correct code

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

char * mystrcpy( char *string2 ,char *string1 )
{

string2 = (char*)malloc(20);
char * temp = string2;
while( *string1 != '\0')
{
*temp= *string1;
temp++;
string1++;
}

return string2;
}
int main()
{
printf("****************program starts*************************");
char *str1 ="hello world";
char *str2 =NULL;
char *str3 = mystrcpy( str2,str1);
printf("string1 = %s string2 =%s, str3 = %s ",str1,str2,str3) ;
getchar();
return 0;
}

correct code :
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

char* mystrcpy( char *string2 ,char *string1 )
{
int l,i;
string2 = (char*)malloc(20);
l = strlen(string1);

i=0;
while(i<l)
{
*(string2+i)=*(string1+i);
i++ ;
}
*(string2+i)='\0';
//string2 = "hiii";
return(string2);

}
int main()
{
printf("****************program starts*************************\n");
char *str1 ="hello world";
char *str2 =NULL;
str2 = mystrcpy( str2,str1);
printf("string1 = %s\nstring2 =%s ",str1,str2) ;
return 0;
}

because you are incrementing str2 thats why it is going upto last and it is printing NULL here.....so you take another string and then copy the base address of str2 in that siring the print using that string... :)