string's null character

[geekocoder]

do i need to add
*t='\0';
after line 10....or is it copied from str1 itself??

#include <stdio.h>
int main()
{
    char str1[]="hello";
    char str2[10];
    char *t,*s;
    s=str1;
    t=str2;
    while(*t=*s)
    *t++=*s++;
    printf("%s\n",str2);
    getchar();
}

[snow 56767]

No, you shouldn't have to. But your code probably won't run as expected.

I have reformatted your code so it makes a little more sense:

#include <stdio.h>

int main()
{
    char *pStr1 = "hello";
    char vStr2[10];
    char *pS = pStr1;
    char *pT, = vStr2;

   while(*t != NULL || *s != NULL)
   {
        ++pT;
        ++pS;
        *pT = *pS;
   }

   printf("%s\n", vStr2);
   getc();

   return 0;
}

Now I'm going to explain my changes and why I made them.
Line 5: I changed your array of undefined size to a pointer. No need to worry, they are essentially the same and so many operations still work on a pointer that do an array.

Line 10: Saying while(*t=*s) will always be true and will always set *t to *s. I also did bound checking (crude and rusty) in this line making it so you didn't corrupt memory.

IN THE LOOP: I have made it so each loop iteration will increment your pointers, moving them along the array in memory, and setting the destination to the source *pT = *pS;

NOTE: This code may seem confusing at first, but that is because I am using a coding convention that makes sense to me. What is important here are the principles at work.

[geekocoder]

dont you think you arent copying the 1st letter of pSTR1???

[JLBorges]

Canonical:

const char str1[] = "hello" ;
char str2[ sizeof(str1) ] ;

// copy characters in str1 (including the null character) to str2 
char* dest = str2 ;
const char* srce = str1 ; // EDIT: made const-correct
while( ( *dest++ = *srce++ ) ) ;

[ne555]

> I changed your array of undefined size to a pointer.
It was not undefined, it was `as big to hold the string and terminator character'.
And string literals are const char*

> Saying while(*t=*s) will always be true
No, it fails when the assignment is zero.

> I also did bound checking
*t is not initialized at the start of the program.

[geekocoder]

can any 1 plz tell me when will while(*t=*s) return 0 in the above code??

[vlad from moscow]

I think you meant while ( *t++ = *s++ ) as JLBorges showed.

At first *t is assigned the value of *s and this value before incrementing pointer t becomes the result of the assignment operator which is tested in the while statement So if *s is equal to '\0' then at first it is assigned to *t and becomes the result of the assignment expression.