Kunal Agarwal Kunal Agarwal - 1 month ago 5
C Question

How does a char pointer differ from an int pointer in the below code?

1)
int main()
{
int *j,i=0;
int A[5]={0,1,2,3,4};
int B[3]={6,7,8};
int *s1=A,*s2=B;
while(*s1++ = *s2++)
{
for(i=0; i<5; i++)
printf("%d ", A[i]);
}
}


2)
int main()
{
char str1[] = "India";
char str2[] = "BIX";
char *s1 = str1, *s2=str2;
while(*s1++ = *s2++)
printf("%s ", str1);
}


The second code works fine whereas the first code results in some error(maybe segmentation fault). But how is the pointer variable s2 in program 2 working fine (i.e till the end of the string) but not in program 1, where its running infinitely....
Also, in the second program, won't the s2 variable get incremented beyond the length of the array?

Answer

The thing with strings in C is that they have a special character that marks the end of the string. It's the '\0' character. This special character has the value zero.

In the second program the arrays you have include the terminator character, and since it is zero it is treated as "false" when used in a boolean expression (like the condition in your while loop). That means your loop in the second program will copy characters up to and including the terminator character, but since that is "false" the loop will then end.

In the first program there is no such terminator, and the loop will continue and go out of bounds until it just randomly happen to find a zero in the memory you're copying from. This leads to undefined behavior which is a common cause of crashes.

So the difference isn't in how pointers are handled, but in the data. If you add a zero at the end of the source array in the first program (B) then it will also work well.