Question

consider the following code:

t[7] = "Hellow\0";
s[3] = "Dad";
//now copy t to s using the following strcpy function:

void strcpy(char *s, char *t) {
    int i = 0;

    while ((s[i] = t[i]) != '\0')
        i++;
}

the above code is taken from “The C programming Language book”.
my question is – we are copying 7 bytes to what was declared as 3 bytes.
how do I know that after copying, other data that was after s[] in the memory
wasn’t deleted?

and one more question please: char *s is identical to char* s?

Thank you !

Answer 1

As you correctly point out, passing s[3] as the first argument is going to overwrite some memory that could well be used by something else. At best your program will crash right there and then; at worst, it will carry on running, damaged, and eventually end up corrupting something it was supposed to handle.

The intended way to do this in C is to never pass an array shorter than required.

By the way, it looks like you’ve swapped s and t; what was meant was probably this:

void strcpy(char *t, char *s) {
    int i = 0;

    while ((t[i] = s[i]) != '\0')
        i++;
}

You can now copy s[4] into t[7] using this amended strcpy routine:

char t[] = "Hellow";
char s[] = "Dad";

strcpy(t, s);

(edit: the length of s is now fixed)