How does scanf work in char array initializing?

Asked Jun 29 '20 at 12:50

Active Jun 29 '20 at 12:50

Viewed 31 times

I learned that char array for string like char s1[10] = "Hello"; has to be initialized with value when it's declared.

But, with scanf, it seems like I can declare it without initializing.

I'm wondering how it works.

e.g.

#define _CRT_SECURE_NO_WARNINGS    
#include <stdio.h>
    
int main() {
    char s1[6];
    s1 = "Hello" // It does not work.
    scanf("%s", s1); // And write Hello, it works. Why does it work?

    return 0;
}

Thank you for reading

asked Jun 29 '20 at 12:50

lu_ucky

Your learning is wrong. Use `strcpy()` to copy strings. – MikeCAT Jun 29 '20 at 12:52
You can't assign to arrays, only copy to them. – Some programmer dude Jun 29 '20 at 12:53
1

As for `scanf` and how it "initializes" the array, it copies one character at a time into the array, beginning by the pointer to the first element (which you pass with the argument). – Some programmer dude Jun 29 '20 at 12:54
It should be noted that using `scanf` on strings is dangerous as it doesn't protect against overrunning memory. – Fiddling Bits Jun 29 '20 at 12:59

How does scanf work in char array initializing?

0 Answers0