When using scanf with a char array in C why do both forms work whether you use & (address of) or not for the array name? E.g.

Question

#include <stdio.h>
int main() {
char text[10];
char text1[10];
scanf("%s", text);
printf("%s", text);
scanf("%s", &text1);
printf("%s", text1);
}

Answer 1

Actually KrOW you're right!!
I verified this using my code:
https://code.sololearn.com/cVbuSnY7BOOM/?ref=app

Answer 2

KrOW What if I had said char **arrptr=&text1 ? In this case &text1 which is arrptr would not be text1 would it? So how can we say that the address of an array is the same as the address of its first element?

Answer 3

Because array would be seen like constant pointers that point to itself then

text

is the value of this pointer (adress of itself)

&text

is the adress of itself then text and &text have same value

Answer 4

Yes... Adress of an array is same as adress of first element ;)

Answer 5

KrOW basically answered the question XD. The name of a one-dimensional array is a constant pointer.

Although you are right with text being equal to the address stored in the pointer.

&text is the address in memory of the array pointer, and the compiler should have an error or atleast have a warning shown

Answer 6

When you pass a variable into a function, the function creates a copy of that variable and anything that the function does is applied ONLY to the copy. But when you pass an array to a function, it's different; the address to the array is always passed and no copy is created. Anything the function does to the array is done to the original. So if you have a very large array, all of that copying that would otherwise have to be done is eliminated.