C 中的 char (*)[20] 是什么意思？

Question

I know that the below code is not valid since %s expects an argument of type char * but I have given an argument of type char ** .

#include <stdio.h>
struct employee
{
    int id;
    char name[20];
    float salary;
};
int main(void)
{
    struct employee e;
    scanf("%19s", &e.name);//Invalid line
    printf("%s", e.name);
}

But I get a different warning:

./ Playground / file0. c: In function main:. / Playground / file0. c: 11: 15: warning: format '% s' expects argument of type' char * ', but argument2 has type char (*) [20]' [- Wformat =] 11 | scanf ("% 195", & e. name); | char (*) [20] char *

I am pretty much sure that char ** and char (*)[20] are somewhat equivalent but not sure how they are.

char ** : pointer to a pointer to a char.

char (*)[20] : An array of pointers to a char. — Not sure if I am totally right here.

I am not getting it. How are they equivalent? Please enlighten me.

Answer 1

The type char (*)[20] is read as "pointer to array of size 20 of char . It is not an array of pointers to char .

An array (of size 20) of pointers to char would be char *[20] . Because this is an array, it can decay to a pointer to the first element of the array. Since the array's elements are of type char * , such a pointer would have type char ** .

Going back to your code, e.name has type char [20] , ie array of size 20 of char . From there it follows that &e.name has type char (*)[20] , ie pointer to array of size 20 of char .