CODEWITHSUNDEEP
cpointerschar
user13469230
user13469230

Reputation:

What does char (*)[20] in C mean?

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.

Upvotes: 1

Views: 5941

Answers (1)

dbush
dbush

Reputation: 225807

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

Upvotes: 1

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