CODEWITHSUNDEEP
c
jhonnna
jhonnna

Reputation: 107

equivalent expressions memset and array

#include <string.h>
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv){

    struct name{

         char array[8];

    };

    struct name na;
    memset(&na.array,0,8); 


 }

It is the same memset(&na.array,0,8); that memset(na.array,0,8); ?

and it is the same na.array that &na.array[0] ?

Upvotes: 1

Views: 47

Answers (3)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

The three expressions that you mention produce identical results when passed as parameters to a function taking void*. The reasons why it is so are different:

  • &na.array is the address of the array. It matches the address of its initial element, but it has a different type. However, you are passing it to a function taking void*, so the type does not matter.
  • na.array is the array itself. It "decays" to a pointer to array's initial element when passed to a void* parameter.
  • &na.array[0] is the address of array's initial element, taken directly. It gets converted to void* when you pass it as a parameter to memset.

Upvotes: 1

cadaniluk
cadaniluk

Reputation: 15229

It is the same memset(&na.array,0,8); that memset(na.array,0,8); ?

Yes.

However, this only means that both variants behave equivalently. &na.array and na.array have very different types.

na.array is of type char[8], an array of eight chars. &na.array is of type char(*)[8], a pointer to an array of eight chars.

and it is the same na.array that &na.array[0] ?

Strictly speaking, no. na.array is actually of type char[8] while &na.array[0] is of type char *. However, there is an implicit conversion rule from char[8] to char *, which is called decay. So, na.array can decay to a char *, which is precisely the type of &na.array[0].

Upvotes: 1

tdao
tdao

Reputation: 17678

It is the same memset(&na.array,0,8); that memset(na.array,0,8); ?

Same thing, and you should use sizeof instead of hardcode the array size.

memset(&na.array,0,sizeof(na.array));

Personally I would prefer using & to make the address of the array explicit.

Upvotes: 1

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