Passing an array as a parameter in C

why does this code work?

#include <stdio.h>

void func(int v[]){
    v[0] = 1;
}

int main(){
    int v[5] = {0};
    func(v);
    for (int i = 0; i < 5; i++)
    {
        printf("%d ", v[i]);
    }
}

The output I get from this is ‘1 0 0 0 0’ but why? I’m not passing a pointer, why can the function change the array in my main?

>Solution :

This function declaration

void func(int v[]){
    v[0] = 1;
}

is adjusted by the compiler to the declaration

void func(int *v){
    v[0] = 1;
}

From the C Standard (6.7.6.3 Function declarators (including prototypes))

7 A declaration of a parameter as ‘‘array of type’’ shall be
adjusted to ‘‘qualified pointer to type’’, where the type qualifiers
(if any) are those specified within the [ and ] of the array type
derivation. If the keyword static also appears within the [ and ] of
the array type derivation, then for each call to the function, the
value of the corresponding actual argument shall provide access to the
first element of an array with at least as many elements as specified
by the size expression

On the other hand, in this call

func(v);

the array designator v is implicitly converted to a pointer to its first element.

The C Standard (6.3.2.1 Lvalues, arrays, and function designators)

3 Except when it is the operand of the sizeof operator or the unary &
operator, or is a string literal used to initialize an array, an
expression that has type ‘‘array of type’’ is converted to an
expression with type ‘‘pointer to type’’ that points to the initial
element of the array object and is not an lvalue. If the array object
has register storage class, the behavior is undefined.

That is the function call is equivalent to

func( &v[0] );

So in fact the first element of the array is passed to the function by reference through a pointer to it. Dereferencing the pointer by means of the subscript operator (the expression v[0] is equivalent to the expression *v)

v[0] = 1;

the referenced first element of the array is changed.