Why can a function receive a char array as a char pointer and why can the function change the value of an element?

Question

I declare char pointer as a parameter of a function and put a char array as an argument.

I was told that char pointer and char array are different types.

But the function can have a char array as a char pointer.

In addition,an element of char pointer can be changed, although the parameter is char pointer.

Here's the code to show the case.

#include<stdio.h>
void changeelement(char *p) {
  p[0]='v';
}
int main() {
    char array[]="boice";
    changeelement(array);
  printf("%s", array);
    return 0;
}

And this is the result. enter image description here

There are many answers about the difference of char pointer and char array, but there is no answer of this case.

Thank you for taking time out of your time.

Your function doesn't change the variable `p` itself, it changes the memory where `p` is *pointing*. Which is the first element of the array `array` inside the `main` function. — Some programmer dude, Apr 03 '23 at 12:21
Regarding the relationship between arrays and pointers, an array can *decay* to a pointer to its first element. So the call `changeelement(array)` is really the same as `changeelement(&array[0])`. — Some programmer dude, Apr 03 '23 at 12:22
Does this answer your question? [What is array to pointer decay?](https://stackoverflow.com/questions/1461432/what-is-array-to-pointer-decay) — BoP, Apr 03 '23 at 13:14

Vlad from Moscow · Answer 1 · 2023-04-03T13:26:40.330

The compiler adjusts parameters having array types to pointers to array element types.

So these function declarations

void changeelement(char p[100] );
void changeelement(char p[10] );
void changeelement(char p[] );

and

void changeelement(char *p );

are equivalent and declare the same one function.

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.

You may write for example in the same program

void changeelement( char *p ); 

void changeelement( char p[] ) 
{
    p[0] = 'v';
}

On the other hand, an array used as an argument expression is implicitly converted to pointer to its first element.

From 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.

In C passing an object by reference means passing the object indirectly through a pointer to it. From the C Standard: "A pointer type describes an object whose value provides a reference to an entity of the referenced type." Thus dereferencing the pointer the function has a direct access to the object pointed to by the pointer and can change it.

For example

#include <stdio.h>

void f( int *p )
{
    *p = 10;
}

int main( void )
{
    int x = 0;

    f( &x );

    printf( "x = %d\n", x );
}

The program output is

x = 10

Similarly when a function deals with a pointer to the first element of an array then using the pointer arithmetic and dereferencing the pointer the function can change any element of the array.

By the way the above function may be rewritten like

void f( int p[] )
{
    p[0] = 10;
}

though the argument expression does not use an array. In any case even you use an array as an argument expression it is implicit;y converted to a pointer to a single element: the first element of the array.

Why can a function receive a char array as a char pointer and why can the function change the value of an element?

1 Answers1