指向整数数组和正常整数数组的指针

Question

In KR C book page 112 it says that following:

int (*arr1)[10];

is a pointer to an array of 10 integers. I don't get what's difference between above and:

int arr2[10];

1- Isn't arr2 itself a pointer to array of 10 integers? (Because name of an array is a pointer itself.)

2- If the name of an array is the array address and pointer to that array, then both arr1 and arr2 are pointer to array of integers, isn't this true?

Answer 1

Isn't arr2 itself a pointer to array of 10 integers?

No, it's an array.

Isn't the name of an array the array address and pointer to that array?

Array names can be/are converted to pointers to their 0th element (not the entire array).

So both arr1 and arr2 are pointer to array of integers?

No.

1. arr1 is a pointer to an array of 10 integers.
2. arr2 is an array of 10 integers. In most contexts it converts to a pointer to an integer (not a pointer to an array).

---

Check this wrong example for instance:

#include <stdio.h>

int main(void)
{
    int arr2[10] = {0};
    arr2[5] = 747;

    int (*arr1)[10] = {0};
    arr1[5] = 747;

    return 0;
}

Here I am treating both arr1 and arr2 as the "same thing", and I got this error:

C02QT2UBFVH6-lm:~ gsamaras$ gcc -Wall main.c 
main.c:9:13: error: array type 'int [10]' is not assignable
    arr1[5] = 747;
    ~~~~~~~ ^
1 error generated.

But if I do:

arr1[0][5] = 747;

it will pass compilation! Same with (*arr1)[5] = 747; of course.

Answer 2

The relationship between arrays and pointers is one of the more confusing aspects of C. Allow me to explain by way of example. The following code fills and displays a simple one-dimensional array:

void showArray( int *ptr, int length )
{
    for ( int i = 0; i < length; i++ )
        printf( "%d ", ptr[i] );
    printf( "\n" );
}

int main( void )
{
    int array[10];
    for ( int i = 0; i < 10; i++ )
        array[i] = i;
    showArray( array, 10 );
}

You can see that when an array is passed to a function, the array name is taken as a pointer to the first element of the array. In this example, the first element is an int , so the pointer is an int * .

Now consider this code that fills and prints a two-dimensional array:

void showArray( int (*ptr)[10], int rows, int cols )
{
    for ( int r = 0; r < rows; r++ )
    {
        for ( int c = 0; c < cols; c++ )
            printf( "%2d ", ptr[r][c] );
        printf( "\n" );
    }
}

int main( void )
{
    int array[5][10];
    for ( int row = 0; row < 5; row++ )
        for ( int col = 0; col < 10; col++ )
            array[row][col] = row * 10 + col;
    showArray( array, 5, 10 );
}

The array name is still a pointer to the first element of the array. But in this example the first element of the array is itself an array, specifically an array of 10 int . So the pointer in the function is a pointer to an array of 10 int .

What I hope to impress upon you is that a pointer of the form (int *ptr)[10] has some correspondence to a two-dimensional array, whereas a pointer of the form int *ptr has some correspondence to a one-dimensional array.

Answer 3

cdecl.org will show you what C interprets your variable declaration as. Quite handy as you start getting into more complicated variable declarations.

int arr2[10]; declare arr2 as array 10 of int

int (*arr1)[10]; declare arr1 as pointer to array 10 of int