standard conversions: Array-to-pointer conversion

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予麋鹿
予麋鹿 2020-11-29 10:33

This is the point from ISO :Standard Conversions:Array-to-pointer conversion: $4.2.1

An lvalue or rvalue of type “array of N T” or “array o

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  • 2020-11-29 11:10

    One example of this is that any array variable will automatically degenerate into a pointer to it's first element when passed to a function which takes a pointer of the array's type.

    Take a look at this section from the C-Faq on Arrays and Pointers. This is equally applicable in C++.

    void foo(int *a) {
        a[0] = 1;
    }
    
    int main(void) {
        int b[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    
        foo(b);
    
        printf("b[0] == %d\n", b[0]);
    }
    
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  • 2020-11-29 11:15
    int a[6];
    int *b = a;
    

    Pointer b points to the a[0], i.e. contains the address of the element a[0].

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  • 2020-11-29 11:23

    In both C and C++, an array can be used as if it were a pointer to its first element. Effectively, given an array named x, you can replace most uses of &x[0] with just x.

    This is how subscripting is able to be used with array objects:

    int x[5];
    x[2];     // this is the same as (&x[0])[2]
    

    This is also how an array can be passed to a function that has a parameter of pointer type:

    void f(int* p);
    
    int x[5];
    f(x);     // this is the same as f(&x[0])
    

    There are several contexts in which the array-to-pointer conversion does not take place. Examples include when an array is the operand of sizeof or the unary-& (the address-of operator), when a string literal is used to initialize an array, and when an array is bound to a reference to an array.

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  • 2020-11-29 11:29

    This means, that you can have the following situation:

    int arr[100];
    arr[ 0 ] = arr[ 1 ] = 666;
    // ..
    

    You can use arr as pointer to int, which points to the first element of the array, for example:

    *arr = 123;
    

    and then the array will be: arr = { 123, 666, ... }

    Also, you could pass the array to a function, that takes int*:

    void f( int* a ) { /* ... */ }
    

    and call it:

    f( arr );
    

    It's absolutely the same as calling it like this:

    f( &arr[ 0 ] );
    

    That is what The result is a pointer to the first element of the array. means.


    Another way, you could use the address of the first element is:

    *( &arr[ 0 ] + 1 ) = 222;
    

    this will make the second element in the array with value 222; It's the same as

    arr[1] = 222;
    

    and

    *( arr + 1 ) = 222;
    
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