数组的值和地址相同，但传递给函数时除外？

Question

void pass_arr(int arr[]);

void pass_arr_test()
{
    int arr[5] = {1,2,3,4,5};

    printf( "arr  = %p\n"
            "&arr = %p\n\n", arr, &arr);

    pass_arr(arr);
}

void pass_arr(int arr[])
{
    printf( "passed arr  = %p\n"
            "passed &arr = %p\n\n", arr, &arr);
}

---

Output:
arr = 0x28ccd0
&arr = 0x28ccd0

passed arr = 0x28ccd0
passed &arr = 0x28ccc0

---

Can someone explain why the value and adress of arr points to the same adress when evaluated in the block where arr was created, but when passed the value and adress point to two different adresses?

Answer 1

That's because in the function arr is actually a pointer , not an array. Taking the address of a pointer does not yield the same address, the way it does for an array.

Answer 2

Except when it is the operand of the sizeof , _Alignof , or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T " will be converted ("decay") to type "pointer to T ", and the value of the expression will be the address of the first element of the array ¹ .

When you call

`pass_arr(arr)`;

the expression arr is converted from type "5-element array of int " to "pointer to int ", or int * .

Note that the address of the first element of the array is the address of the array itself; that's why you get the same value when you print the results of arr and &arr int pass_arr_test , but remember that the types are different; the expression arr is converted to type int * , but &arr has type int (*)[5] ; this matters for things like pointer arithmetic.

Secondly, in the context of a function prototype, declarations of the form T a[] and T a[N] are interpreted as T *a ; a is actually declared as a pointer instead of an array. ²

The important thing to remember is that arrays are not pointers . Rather, in most contexts, array expressions are converted to pointers as necessary.

---

^{1 - N1570 , 6.3.2.1 Lvalues, arrays, and function designators, ¶ 3
2 - N1570 , 6.7.6.3 Function declarators (including prototypes), ¶ 7}

Answer 3

That is due to pass by value semantics where the arr in pass_arr method is a local pointer variable on the stack whose value is the location of the arr passed from pass_arr_test method. So, arr variable in both pass_arr and pass_arr_test are point to the same location and hence the value is same. Since they are two different pointers, their memory address is different. In case of array definition, arr is just an alias to the start of the array and hence it's location and it's value are the same.