Why cannot use pointer of pointer as parameter to declare a function receive pointer of array

Question

I have a c function receiving pointer of array as its one parameter.

Since pass an array is actually the pointer of its first element, so the pointer of an array should be a pointer of pointer.

int arr[]={0,1,2,3};
int main(){
    receiveArray(arr);
    receiveArrayPtr(&arr);
}
int receiveArray(int *arrPara){....}
int receiveArrayPtr(int **arrPtrPara){....} //Why cannot do this?

The Error message is

"expected ‘int **’ but argument is of type ‘int ()[4]’
void receiveArrayPtr(int*);"

Wish to know why

int **p; : type of *p is int * but address of array top is int value of 0, not pointer to int. — BLUEPIXY

John Bode John Bode · Accepted Answer · 2014-07-18T14:21:51

Since pass an array is actually the pointer of its first element, so the pointer of an array should be a pointer of pointer.

That's not how it works.

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

The type of the expression arr in the call to receiveArray is "4-element array of int". Since it is not the operand of the sizeof or unary & operators, it is converted to an expression of type int *, which is what the function receives.

In the call to receiveArrayPtr, the expression arr is the operand of the unary & operator, so the conversion rule doesn't apply; the type of the expression &arr is "pointer to 4-element array of int", or int (*)[4].

This conversion is called array name decay.

Why cannot use pointer of pointer as parameter to declare a function receive pointer of array

2 Answers