函数头中的C / C ++数组变量

Question

We can pass an array as a variable in a C/C++ function header, as in

int func(int arr[]) { ... }

I'm wondering: Is it ever possible that something goes inside the [] in a variable that's passed into the function header, or is it always empty?

Answer 1

For any (non-reference) type T , the function signatures R foo(T t[]) and R foo(T t[123]) (or any other number) are identical to R foo(T * t) , and arrays are passed by passing the address of the first element.

Note that T may itself be an array type, such as T = U[10] .

Answer 2

for a one-dimensional array, it will always be empty, the brackets are another way of writing:

int fun(int * arr)
{

}

As for a two-dimensional array, you need to specify how many elements each element itself holds

int fun(int arr[][3])
{

}

Answer 3

int func(int arr[]) { ... } is an invalid decleration of an array passed to a function.

An array name is a pointer variable. so it is enough that we just pass the array name (which itself is a pointer )

int func(int *arr) { ... } will pass the starting address of the array to the function so that it can use the array.

if the original array needs to be kept intact, a copy of the array can be created & used within the function.

Answer 4

The name of an array decays into a pointer to its first element in most contexts. So when you write

void f(int arr[]);
void g(int arr[42]);

the name arr decays into a pointer to int. The two declarations are equivalent to these:

void f(int *arr);
void g(int *arr);

Two places where the name does not decay are in the definition of an array and as an argument to sizeof . So this declaration at global scope does not define a pointer to int:

int arr[];

I mention this particular one because it's an easy mistake to make and one that's hard to track down. This defines an array, not a pointer, even though the number of elements is not specified. It is an error (but one that doesn't have to be diagnosed) to refer to arr from another source file as int *arr; .