C++ * vs [] as a function parameter

问题

What is the difference between:

void foo(item* list)
{
    cout << list[xxx].string;
}

and

void this(item list[])
{
    cout << list[xxx].string;
}

Assuming item is:

struct item
{
    char* string;
}

With the pointer pointing to the first of an array of chars

and list is just an array of items...

回答1:

To the compiler, there is no difference.

It reads different though. [] suggests you want to pass an array to the function, whereas * could also mean just a simple pointer.

Note that arrays decay to pointers when passed as parameters (in case you didn't already know).

回答2:

They are the same - completely synonymous. And the second is item list[], not item[]list.

However it is customary to use [] when the parameter is used like an array and * when it's used like a pointer.

回答3:

FYI:

void foo(int (&a)[5]) // only arrays of 5 int's are allowed
{
}

int main()
{
  int arr[5];
  foo(arr);   // OK

  int arr6[6];
  foo(arr6); // compile error
}

but foo(int* arr), foo(int arr[]) and foo(int arr[100]) are all equivalent

来源：https://stackoverflow.com/questions/10760893/c-vs-as-a-function-parameter