When a function has a specific-size array parameter, why is it replaced with a pointer?

Question

Given the following program,

#include 

using namespace std;

void foo( char a[100] )
{
    cout << \"foo() \" << sizeof( a ) <         


        

Answer 1

There is magnificent word in C/C++ terminology that is used for static arrays and function pointers - decay. Consider the following code:

int intArray[] = {1, 3, 5, 7, 11}; // static array of 5 ints
//...
void f(int a[]) {
  // ...
}
// ...
f(intArray); // only pointer to the first array element is passed
int length = sizeof intArray/sizeof(int); // calculate intArray elements quantity (equals 5)
int ptrToIntSize = sizeof(*intArray); // calculate int * size on your system