Array Assignment

Question

Let me explain with an example -

#include 

void foo( int a[2], int b[2] ) // I understand that, compiler doesn\'t bother about the
                  


        

Answer 1

You answered your own question.

Because these

int a[] = { 1,2 };
int b[] = { 3,4 };

have type of int[2]. But these

void foo( int a[2], int b[2] )

have type of int*.

You can copy pointers but cannot copy arrays.

Answer 2

The answer is in the concept "pass by value", which means that the called function receives copies of the arguments -- which are pointers to ints. So a and b are local copies of those pointers (which don't exist in the caller; they were the results of conversions from the arrays, that is, the addresses of their first elements). It would be no different if you wrote

void foo( int aparam[2], int bparam[2] )
{
    int* a = aparam;
    int* b = bparam;

    a = b ;
}

Dennis Ritchie has acknowledged that the array syntax for parameters is a wart on the language, and was only there to ease conversion of B programs -- ancient history! It also had a deleterious effect on the design of C++, because arrays cannot be passed by value. This syntax is a constant source of confusion. So ... don't use it; pretend it's not legal. If everyone does that, it can fade away and maybe in a few decades can be given proper semantics.

Update: For more information on call-by-value (the only form of call in C and Java; see my comments below), call-by-reference (added in C++), and other evaluation strategies, see http://en.wikipedia.org/wiki/Evaluation_strategy

Answer 3

In the main function
a and b are constant pointers actually they are the address of the first elements. They are like l-value. you cannot copy to l-value but you can change the value of the integers they point to.

In the foo function
a and b are pointers. so you can change their values.