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Home/ Questions/Q 4058560
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T15:05:41+00:00 2026-05-20T15:05:41+00:00

Let me explain with an example – #include <iostream> void foo( int a[2], int

  • 0

Let me explain with an example –

#include <iostream>

void foo( int a[2], int b[2] ) // I understand that, compiler doesn't bother about the
                               // array index and converts them to int *a, int *b
{
    a = b ;  // At this point, how ever assignment operation is valid.

}

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

    foo( a, b );

    a = b; // Why is this invalid here.

    return 0;
}

Is it because, array decays to a pointer when passed to a function foo(..), assignment operation is possible. And in main, is it because they are of type int[] which invalidates the assignment operation. Doesn’t a,b in both the cases mean the same ? Thanks.

Edit 1:

When I do it in a function foo, it’s assigning the b's starting element location to a. So, thinking in terms of it, what made the language developers not do the same in main(). Want to know the reason.

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  1. Editorial Team
    Editorial Team
    2026-05-20T15:05:42+00:00Added an answer on May 20, 2026 at 3:05 pm

    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.

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