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Home/ Questions/Q 1106471
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T01:47:55+00:00 2026-05-17T01:47:55+00:00

Consider the following code: void Increment(int *arr) { arr++; } int main() { int

  • 0

Consider the following code:

void Increment(int *arr) {
    arr++;
}

int main() {
    int arr[] = {1,2,3,4,5};
    // arr++  // illegal because its a const pointer
    Increment(arr);   // legal
}

My question is if arr is a const pointer, how come I can send it to a function that doesn’t receive a const pointer?

The code compiles without the warning of discarding const qualifiers.

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  1. Editorial Team
    Editorial Team
    2026-05-17T01:47:55+00:00Added an answer on May 17, 2026 at 1:47 am

    Don’t get fooled by the pointer. The same holds for plain ints:

    const int a = 42;
    int b = a; // How can I assign a const int to a non-const int?
    int c = 4; // Come to think of it, the literal 4 is a constant too
    void foo (int x) { std::cout << x; }
    foo(a); // How come I can call foo with a const int?
    

    In summary, const applies to each object individually. A copy of a const object doesn’t need to be const too.

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