Does this C++ code produce an undefined behavior?

 

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  • With arrays, why is it the case that a[5] == 5[a]? 17 answers
  • This doesn't look like a function. What is this? 5 answers

    • It's perfectly legal. With pointer arithmetic, a[42] is equivalent to *(a + 42) , which (addition being commutative) is equivalent to *(42+ a) , which (by definition of [] ) is equivalent to 42[a] .

    So it's obscure, but well-defined.

    The array operator is commutative.

    a[n] == *(a + n) == *(n + a) == n[a]

    And it's perfectly legal.

    a[i] is defined as *(a+i) .

    So 42[a] = a[42]; and it is perfectly safe

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