v it's an array of ints and a it's an int:
#include <iostream>
using namespace std;
int main() {
int v[10], a;
cout << v[a] << endl;
cout << a[v] << endl;
return 0;
}
returns the same value: 0 0
Why is that ?
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Mardaloescu Serban
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1Because they're both translated to `*(a+v)`. – Maroun Jun 13 '13 at 11:17
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1There is no declaration for `a` here. – John Dibling Jun 13 '13 at 11:17
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sorry John Dibling, I edited now. – Mardaloescu Serban Jun 13 '13 at 11:19
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This is a duplicate of a question which already has been answered. – user2448027 Jun 13 '13 at 11:21
1 Answers
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Because the indexer syntax means "the value in the address denoted by the beginning of the array plus an offset". Or, to put it another way:
v[a] == *(v + a) == *(a + v) == a[v]
Theodoros Chatzigiannakis
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