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vector_math: Return by const reference for const operator[]
Makes behavior between both overloads consistent.
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@ -131,7 +131,7 @@ public:
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{
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return *((&x) + i);
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}
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T operator[](const int i) const {
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const T& operator[](const int i) const {
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return *((&x) + i);
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}
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@ -288,7 +288,7 @@ public:
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{
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return *((&x) + i);
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}
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T operator[](const int i) const {
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const T& operator[](const int i) const {
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return *((&x) + i);
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}
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@ -502,7 +502,7 @@ public:
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{
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return *((&x) + i);
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}
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T operator[](const int i) const {
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const T& operator[](const int i) const {
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return *((&x) + i);
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}
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