If as an example a = 2, then arr[a++] requires the storage of the old value of a since after evaluating a++, a would be equal to 3, while it needs to access arr[2].
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I didn't get why compiler is storing value temporarily? Am I missing something?
If as an example
a = 2, thenarr[a++]requires the storage of the old value ofasince after evaluatinga++,awould be equal to3, while it needs to accessarr[2].