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int b= a++ + a++;is undefined behaviour in C and C++, sinceais modified twice without intervening sequence point (ditto for theint c = ...thing).abort()call for the undefined behaviour. But of course no serious implementation does such things. Variations closer to the code include different orders of evaluation (right-to-left, middle-to-outer, invent-your-own) and the question of when the incremented value is stored, whetherais read only once or several times. So it would be legal for the value ofato be read three times, and the incremented value ofastored after evaluating the expression.int c = ++a + a++ + a++;being evaluated as5 + 4 + 4, and then three times storing4+1intoa, thusc == 13anda == 5afterward. Or the middle term evaluated first, and the incremented value not stored yet, then the++awith immediate storing of the incremented value, then the righta++with a new reading of the value, leading toc = 5 + 4 + 5andacould be 5 or 6, depending on which pending store from the twoa++is done last.