Int a = 5; a = a++ + ++a; a =? (2011)

Re:Int a = 5; a = a++ + ++a; a =? (2011) by 1718627440 on Friday May 15, 2026 @08:51AM (#48147960)

And a slightly longer version is, that there are three types involved: the type of access, the effective type of the object[0], and the type of the variable. The type of the variable is only for the compiler to emit warnings, as long as the effective type and the type of access are equal, it isn't UB.

[0] the C meaning of an object, not the C++ one

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