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ijabbott · (This post was last modified: 07-11-2022 07:56 PM by ijabbott.)

(07-10-2022 07:38 PM)Wes Loewer Wrote: While the C++17 documentation has this

Code:

i = i++ + 1; // the value of i is incremented i = i++ + i; // the behavior is undefined i = i + 1; // the value of i is incremented

So if (i = i++ + 1) increments just like (i = i + 1), then it stands to reason that (i=i++) behaves like (i = i).

Not really. For `i = i++ + 1;`, the `i++` part says that the old value of `i` plus 1 will be be stored in `i`, and the `i = i + 1` part also says that the old value of `i` plus 1 will be stored in `i`, so there is no conflict (in C++17, but undefined in C17). However, for `i = i++;`, the `i++` part says that the old value of `i` plus 1 will be stored in `i`, and the `i = i` part says that the old value of `i` will be stored in `i`, so there is a conflict. That's also one reason why `i = i++ + i;` is undefined - the `i++` part says the old value of `i` plus 1 will be stored in `i`, and the `i = i + i` part says the old value of `i` plus the old value of `i` will be stored in `i`. (The other reason is that the storage to `i` by `i++` is unsequenced relative to the evaluation of `i` in the `... + i`.)