Why chained prefix increment/decrement for builtin type is not UB for C++?

Question

In cpprefernce.com example for prefix increment there is such code:

int n1 = 1;
...
int n3 = ++ ++n1;

Why chained increment in this case does not lead to UB? Is rule for at most once modified not violated in this case?

Answer 1

In C++11 and later, UB occurs when there are two writes or a write and a read that are unsequenced and access the same memory location. But ++x is equivalent to x+=1, so ++ ++n1 is equivalent to (n1+=1)+=1, and here the reads and writes happen in a strict sequence because of the properties of assignment and compound assignment operators: first n1 is read, then one plus the original value is written, then the resulting value is read again, then one plus that value is written back.

In C++03, this was UB, because of the old rule you allude to: there is no sequence point between the two modifications. But in C++11 there are no longer any sequence points; instead there is the "sequenced before" partial order.