在C ++ 11及更高版本中，UB发生在是两个写，或者一个写和一个读不顺序并访问相同的存储器位置。但 ++ x 等效于 x + = 1 ，因此 ++ ++ n1 等价于（n1 + = 1）+ = 1 ，这里的读写操作由于赋值和化合物的属性赋值运算符：首先读取 n1 ，然后写入一个加上原始值，然后再次读取结果值，然后写回一个值。

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.

在C ++ 03中，此是 UB，因为您引用的旧规则：两个修改之间没有序列点。但在C ++ 11中不再有任何序列点;而是存在顺序之前部分顺序。

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.