问题描述
代码级别，postincrement ++运算符何时被执行？
In C++ on a machine code level, when does the postincrement++ operator get executed?
表示后缀++运算符是2级：这意味着
The precedence table indicates that postfix++ operators are level 2: which means in
int x = 0 ;
int y = x++ + x++ ; // ans: y=0
后缀++的第一个执行 >
The postfix ++'s execute first.
但是，这行的逻辑操作似乎是第一次发生的加法（0 + 0），但是如何发生呢？
However, it would seem that the logical operation of this line is the addition happens first (0+0), but how does that happen?
我想象的是以下内容：
// Option 1:
// Perform x++ 2 times.
// Each time you do x++, you change the value of x..
// but you "return" the old value of x there?
int y = 0 + x++ ; // x becomes 1, 0 is "returned" from x++
// do it for the second one..
int y = 0 + 0 ; // x becomes 2, 0 is "returned" from x++... but how?
// if this is really what happens, the x was already 1 right now.
因此，另一个选项是虽然x ++在x + x的优先级表上更高，由于x ++插入添加操作
So, the other option is although x++ is higher on the precedence table that x + x, the code generated due to x++ is inserted below the addition operation
// Option 2: turn this into
int y = x + x ; //
x++ ;
x++ ;
第二个选项似乎更有意义，但我对这里的操作顺序感兴趣。具体来说，何时更改？
That second option seems to make more sense, but I'm interested in the order of operations here. Specifically, when does x change?
推荐答案
UB，我将讨论以下完全精确的示例：
Instead of jumping on the details of the example that is UB, I will discuss the following example that is perfectly fine:
int a = 0, b = 0;
int c = a++ + b++;
现在，运算符的优先级意味着最后一行等效于：
Now, the precedence of operators means that the last line is equivalent to:
int c = (a++) + (b++);
而不是：
int c = (a++ + b)++; // compile time error, post increment an rvalue
另一方面，post增量的语义相当于两个单独的指令（从这里只是一个心理图片）：
On the other hand, the semantics of the post increment are equivalent to two separate instructions (from here on is just a mental picture):
a++; // similar to: (__tmp = a, ++a, __tmp)
// -- ignoring the added sequence points of , here
也就是说，编译器将原始表达式解释为：
That is, the original expression will be interpreted by the compiler as:
auto __tmp1 = a; // 1
auto __tmp2 = b; // 2
++a; // 3
++b; // 4
int c = __tmp1 + __tmp2; // 5
但是编译器允许重新排序5个指令，只要满足以下约束（ x> y 表示 x 必须在 y 或 x 之前 y ）：
But the compiler is allowed to reorder the 5 instructions as long as the following constraints are met (where x>y means x must be executed before y, or x precedes y):
1 > 3 // cannot increment a before getting the old value
2 > 4 // cannot increment b before getting the old value
1 > 5, 2 > 5 // the sum cannot happen before both temporaries are created
执行不同的指令，因此以下都是有效的序列：
There are no other constraints in the order of execution of the different instructions, so the following are all valid sequences:
1, 2, 3, 4, 5
1, 2, 5, 3, 4
1, 3, 2, 4, 5
...
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