复合赋值 E1 op= E2 不等同于 E1 = E1 op E2

Question

the behavior of every builtin compound-assignment expression E1 op= E2 (where E1 is a modifiable lvalue expression and E2 is an rvalue expression or a braced-init-list (since C++11)) is exactly the same as the behavior of the expression E1 = E1 op E2, except that the expression E1 is evaluated only once and that it behaves as a single operation with respect to indeterminately-sequenced function calls (e.g. in f(a += b, g()), the += is either not started at all or is completed as seen from inside g())."

我想知道这个解释是错误的（不充分的）还是我理解有误。

我知道 E1 = E1 + E2 和 E1 += E2 之间存在内在差异，这是解释 here:

#include<iostream>
int main() {

    int x;

    x = 1;
    x += (-1) ? 2 : 2; 
    std::cout << x << std::endl; //prints 3

    x = 1;
    x = x + (-1) ? 2 : 2;
    std::cout << x << std::endl; //prints 2

    x = 2;
    x += (-2) == 0;
    std::cout << x << std::endl; //prints 2

    x = 2;
    x = x + (-2) == 0; // prints 1

}

我的猜测是 E1 op= E2 具有以下行为：

对E1和E2进行求值（顺序不确定），将两次求值之间的运算结果赋值给E1，即(E1) = (E1) op (E2).

那么对复合赋值操作行为的更好解释是 (E1) = (E1) op (E2)？（或 E1 = E1 op (E2)，因为 E1 只能有优先级高于赋值运算符和优先级低于 op 运算符的运算符，没有括号，如果 E1 op (E2) 想产生与 [=30= 不同的结果]. 这种结果是可修改左值的运算符不存在。）

Answer 1

这是关于运算符优先级的。 + 的优先级高于三元条件（以及 ==）。所以这个：

x = x + (-1) ? 2 : 2;

是这样评价的：

x = (x + (-1)) ? 2 : 2;

这就是为什么您得到 2 而不是 3 的原因。与另一个样本相同：

x = x + (-2) == 0

被评估为

x = (x + (-2)) == 0 // (x + (-2)) is 0, 0 == 0 is 1

以第一个例子为例：

x = x + (-1) ? 2 : 2;

这里A就是x，B就是(x + (-1) ? 2 : 2)（x加上三元运算的结果）

然而这里

x += (-1) ? 2 : 2;

A是x，B只是三元运算的结果。由于 += 的优先级低于 =，A 和 B 发生了变化。所以关于 E1 = E1 + E2 和 E1 += E2 相同的说法是正确的，但这并不意味着将 E1 = E1 + E2 转换为 E1 += E2 永远不会改变 E1 和 E2的意思。这似乎有点误导。

Answer 2

根据 C++ 标准（8.5.18 赋值和复合赋值运算符）

7 The behavior of an expression of the form E1 op = E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once. In += and -=, E1 shall either have arithmetic type or be a pointer to a possibly cv-qualified completely-defined object type. In all other cases, E1 shall have arithmetic type.

和（8.5.6 加法运算符）

1 The additive operators + and - group left-to-right. The usual arithmetic conversions (8.3) are performed for operands of arithmetic or enumeration type.

这些表达式语句

x = x + (-1) ? 2 : 2;
x += (-2) == 0;
x = x + (-2) == 0;

相当于

x = ( x - 1 ) ? 2 : 2;
x += ( -2 == 0 );
x = ( ( x - 2 ) == 0 );

Answer 3

cppreference 的引用直接来自 C++ 标准：

[expr.ass]/6

The behavior of an expression of the form E1 op= E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once.
Such expressions are deprecated if E1 has volatile-qualified type; see [depr.volatile.type].
For += and -=, E1 shall either have arithmetic type or be a pointer to a possibly cv-qualified completely-defined object type.
In all other cases, E1 shall have arithmetic type.

在这种情况下，the term expression has already been defined 和 E1 op= E2 等同于 E1 = E1 op E2 显然并不意味着表达式在它们的文本表示中是等价的，但在它们的解析（类型，值和副作用）。

[A] better explanation for compound assignment operation's behavior could be (E1) = (E1) op (E2)?

我只能表达我的意见：我认为 cpprederence 页面在这里引用标准是正确的，但可以添加注释以确保读者不会理解错.

复合赋值 E1 op= E2 不等同于 E1 = E1 op E2

Compound assignment E1 op= E2 is not equivalent to E1 = E1 op E2

c++

operators

variable-assignment

`[expr.ass]/6`