c - 为什么++variable 在 C 中不被视为左值？

Question

考虑以下 C 代码:

#include <stdio.h>

int main(void) {
    int a = 0;
    printf("%d\n", ++a += 1);
}

++ 优先于 += 所以它应该首先被评估。所以我们在左边有 a，在右边有 1。现在根据我的理解 ++a 应该导致 a (只改变它的值)这是左值所以为什么这个语句给出以下错误:

lvalue required as left operand of assignment

Answer 1

前缀增量运算符 ++ 导致其操作数的增量值，但它不是左值。 C standard 的第 6.5.3.1p2 节: 语义描述如下:

The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++E is equivalent to (E+=1). See the discussions of additive operators and compound assignment for information on constraints, types, side effects, and conversions and the effects of operations on pointers.

然后第 6.5.16.2p3 节关于复合赋值运算符指出:

A compound assignment of the form E1 op = E2 is equivalent to the simple assignment expression E1 = E1 op (E2), except that the lvalue E1 is evaluated only once, and with respect to an indeterminately-sequenced function call, the operation of a compound assignment is a single evaluation.

而 6.5.16p3 关于赋值运算符进一步指出:

An assignment operator stores a value in the object designated by the left operand. An assignment expression has the value of the left operand after the assignment, but is not an lvalue.

所以这是明确不允许的。即使是这样，像 ++a += 1 这样的表达式也会导致 a 在没有中间序列点的情况下被修改多次，这会触发未定义的行为。

这是 C 和 C++ 不同的地方之一。 C++ 实际上允许 = 运算符的结果，并通过扩展复合赋值和前缀 ++/--，成为左值.