C#自增运算符错误: Operand is not syntactically correct?

Question

我正在查看 the docs并尝试了解运算符的实际工作方式。

The increment operator (++) increments its operand by 1. The increment operator can appear before or after its operand: ++variable and variable++.

The first form is a prefix increment operation. The result of the operation is the value of the operand after it has been incremented.

The second form is a postfix increment operation. The result of the operation is the value of the operand before it has been incremented.

我希望以下操作返回 3，但它无法编译，并声明运算符必须是变量、属性或索引器:

int x = 0;
Console.WriteLine(x++ ++ ++);
/*Expected output: 3*/

为什么错了？我是否应该假设 x++ 不会为下一个 ++ 运算符返回相同类型的值？

Answer 1

来自draft C# 6 language specification :

The operand of a postfix increment or decrement operation must be an expression classified as a variable, a property access, or an indexer access. The result of the operation is a value of the same type as the operand.

x 是一个变量，但 x++ 不是“变量、属性访问或索引器访问”。

现在，让我们想象这样一个世界，在这个世界中，这样的后缀增量运算符调用是合法的。给定一个 int x = 42;，x++ 将 x 递增到 43，但它的计算结果为 42。如果 x++ code> 是合法的，它将应用于 x++，这不是您想要的(它会增加临时值，而不是 x)