C#自动增量运算符错误：操作数在语法上是不正确的？

发布时间：2020-12-15 23:41:53 所属栏目：百科来源：网络整理

导读：我正在寻找 the docs并试图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 ope

我正在寻找 the docs并试图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,但它不编译,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*/

为什么这是错的？我是否应该假设x不为下一个运算符返回相同类型的值？

解决方法

从 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,让我们想象一个这样的后缀增量运算符调用合法的世界.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是合法的,它将应用于x,这不是你想要的(它会增加临时值,而不是x)

（编辑：李大同）

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