奇怪的运营商precedence用？ （空合并运算符） [英] Weird operator precedence with ?? (null coalescing operator)

问题描述

最近，我有一个奇怪的错误我在那里用串联的字符串 INT？并在那之后增加一个字符串。

Recently I had a weird bug where I was concatenating a string with an int? and then adding another string after that.

我的code基本上这是等效的：

My code was basically the equivalent of this:

int? x=10;
string s = "foo" + x ?? 0 + "bar";

令人吃惊的是，这将运行并没有警告或不兼容的类型编译错误，因为这会：

Amazingly enough this will run and compile without warnings or incompatible type errors, as will this:

int? x=10;
string s = "foo" + x ?? "0" + "bar";

然后这会导致意外的类型不兼容的错误：

And then this results in an unexpected type incompatibility error:

int? x=10;
string s = "foo" + x ?? 0 + 12;

由于将这个简单的例子：

As will this simpler example:

int? x=10;
string s = "foo" + x ?? 0;

有人能解释这是如何工作对我？

Can someone explain how this works to me?

推荐答案

空合并经营者具有非常低的 precedence 所以你的code正在跨preTED为：

The null coalescing operator has very low precedence so your code is being interpreted as:

int? x = 10;
string s = ("foo" + x) ?? (0 + "bar");

在这个例子中，两个前pressions是字符串，因此编译，但你想要什么没有做。在你的下一个例子中， ?? 运算符的左侧是一个字符串，而右手边是一个整数，所以它不会编译：

In this example both expressions are strings so it compiles, but doesn't do what you want. In your next example the left side of the ?? operator is a string, but the right hand side is an integer so it doesn't compile:

int? x = 10;
string s = ("foo" + x) ?? (0 + 12);
// Error: Operator '??' cannot be applied to operands of type 'string' and 'int'

课程的解决方案是增加括号

The solution of course is to add parentheses:

int? x = 10;
string s = "foo" + (x ?? 0) + "bar";

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