奇怪的运算符优先级 ??(空合并运算符) [英] Weird operator precedence with ?? (null coalescing operator)

问题描述

最近我遇到了一个奇怪的错误，我将一个字符串与一个 int? 连接起来，然后在此之后添加另一个字符串.

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

我的代码基本上是这样的:

My code was basically the equivalent of this:

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

令人惊讶的是，这将在没有警告或不兼容的类型错误的情况下运行和编译，就像这样:

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错误:

And then this results in an unexpected type incompatibility error:

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

这个更简单的例子也一样:

As will this simpler example:

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

有人可以向我解释一下这是如何工作的吗?

Can someone explain how this works to me?

推荐答案

空合并运算符具有非常低的 优先级 所以你的代码被解释为:

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，两个表达式都是字符串，所以它可以编译，但不会做你想要的.在您的下一个示例中，?? 运算符的左侧是一个字符串，但右侧是一个整数，因此无法编译:

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'

解决办法当然是加括号:

The solution of course is to add parentheses:

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

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