javascript - 有没有一种更简单的方法可以在不捕获的情况下进行单行尝试？

Question

当我做这样的事情时:

let resolved = null;
try {
    resolved = require.resolve(modulePath)
} catch (e) {

}

我想知道是否有更短的语法，例如:

let resolved = null;
try resolved = require.resolve(modulePath)

有没有办法在不打开 catch block 的情况下原谅这一行？

有这样的东西，但我正在寻找更自然的东西:

function t (fn, def) {
    let resolved = def;
    try {
        resolved = fn()
    } catch (e) {

    }
    return resolved;
}

Answer 1

没有； JavaScript try block 必须有 catch 或 finally 是有意为之的。来自 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/try...catch#Description :

The try statement consists of a try block, which contains one or more statements. {} must always be used, even for single statements. At least one catch clause, or a finally clause, must be present.

(注意 try { ... } finally { }，没有 catch，相反你想要的:你希望一切都被捕获和吞噬，而 try { ... } finally { } 没有捕获任何东西。)

---

在你的情况下，我认为最好的写法是:

let resolved;
try {
    resolved = require.resolve(modulePath);
} catch (e) {
    resolved = null;
}

清楚地表明 resolved == null 是错误情况。

(更好的是 - 添加一些逻辑以确保您捕获的异常确实是您所期望的异常。您可能不想静静地吞下由意外错误导致的异常!)