为什么`defer recovery()`不会引起恐慌? [英] Why does `defer recover()` not catch panics?

问题描述

为什么调用defer func() { recover() }()成功恢复了恐慌的goroutine，但是调用defer recover()却没有成功?

Why does a call to defer func() { recover() }() successfully recover a panicking goroutine, but a call to defer recover() not?

作为一个简单的示例，此代码不会惊慌

As an minimalistic example, this code doesn't panic

package main

func main() {
    defer func() { recover() }()
    panic("panic")
}

但是，将匿名函数替换为直接恢复紧急情况

However, replacing the anonymous function with recover directly panics

package main

func main() {
    defer recover()
    panic("panic")
}

推荐答案

处理恐慌部分提到

两个内置函数panic和recover有助于报告和处理运行时紧急情况

Two built-in functions, panic and recover, assist in reporting and handling run-time panics

recover功能允许程序管理紧急恐慌例程的行为.

The recover function allows a program to manage behavior of a panicking goroutine.

假设函数G 推迟了调用recover的函数D ，并且panic在执行G的同一goroutine上的函数中出现.

Suppose a function G defers a function D that calls recover and a panic occurs in a function on the same goroutine in which G is executing.

当递延函数的运行达到D时，D调用recover的返回值将是传递给panic调用的值.
如果D正常返回而没有开始新的恐慌，恐慌序列将停止.

When the running of deferred functions reaches D, the return value of D's call to recover will be the value passed to the call of panic.
If D returns normally, without starting a new panic, the panicking sequence stops.

这说明recover应该在延迟函数中调用，而不是直接调用.
出现紧急情况时，递延函数"不能是内置的recover()，而是在 defer语句中指定的函数. .

That illustrates that recover is meant to be called in a deferred function, not directly.
When it panic, the "deferred function" cannot be the built-in recover() one, but one specified in a defer statement.

DeferStmt = "defer" Expression .

表达式必须是函数或方法调用；不能用括号括起来.
对内置函数的调用与表达式语句 相同.

The expression must be a function or method call; it cannot be parenthesized.
Calls of built-in functions are restricted as for expression statements.

除了特定的内置函数，函数和方法调用以及接收操作可以出现在语句上下文中.

With the exception of specific built-in functions, function and method calls and receive operations can appear in statement context.

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