英文:
Run a Go routine indefinitely (restart when done)
问题
我正在将我的TypeScript项目重写为Golang，并遇到了一个问题：
我正在运行一个for循环，在程序加载时启动异步工作器。如果我理解正确，Go协程是一种同时运行异步代码的方式。我想要做的是在函数完成后无限次地重新启动该函数。
在TypeScript中，代码看起来像这样：
async function init() {
const count = await Users.getCount();
// 同时运行所有工作器
for (let i = 0; i < count; i += PER_WORKER) {
const id = i / PER_WORKER;
worker(id);
}
}
async function worker(id: number) {
await new Promise((resolve) => {
// 做一些事情
resolve();
})
// 重新启动函数
worker(workerId);
}
在Go中，我有非常相似的代码，但在内部重新调用函数时会出现问题。我考虑过按时间间隔运行函数，但我无法预先知道它需要多长时间...
谢谢
NVH
英文:
I am rewriting my TypeScript projects to Golang and I encountered a problem:
I am running a for loop which starts up async workers on program load. If I understood correctly Go routines are a way to run async code concurrently. What I'd like to do is restart the function once it is done, indefinitely.
In TypeScript it looks something like
async function init() {
const count = async Users.getCount();
// run all workers concurrently
for (let i = 0; i < count; i += PER_WORKER) {
const id = i / PER_WORKER;
worker(id);
}
}
async worker(id: number) {
await new Promise((resolve) => {
// do stuff
resolve();
})
// restart function
worker(workerId);
}
in Go I have pretty similar stuff, but when re-calling the function inside it just makes a mess? I thought of running the function by interval but I cannot know in advance the time it takes...
Thank you
NVH
答案1
得分: 6
我想做的是在函数完成后无限重启该函数。
你需要一个无限循环：
for {
f()
}
如果你想要将无限循环分离出来，可以将其放在一个函数中，并使用go关键字调用它：
go func() {
for {
f()
}
}()
也就是说，我们将for循环本身分离出来；for循环每次调用我们的函数一次。这样做：
for {
go f()
}
是错误的，因为它会无限次地快速启动f()，而不等待f()返回。
英文:
> What I'd like to do is restart the function once it is done, indefinitely.
You need a forever loop, of course:
for {
f()
}
If you want to spin off the forever loop itself, put that in a function and invoke it with go:
go func() {
for {
f()
}
}()
That is, we spin off the for loop itself; the for loop calls our function once each time. Doing this:
for {
go f()
}
is wrong because that spins off f() an infinite number of times, as fast as it can, without waiting for f() to return.
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