问题
Please ignore the fact that this code achieves nothing and apologies for what's probably an inane question!
I understand that I cannot pass a function call into setTimeout() as the first argument, but why can I not do that?
let names = ['Andy', 'Ross', 'David'];
function printer (name) {
console.log(name);
}
names.forEach(name => setTimeout(printer(name), 1000);
Result:
Andy
timers.js:327
throw new TypeError('"callback" argument must be a function');
^
I can solve the problem by instead using a reference to printer and using bind() to send name along with it, but why must I take these extra steps?
let names = ['Andy', 'Ross', 'David'];
function printer (name) {
console.log(name);
}
names.forEach(name => setTimeout(printer.bind(null, name), 1000));
Result:
Andy
Ross
David
回答1:
This is because of the order of execution. If you pass a function call to setTimeout, the function will be executed immediately, i.e. the function is put on javascript's execution stack immediately.
If you pass a function name, i.e. a reference to a function, the function is only put in the javascript thread's execution stack once the timer finishes.
回答2:
you can try this:
setTimeout(function(){printer(name)}, 1000)
回答3:
setTimeout should take a function as its first argument.
Please refer:
https://www.w3schools.com/jsref/met_win_settimeout.asp
Here you are passing the result of the function as the first argument which is undefined.
回答4:
You can do it like this:
setTimeout(printer, 1000, name)
回答5:
The correct way of passing a function reference is to use callbacks.
names.forEach(name => setTimeout(function() {
printer(name);
}, 1000));
callbacks contains reference to the function.
setTimeout(callbackFunction, milliseconds);
来源:https://stackoverflow.com/questions/43317206/why-can-i-not-pass-a-function-call-rather-than-a-function-reference-or-an-anony