javascript - 替换在javascript中通过引用传递的函数

Question

我的理解是在 javascript 中，函数是通过引用传递的……在这种情况下，我试图用另一个函数替换一个函数。它不工作。我确定我错过了一些简单的东西。但是什么？

var spit = {
	spittle: function (x) {
		console.log('i spittle at ' + x);
	}
}

replaceFunc = {
	replace: function (func) {
		func = console.log;
	}
}

replaceFunc.replace(spit.spittle);

spit.spittle('joe'); /// should be just 'joe' instead the spittle func is not replaced and it outputs 'i spittle at joe'

Answer 1

JavaScript 中没有引用传递。一个选项是传递整个 spit 对象并修改它。

var spit = {
    spittle: function (x) {
        console.log('i spittle at ' + x);
    }
}

var replaceFunc = {
    replaceSpittle: function (funcContainer) {       
        funcContainer.spittle = function() {
            console.log.apply(console, arguments);
        }
    }
}

replaceFunc.replaceSpittle(spit);

spit.spittle('joe');