javascript 递归函数的javascript返回
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javascript return of recursive function
提问by MirrorMirror
Hate to open a new question for an extension to the previous one:
讨厌打开一个新问题来扩展上一个问题：
function ctest() {
this.iteration = 0;
this.func1 = function() {
var result = func2.call(this, "haha");
alert(this.iteration + ":" + result);
}
var func2 = function(sWord) {
this.iteration++;
sWord = sWord + "lol";
if ( this.iteration < 5 ) {
func2.call(this, sWord);
} else {
return sWord;
}
}
}
this returns iteration = 5 but result UNDEFINED ? how is that possible ? i explicitly return sWord. It should have returned "hahalollollollollol" and just for doublecheck, if i alert(sWord) just before the return sWord it displays it correctly.
这将返回迭代 = 5 但结果 UNDEFINED ？这怎么可能？我明确返回 sWord。它应该返回“hahalollollollollol”并且只是为了仔细检查，如果我在返回 sWord 之前警告（sWord）它会正确显示它。
回答by Quentin
You have to return all the way up the stack:
您必须一直返回堆栈：
func2.call(this, sWord);
should be:
应该：
return func2.call(this, sWord);
回答by Rich O'Kelly
You need to return the result of the recursion, or else the method implicitly returns undefined. Try the following:
您需要返回递归的结果，否则该方法隐式返回undefined. 请尝试以下操作：
function ctest() {
this.iteration = 0;
this.func1 = function() {
var result = func2.call(this, "haha");
alert(this.iteration + ":" + result);
}
var func2 = function(sWord) {
this.iteration++;
sWord = sWord + "lol";
if ( this.iteration < 5 ) {
return func2.call(this, sWord);
} else {
return sWord;
}
}
}
回答by shareef
keep it simple :)
把事情简单化 ：）
your code modified in JSFiddle
iteration = 0;
func1();
function func1() {
var result = func2("haha");
alert(iteration + ":" + result);
}
function func2 (sWord) {
iteration++;
sWord = sWord + "lol";
if ( iteration < 5 ) {
func2( sWord);
} else {
return sWord;
}
return sWord;
}
回答by alex
Your outer function doesn't have a returnstatement, so it returns undefined.
您的外部函数没有return语句，因此它返回undefined.
