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帮忙看下怎么会返回null

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Release: 2016-06-23 14:13:21
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本帖最后由 snowlove 于 2013-06-17 18:27:34 编辑

<?phpfunction newtripos($str,$findstr,$count,$off=0){	$pos=stripos($str,$findstr,$off);	$count--;	if($count>0 && $pos!=false){		$pos=newtripos($str,$findstr,$count,$pos+1);	}else{		var_dump($pos);		return $pos;	}}$a="456123456455654466";$b=newtripos($a,'6',4);var_dump($b);?>	
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执行后显示$b是null。而在函数返回前,执行的var_dump($pos),显示是int(16)。
为什么var_dump($b)会是null?


回复讨论(解决方案)

你使用了递归,return $pos; 在进入递归后会作用于 $pos=newtripos($str,$findstr,$count,$pos+1);
而你并没有在这个分支里对 $pos 作处理(也就是将结果返回到上一级递归)
我说的可能连我自己都不好理解,看代码:

function newtripos($str,$findstr,$count,$off=0){    $pos=stripos($str,$findstr,$off);    $count--;    if($count>0 && $pos!=false){        $pos=newtripos($str,$findstr,$count,$pos+1);    }    return $pos;}
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你的代码没问题。
function newtripos($str,$findstr,$count,$off=0){
$pos=stripos($str,$findstr,$off);
$count--;
if($count>0 && $pos!=false){
$pos=newtripos($str,$findstr,$count,$pos+1);
}else{
var_dump($pos);//程序执行到这里可以正确的打出$pos的值。继续执行就应该结束函数,返回$pos
return $pos; //这里就应该返回$pos的值了。
}
}
$a="456123456455654466";
$b=newtripos($a,'6',4); //可是这里$b 怎么没接收到$pos,是个null
var_dump($b);
?>

把else体去掉,和没去掉有这么大差别?

差别大多了,因为没有去掉else时,相当于

<?php	if($count>0 && $pos!=false){		$pos=newtripos($str,$findstr,$count,$pos+1);                return null;	}else{		var_dump($pos);		return $pos;	}}?>	
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而去掉else以后,则是

<?php	if($count>0 && $pos!=false){		$pos=newtripos($str,$findstr,$count,$pos+1);                return $pos;	}else{		var_dump($pos);		return $pos;	}}?>	
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