dtd5644 2016-03-08 09:42
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重构参数未在功能问题中定义

I am working on some old PHP, and I am not a PHP top notch expert (yet).

Some piece of code in a class calls a method with an array parameter:

$saleServers = $this->getSaleServers();
// Getting prices from cache or request
$prices = $this->getProductPrices($saleServers);

Yet, the method does not define an array parameter:

public function getProductPrices($getLicenseLifetime = true) { ... }

My questions are:

  1. How is PHP going to handle $saleServers, is it going to ignore it? If yes, I guess I can remove it from the function call.

  2. How is PHP going to handle $getLicenseLifetime in regards to $saleServers? Is it going to mix them up or keep them separate (i.e., PHP is not going to assign $saleServers to $getLicenseLifetime, correct)?

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1条回答 默认 最新

  • doupiao1893 2016-03-08 10:02
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    $getLicenseLifetime will be set to the contents of $saleServers, if you call it that way, and will be true, if you omit the parameter.

    PHP is weakly typed, so you can call the function with variables of any type and inside the function you can check for the type of the variable which holds the parameter with functions like is_array() etc. PHP5 knows the concept of type hinting, which mainly works on objects (does not support scalars for example) and with PHP7 you can declare argument types.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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