通过引用返回大数据还是作为函数返回？
今天在工作中，我与拼贴画就在范围之间传递大数据进行了争论。 误解是，在两个作用域之间传递时，引用使用的内存/CPU 使用量较少。我们构建了一个谁是对的概念证明......所以：
function by_return($dummy=null) {
$dummy = str_repeat("1",100 * 1024 * 1024);
return $dummy;
}
function by_reference(&$dummy) {
$dummy = null;
$dummy = str_repeat("1",100 * 1024 * 1024);
}
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
//1 always returns: 105493696/105496656
$nagid = by_return();
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
unset($nagid);
//2 always returns: 105493696/210354184 even if we comment 1st part
by_reference($dummy);
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
unset($dummy);
但是根据函数“memory_get_peak_usage()”，通过引用似乎会消耗更多内存
如您所见，使用大数据进行返回比使用作为引用要聪明得多，但是问题是，为什么？ 欢迎任何启发:)
On the job today I had an arguement with a collage about passing large data between scopes.
The myth was that reference uses less memory/CPU usage when passing between 2 scopes. We build a proof of concept who was right... so:
function by_return($dummy=null) {
$dummy = str_repeat("1",100 * 1024 * 1024);
return $dummy;
}
function by_reference(&$dummy) {
$dummy = null;
$dummy = str_repeat("1",100 * 1024 * 1024);
}
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
//1 always returns: 105493696/105496656
$nagid = by_return();
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
unset($nagid);
//2 always returns: 105493696/210354184 even if we comment 1st part
by_reference($dummy);
echo memory_get_usage()."/".memory_get_peak_usage()."\n";
unset($dummy);
But it seems that by reference it consumes more memory according to function "memory_get_peak_usage()"
As you see, using large data for returning is much smarter than using as reference but the question is, why?
Any enlightening is welcomed :)
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这是由于 php 处理变量的方式造成的，对于使用过 C 或 C++ 的人来说有点违反直觉。
不建议通过引用传递来比 PHP 更智能。 PHP 实际上不会复制数据，除非需要（即当有超过 1 个引用时更改变量的值），这是一种与共享内存页的写时复制非常相似的优化策略。
因此，假设您有一个变量，您在给定脚本中多次按值传递该变量。如果您随后获取该变量并通过引用传递它，那么您实际上是在复制该变量，而不仅仅是获取指向该对象的指针。
这是因为，在内部，PHP zvals（PHP 用于存储变量的数据结构）只能是引用变量或非引用变量。因此，zval 的 ref_count 字段是什么并不重要，因为它不是引用变量（zval 结构的 is_ref 字段）。因此，在内部，PHP 被迫创建一个新的 zval 并将其 is_ref 字段设置为 true，从而使内存加倍。
告诉您的同事不要再试图智取 PHP。除非在整个代码中 100% 完美地完成，否则通过引用传递将导致大量开销并使内存使用量加倍。
有关更详细的讨论，请参阅此链接：http:// /porteightyeight.com/2008/03/18/the-truth-about-php-variables/
This is due to the way php handles variables, and is a bit counter-intuitive to anyone who has worked in C or C++.
Passing by reference to be smarter than PHP isn't advised. PHP doesn't actually make copies of data unless it needs to (i.e. you change a variable's value when there's more than 1 reference to it), an optimization strategy very similar to copy-on-write for shared memory pages.
So, let's say you have a variable that you pass by value several times in a given script. If you then take this variable and pass it by reference, you're actually duplicating the variable rather than just getting a pointer to the object.
This is because internally, PHP zvals (the data structure PHP uses to store variables) can only be reference variables or non-reference variables. So it doesn't matter what the zval's ref_count field is, because it's not a reference variable (the is_ref field of the zval structure). So internally, PHP is forced to create a new zval and set its is_ref field to true, thus doubling the memory.
Tell your co-worker to stop trying to outsmart PHP. Passing by reference unless done 100% perfectly throughout the code will cause a lot of overhead and double the memory usage.
For a more detailed discussion, please see this link: http://porteightyeight.com/2008/03/18/the-truth-about-php-variables/