Why does the function of PHP add 2 to the formal parameter refcount?

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Well, the Lazyboy asked me to find out the answer: the refcount of the parameter passing value

Example content from Http://julien-pauli.developpez.com/tutoriels/php/internals/variables/?page=page_6

A RefCount of 2, here, is extremely non-obvious. Especially considering the above examples. So what ' s happening?

When a variable have a single reference (as did $var 1 before it is used as an argument to debugzvalDump ()), PHP ' s Engine optimizes the manner in which it's passed to a function. Internally, PHP treats $var 1 like a reference (in the. The refcount is increased for the scope of this function), with the Caveat that if the passed reference happens to being written to, a copy was made, but only at the moment of writing. This is known as "Copy on Write."

So, if debugzvalDump () happened to write to its sole parameter (and it doesn ' t), then a copy would is made. Until then, the parameter remains a reference, causing the refcount to being incremented to 2 for the scope of the function C All.

The Xdebugdebugzval is transmitted in a quoted manner.

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