php - 输出中的一些奇怪的行为？

Question

I was just playing around with PHP when this happened. Look at the commented code.

    <?php
     error_reporting(0); //turns of errors and notices (which will be shown otherwise)

     //here we would've gotten a notice saying $_ has no value, which is true. But PHP        automatically gives it the value 0, then we add one to it. ($_++), add $_ and add $_. So it's    1 + 1 + 1 which is two somehow
     echo ($_++ + $_ + $_);

So my question is... why does it output 2?

Answer 1

Most of the answer is contained in your code.

$_ is 0 initially. By $_++, you increment $_, setting it to one. Therefore, $_ is 1, but the value of the postincrement (!) $_++ is still 0. The value of ++$_ would be 1.

Then, you add two times $_ (which is 1), yielding 2 overall.

See this SO post for a detailed preincrement/postincrement comparison.