Integer is not being shown as die() argument?

I have this strange problem. When debugging, I have sometimetimes code looking like this

<?php
$var = 15;
die($var);

开发者_运维知识库

die() function works, but outputs nothing

However, this one works

<?php
$var = 15;
die($var."<-");

http://sandbox.phpcode.eu/g/81462.php

How is it possible? Did I miss something? or is it bug?

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See http://www.php.net/manual/en/function.exit.php (die() is equivalent to exit())

If status is a string, this function prints the status just before exiting.

If status is an integer, that value will be used as the exit status and not printed. Exit statuses should be in the range 0 to 254, the exit status 255 is reserved by PHP and shall not be used. The status 0 is used to terminate the program successfully.

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die() is the same as exit(), looking at the exit docs it takes 1 parameter, $status, the parameter information states

If status is a string, this function prints the status just before exiting.

If status is an integer, that value will be used as the exit status and not printed. Exit statuses should be in the range 0 to 254, the exit status 255 is reserved by PHP and shall not be used. The status 0 is used to terminate the program successfully.

Note: PHP >= 4.2.0 does NOT print the status if it is an integer.

Self-explanatory really, if you want to pass a number you need to type it to a string like so:

die( (string)$code );

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die() function needs a string parameter.

In your second example

die($var."<-");

$var is converted into String before concat with "<-". So this line will print out "15<-". This is normal. There is neither a bug nor any thing wrong.