问题
I have this strange problem. When debugging, I have sometimetimes code looking like this
<?php
$var = 15;
die($var);
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?
回答1:
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.
回答2:
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 );
回答3:
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.
来源：https://stackoverflow.com/questions/6913312/integer-is-not-being-shown-as-die-argument