Ligth Talk

hi… im a newbie… using function_exists() is very very useful…

is it true that print() is a function? if yes, then proceed to this…

with this code:

[CODE]<?php
print(function_exists(“function_exists”));
?>[/CODE]

the output would be:

1

any function test works fine with it except for print()…

with this code:

[CODE]<?php
print(function_exists(“print”));
?>[/CODE]

the result is empty… i know that it is useless testing print()… but is it really the exact explanation behind why print() does not work with function_exists() (that it is useless)?

if it so, then why function_exists() works when function_exists itself is used as a parameter (see the first example)? well, function_exists is also useless because in the first place you cannot invoke it if it does not exist right?

i just couldn’t see which is which…

1) when using function_exists as a parameter works fine is reasonable, then print should work fine too, because it is also a function. otherwise, this is a lack of codes within the function_exists().

2) on the other hand, when using print as a paramater doesn’t work is reasonable, then function_exists shouldn’t work too, because it is useless testing if a function exists when itself is used as a parameter. otherwise, this is a waste of codes within the function_exists().

Note:
– i am using PHP v5.2.3
– this is not a question of the usage of the function but the integrity of the function

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