if ( ($var1 == $var2) == $var3 )
Would be the same as:
if ( $var1 == $var2 && $var1 == $var3 ) {
Wouldn't it?
It wouldn't. You can think of "==" as if it were a normal function (and in Perl 6 it will be), that returns true or false. (both are represented internally by something special, and they are both numeric and string. In string context, true is "1" and false is "", in numeric context true is 1 and false is 0)
So, let's read it as:
if ( ==( ==($var1, $var2), $var3 ) ) {
The first == returns a boolean value, which is put into numerical context because of the other ==. That one compares $var3 to the value returned (0 or 1), so the entire expression is true if:
- $var1 and $var2 are equal and $var3 is 1
... or ...
- $var1 and $var2 are not equal and $var3 is 0
- Yes, I reinvent wheels.
- Spam: Visit eurotraQ.
| [reply]
[d/l] |
Right.
I was meaning more in the hypothetical sense. I realize as Perl is now, these expressions wouldn't work. However, hypothetically or mathematically the expression (($var1 == $var2) == $var3) would be the same as Perl's ( $var1 == $var2 && $var1 == $var3 ), right? In other words, the precedence is the same, isn't it?
_
_
_
_
_
_
_
_
_
_
- Jim
Insert clever comment here...
| [reply]
[d/l] |