Re: Re: 0 illegal modulus?

[Vynce]

well, i actually said $y*1 to force that into a numerical context. if you don't, then what if $x is 5 and $y is 'foo'?

bash-2.04$ perl -e 'print 5 % foo'
Illegal modulus zero at -e line 1.

so...

bash-2.04$ perl -e '$x = 5; $y = "foo"; print ($y ? $x % $y : $x); pri
+nt "\n"'
Illegal modulus zero at -e line 1.
bash-2.04$ perl -e '$x = 5; $y = "foo"; print ($y*1 ? $x % $y : 0); pr
+int "\n"'
0
bash-2.04$ perl -e '$x = 5; $y = "foo"; print (($y*1) ? $x % $y : $x);
+ print "\n"'
5

as you can see, one still generates an error. as for what to return if $y is zero, i had misread your original statement.

.

[Anonymous Monk]

Dont use *1! Surely "foo" *is* an illegal modulus. And "0" is still false.