Do you know where your variables are? | |
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The first argument of vec says what variable it acts on. The third says how many bits make a unit in the vector. The second is the offset in units of the third argument where things will happen. ord applied to a string returns ord of its first character. Thus, vec($foo,0,8) = ord($foo); is a no-op, replacing the first character of $foo by itself. In more depth: $ perl -e'my $foo = "qa"; vec($foo,0,8) = ord $foo;print $foo,$/' qa $ perl -e'my $foo = "qa"; vec($foo,1,8) = ord $foo;print $foo,$/' qq $ perl -e'my $foo = "qa"; vec($foo,2,8) = ord $foo;print $foo,$/' qaqFurther increments of the second argument would place NUL in the intermediate positions of the string. Update: I just noticed that the op is ==, numerical equality. I think that should always be true, but there might be tricky edge cases related to encoding. After Compline, In reply to Re: oct and vec
by Zaxo
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