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It's not really an operator. It's part of the syntax of string literals. "foo" "bar" is an alternative way of writing the string literal "foobar". If foo is not
by housecarpenter 5y ago
It's not really an operator. It's part of the syntax of string literals. "foo" "bar" is an alternative way of writing the string literal "foobar". If foo is not a string literal, foo "bar" is invalid syntax.
- oaiey 5y agoOkay... So it is not a implicit operator. That is good. Some small reputation points are regained. Thanks.