Qiang Li - 1 month ago 5x

Perl Question

I have

`@a = (1,2,3); print (@a ~~ (1,2,3))`

and

`@a = (1,2,3); print (@a == (1,2,3))`

The first one is the one I expect to work, but it does not print anything. The second one does print 1.

Why? Isn't the smart matching operator

`~~`

`@a ~~ (1,2,3)`

Answer

For a second, lets consider the slightly different

```
\@a ~~ (1,2,3)
```

`~~`

evaluates its arguments in scalar context, so the above is the same as

```
scalar(\@a) ~~ scalar(1,2,3)
```

`\@a`

(in any context) returns a reference to`@a`

.`1, 2, 3`

in scalar context is similar to`do { 1; 2; 3 }`

, returning`3`

.

So minus a couple of warnings*, the above is equivalent to

```
\@a ~~ 3
```

What you actually want is

```
\@a ~~ do { my @temp = (1,2,3); \@temp }
```

which can be shortened to

```
\@a ~~ [ 1,2,3 ]
```

Finally, the magic of `~~`

allows `\@a`

to be written as `@a`

, so that can be shortened further to

```
@a ~~ [ 1,2,3 ]
```

* — Always use `use strict; use warnings;`

!

Source (Stackoverflow)

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