# How can I swap two Perl variables without using the third?

I want to swap two variables values without using the third variable in Perl, e. g.:

```my \$first = 10;
my \$second = 20;
```

Please suggest me how we can do this in Perl in a simple way.

The best way that only provide us is like this in one line you can swap the values:

``` (\$first, \$second) = (\$second, \$first);
```

You can write:

```(\$first, \$second) = (\$second, \$first);
```

The Perl-specific methods already listed are best, but here's a technique using XOR that works in many languages, including Perl:

```use strict;

my \$x = 4;
my \$y = 8;

print "X: \$x  Y: \$y\n";

\$x ^= \$y;
\$y ^= \$x;
\$x ^= \$y;

print "X: \$x  Y: \$y\n";

X: 4  Y: 8
X: 8  Y: 4
```

You can do this relatively easy using simple maths.

We know;

```First = 10
Second = 20
```

If we say First = First + Second

We now have the following;

```First = 30
Second = 20
```

Now you can say Second = First - Second (Second = 30 - 20)

We now have;

```First = 30
Second = 10
```

Now minus Second from First, and you get First = 20, and Second = 10.

```\$first = \$first + \$second;
\$second = \$first - \$second;
\$first = \$first-\$second;
```

This will swap two integer variables A better solution might be

```\$first = \$first xor \$second;
\$second = \$first xor \$second;
\$first = \$first xor \$second;
```

```#!/usr/bin/perl

\$a=5;
\$b=6;

print "\n The value of a and b before swap is --> \$a,\$b \n";

\$a=\$a+\$b;
\$b=\$a-\$b;
\$a=\$a-\$b;

print "\n The value of a and b after swap is as follows:";
print "\n The value of a is ---->\$a \n";
print "\n The value of b is----->\$b \n";
```

you can use this logic

```firstValue = firstValue + secondValue;

secondValue = firstValue - secondValue;

firstValue = firstValue - secondValue;
```

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