### Re: shift left syntax?

```
>>
>> Hi All,
>>
>> Is this the only way to shift left?
>>
>> \$i = \$i +< 0x01
>>
>> \$ p6 'my int32 \$i=0x5DAE; say \$i.base(0x10); \$i = \$i +< 0x01; say
>> \$i.base(0x10);'
>>
>> 5DAE
>> BB5C
>>
>>
>> Does we have any of those fancy += ~= ways of doing it?
>>
>> Many thanks,
>> -T

On 2/8/19 10:02 AM, Brad Gilbert wrote:

The `=` infix operator is a meta operator.

That means it takes an infix operator as a sort of "argument".

There is no `+=` operator, it is just the `=` operator combined with
the `+` operator.

\$a += 2;
\$a [+]= 2; # more explicitly take the + operator as an argument to
the = operator

So if you want to know how to use a similar operator to `+=`, start
with the infix operator you want, and add `=`

\$i = \$i +< 0x01;

\$i [+<]= 0x01;
\$i +<= 0x01;

On Fri, Feb 8, 2019 at 12:20 AM Todd Chester via perl6-users
wrote:

Thank you!

```

### Re: shift left syntax?

```The `=` infix operator is a meta operator.

That means it takes an infix operator as a sort of "argument".

There is no `+=` operator, it is just the `=` operator combined with
the `+` operator.

\$a += 2;
\$a [+]= 2; # more explicitly take the + operator as an argument to
the = operator

So if you want to know how to use a similar operator to `+=`, start
with the infix operator you want, and add `=`

\$i = \$i +< 0x01;

\$i [+<]= 0x01;
\$i +<= 0x01;

On Fri, Feb 8, 2019 at 12:20 AM Todd Chester via perl6-users
wrote:
>
> Hi All,
>
> Is this the only way to shift left?
>
>\$i = \$i +< 0x01
>
> \$ p6 'my int32 \$i=0x5DAE; say \$i.base(0x10); \$i = \$i +< 0x01; say
> \$i.base(0x10);'
>
> 5DAE
> BB5C
>
>
> Does we have any of those fancy += ~= ways of doing it?
>
> Many thanks,
> -T

```

### Re: shift left syntax?

```

On 2/7/19 10:36 PM, yary wrote:

perl6 -e 'my \$i = 0x5DAE; \$i +<= 1; say \$i.base(0x10);'

BB5C

-y

Hi Yary,

\$ p6 'my Buf \$x=Buf.new(0xAE,0x5D); my int32 \$i=0x5DAE; say \$x; say
\$i.base(0x10); \$i +< 0x01; say \$i.base(0x10);'

WARNINGS for -e:
Useless use of "+<" in expression "\$i +< 0x01" in sink context (line 1)
Buf:0x
5DAE
5DAE

Ah poop!  I forgot the =

\$ p6 'my int32 \$i=0x5DAE; say \$i.base(0x10); \$i +<= 0x01; say
\$i.base(0x10);'

5DAE
BB5C

Thank you!

-T

```

### Re: shift left syntax?

```perl6 -e 'my \$i = 0x5DAE; \$i +<= 1; say \$i.base(0x10);'

BB5C

-y

>

```

### shift left syntax?

```
Hi All,

Is this the only way to shift left?

\$i = \$i +< 0x01

\$ p6 'my int32 \$i=0x5DAE; say \$i.base(0x10); \$i = \$i +< 0x01; say
\$i.base(0x10);'

5DAE
BB5C

Does we have any of those fancy += ~= ways of doing it?

Many thanks,
-T

```