Perl 6 iterate through command line arguments
I want to write a Perl 6 program that performs the same operation on an undetermined number of files, just like how you can use rm file1.txt file2.txt file3.txt in Linux.
My code looks like this:
#!/usr/bin/env perl6
my @args = @*ARGS.perl;
for @args -> $arg {
say $arg
}
However, when I invoke it as ./loop_args.pl6 file1.txt file2.txt I get the following output: ["file1.txt", "file2.txt"]
What am I doing wrong? I just want it to be an array.
perl6
asked Nov 13 '18 at 1:49
Taylor Lee
432
add a comment |
I want to write a Perl 6 program that performs the same operation on an undetermined number of files, just like how you can use rm file1.txt file2.txt file3.txt in Linux.
My code looks like this:
#!/usr/bin/env perl6
my @args = @*ARGS.perl;
for @args -> $arg {
say $arg
}
However, when I invoke it as ./loop_args.pl6 file1.txt file2.txt I get the following output: ["file1.txt", "file2.txt"]
What am I doing wrong? I just want it to be an array.
perl6
asked Nov 13 '18 at 1:49
Taylor Lee
432
I would like to know what lead you to use.perl, so that we can prevent others from going down this path in the future.
– Brad Gilbert
Nov 16 '18 at 16:07
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41
add a comment |
I want to write a Perl 6 program that performs the same operation on an undetermined number of files, just like how you can use rm file1.txt file2.txt file3.txt in Linux.
My code looks like this:
#!/usr/bin/env perl6
my @args = @*ARGS.perl;
for @args -> $arg {
say $arg
}
However, when I invoke it as ./loop_args.pl6 file1.txt file2.txt I get the following output: ["file1.txt", "file2.txt"]
What am I doing wrong? I just want it to be an array.
perl6
asked Nov 13 '18 at 1:49
Taylor Lee
432
I want to write a Perl 6 program that performs the same operation on an undetermined number of files, just like how you can use rm file1.txt file2.txt file3.txt in Linux.
My code looks like this:
#!/usr/bin/env perl6
my @args = @*ARGS.perl;
for @args -> $arg {
say $arg
}
However, when I invoke it as ./loop_args.pl6 file1.txt file2.txt I get the following output: ["file1.txt", "file2.txt"]
What am I doing wrong? I just want it to be an array.
perl6
perl6
asked Nov 13 '18 at 1:49
Taylor Lee
432
asked Nov 13 '18 at 1:49
Taylor Lee
432
asked Nov 13 '18 at 1:49
Taylor Lee
432
asked Nov 13 '18 at 1:49
Taylor Lee
432
asked Nov 13 '18 at 1:49
Taylor Lee
432
432
I would like to know what lead you to use.perl, so that we can prevent others from going down this path in the future.
– Brad Gilbert
Nov 16 '18 at 16:07
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41
add a comment |
I would like to know what lead you to use.perl, so that we can prevent others from going down this path in the future.
– Brad Gilbert
Nov 16 '18 at 16:07
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41
I would like to know what lead you to use
.perl, so that we can prevent others from going down this path in the future.– Brad Gilbert
Nov 16 '18 at 16:07
I would like to know what lead you to use
.perl, so that we can prevent others from going down this path in the future.– Brad Gilbert
Nov 16 '18 at 16:07
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41
add a comment |
2 Answers
2
active
oldest
votes
For Perl6 the arguments array is @*ARGS, this gives the code:
#! /usr/bin/env perl6
use v6;
for @*ARGS -> $arg
{
print("Arg $argn");
}
So you just need to remove the .perl from your my @args = @*ARGS.perl; line.
EDIT: Updated as per Ralphs's comments
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
You don't need theuse strict;there :)
– Scimon
Nov 13 '18 at 9:04
2
To clarify @Scimon's comment you don't need theuse strict;because P6 defaults to strict. While I'm writing a comment I'll also note that themy $arg;declares a$argbut it's redundant and misleading. The-> $argdeclares another$argthat shadows themyone inside the block.
– raiph
Nov 13 '18 at 19:24
add a comment |
So it depends on what you want to do with the files. If you want to read through them all you might want to take a look at $*ARGFILES this is an IO::CatHandle Object that batches up all the arguments and treats them as a list of files. So to print them all out you could do.
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
$IO.path.say
}
Ok so that's a bit more convoluted than looking at @*ARGS. But what if you wanted to print the first 5 lines for each file?
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
say $IO.lines: 5;
}
Or maybe you just want to read the contents of all the files and print them out?
#! /usr/bin/env perl6
use v6;
.say for $*ARGFILES.lines;
Hope that gives you some ideas.
answered Nov 13 '18 at 9:12
Scimon
1,5541010
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
For Perl6 the arguments array is @*ARGS, this gives the code:
#! /usr/bin/env perl6
use v6;
for @*ARGS -> $arg
{
print("Arg $argn");
}
So you just need to remove the .perl from your my @args = @*ARGS.perl; line.
EDIT: Updated as per Ralphs's comments
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
You don't need theuse strict;there :)
– Scimon
Nov 13 '18 at 9:04
2
To clarify @Scimon's comment you don't need theuse strict;because P6 defaults to strict. While I'm writing a comment I'll also note that themy $arg;declares a$argbut it's redundant and misleading. The-> $argdeclares another$argthat shadows themyone inside the block.
– raiph
Nov 13 '18 at 19:24
add a comment |
For Perl6 the arguments array is @*ARGS, this gives the code:
#! /usr/bin/env perl6
use v6;
for @*ARGS -> $arg
{
print("Arg $argn");
}
So you just need to remove the .perl from your my @args = @*ARGS.perl; line.
EDIT: Updated as per Ralphs's comments
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
You don't need theuse strict;there :)
– Scimon
Nov 13 '18 at 9:04
2
To clarify @Scimon's comment you don't need theuse strict;because P6 defaults to strict. While I'm writing a comment I'll also note that themy $arg;declares a$argbut it's redundant and misleading. The-> $argdeclares another$argthat shadows themyone inside the block.
– raiph
Nov 13 '18 at 19:24
add a comment |
For Perl6 the arguments array is @*ARGS, this gives the code:
#! /usr/bin/env perl6
use v6;
for @*ARGS -> $arg
{
print("Arg $argn");
}
So you just need to remove the .perl from your my @args = @*ARGS.perl; line.
EDIT: Updated as per Ralphs's comments
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
For Perl6 the arguments array is @*ARGS, this gives the code:
#! /usr/bin/env perl6
use v6;
for @*ARGS -> $arg
{
print("Arg $argn");
}
So you just need to remove the .perl from your my @args = @*ARGS.perl; line.
EDIT: Updated as per Ralphs's comments
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
edited Nov 13 '18 at 21:32
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
answered Nov 13 '18 at 2:14
Kingsley
2,26011023
2,26011023
You don't need theuse strict;there :)
– Scimon
Nov 13 '18 at 9:04
2
To clarify @Scimon's comment you don't need theuse strict;because P6 defaults to strict. While I'm writing a comment I'll also note that themy $arg;declares a$argbut it's redundant and misleading. The-> $argdeclares another$argthat shadows themyone inside the block.
– raiph
Nov 13 '18 at 19:24
add a comment |
You don't need theuse strict;there :)
– Scimon
Nov 13 '18 at 9:04
2
To clarify @Scimon's comment you don't need theuse strict;because P6 defaults to strict. While I'm writing a comment I'll also note that themy $arg;declares a$argbut it's redundant and misleading. The-> $argdeclares another$argthat shadows themyone inside the block.
– raiph
Nov 13 '18 at 19:24
You don't need the
use strict; there :)– Scimon
Nov 13 '18 at 9:04
You don't need the
use strict; there :)– Scimon
Nov 13 '18 at 9:04
2
2
To clarify @Scimon's comment you don't need the
use strict; because P6 defaults to strict. While I'm writing a comment I'll also note that the my $arg; declares a $arg but it's redundant and misleading. The -> $arg declares another $arg that shadows the my one inside the block.– raiph
Nov 13 '18 at 19:24
To clarify @Scimon's comment you don't need the
use strict; because P6 defaults to strict. While I'm writing a comment I'll also note that the my $arg; declares a $arg but it's redundant and misleading. The -> $arg declares another $arg that shadows the my one inside the block.– raiph
Nov 13 '18 at 19:24
add a comment |
So it depends on what you want to do with the files. If you want to read through them all you might want to take a look at $*ARGFILES this is an IO::CatHandle Object that batches up all the arguments and treats them as a list of files. So to print them all out you could do.
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
$IO.path.say
}
Ok so that's a bit more convoluted than looking at @*ARGS. But what if you wanted to print the first 5 lines for each file?
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
say $IO.lines: 5;
}
Or maybe you just want to read the contents of all the files and print them out?
#! /usr/bin/env perl6
use v6;
.say for $*ARGFILES.lines;
Hope that gives you some ideas.
answered Nov 13 '18 at 9:12
Scimon
1,5541010
add a comment |
So it depends on what you want to do with the files. If you want to read through them all you might want to take a look at $*ARGFILES this is an IO::CatHandle Object that batches up all the arguments and treats them as a list of files. So to print them all out you could do.
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
$IO.path.say
}
Ok so that's a bit more convoluted than looking at @*ARGS. But what if you wanted to print the first 5 lines for each file?
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
say $IO.lines: 5;
}
Or maybe you just want to read the contents of all the files and print them out?
#! /usr/bin/env perl6
use v6;
.say for $*ARGFILES.lines;
Hope that gives you some ideas.
answered Nov 13 '18 at 9:12
Scimon
1,5541010
add a comment |
So it depends on what you want to do with the files. If you want to read through them all you might want to take a look at $*ARGFILES this is an IO::CatHandle Object that batches up all the arguments and treats them as a list of files. So to print them all out you could do.
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
$IO.path.say
}
Ok so that's a bit more convoluted than looking at @*ARGS. But what if you wanted to print the first 5 lines for each file?
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
say $IO.lines: 5;
}
Or maybe you just want to read the contents of all the files and print them out?
#! /usr/bin/env perl6
use v6;
.say for $*ARGFILES.lines;
Hope that gives you some ideas.
answered Nov 13 '18 at 9:12
Scimon
1,5541010
So it depends on what you want to do with the files. If you want to read through them all you might want to take a look at $*ARGFILES this is an IO::CatHandle Object that batches up all the arguments and treats them as a list of files. So to print them all out you could do.
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
$IO.path.say
}
Ok so that's a bit more convoluted than looking at @*ARGS. But what if you wanted to print the first 5 lines for each file?
#! /usr/bin/env perl6
use v6;
for $*ARGFILES.handles -> $IO {
say $IO.lines: 5;
}
Or maybe you just want to read the contents of all the files and print them out?
#! /usr/bin/env perl6
use v6;
.say for $*ARGFILES.lines;
Hope that gives you some ideas.
answered Nov 13 '18 at 9:12
Scimon
1,5541010
answered Nov 13 '18 at 9:12
Scimon
1,5541010
answered Nov 13 '18 at 9:12
Scimon
1,5541010
answered Nov 13 '18 at 9:12
Scimon
1,5541010
1,5541010
add a comment |
add a comment |
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I would like to know what lead you to use
.perl, so that we can prevent others from going down this path in the future.– Brad Gilbert
Nov 16 '18 at 16:07
Hey Brad, I misread the documentation here; I have a bad habit of only looking at the code: perl6maven.com/parsing-command-line-arguments-perl6
– Taylor Lee
Nov 16 '18 at 23:28
Ok. Short of asking Gabor to change that article, I suppose there's not much to do.
– Brad Gilbert
Nov 16 '18 at 23:41