string til integer

jeg tror ikke det er fordi det er en streng det går galt:

use strict;
use Time::Local;

my $test = 0;
my $string = "1970";

$test += $string;
my @gmtTime = qw(0 0 0 1 0 $test);
my $time_gm = timegm(@gmtTime);

print "gmt: $time_gm\n";
print "$test\n";
exit 0;

giver:

gmt: 946684800
1970

use strict;
use Time::Local;

my $test = 0;
my $string = "1970";

$test += $string;
my @gmtTime = qw(0 0 0 1 0 $test);
my $time_gm = timegm(@gmtTime);

print "gmt: $time_gm\n";
print "$test\n";

my $local_time = localtime($time_gm);
print "$local_time\n";
exit 0;

giver:
gmt: 946684800
1970
Sat Jan  1 01:00:00 2000

jeg tror at fejlen ligger i my @gmtTime = qw(0 0 0 1 0 $test);

måske skal du bruge my @gmtTime = qq(0 0 0 1 0 $test);

use strict;
use Time::Local;

my $test = 0;
my $string = "1970";

$test += $string;
my @gmtTime = qw(0 0 0 1 0 $test);
my $time_gm = timegm(@gmtTime);

my ($sec,$min,$hour,$mday,$mon,$year) = @gmtTime;


print "gmt: $time_gm\n";

print "$sec,$min,$hour,$mday,$mon,$year\n";
exit 0;


giver:
gmt: 946684800
0,0,0,1,0,$test

mhhh ...
så læser perl det, som står i paranteserne måske som en string? ... og så er

sub
{
    $string = $_[0];

}

jeps - den tager ikke værdien af variablen men variablen-navnet

use strict;
use Time::Local;

my $test = 0;
my $string = "1970";

$test += $string;
my @gmtTime = (0, 0, 0, 1, 0, $test);
my $time_gm = timegm(@gmtTime);

my ($sec,$min,$hour,$mday,$mon,$year) = @gmtTime;
my $time2 = timegm($sec,$min,$hour,$mday,$mon,$year);

print "gmt: $time_gm\n";
print "gmt : $time2\n";
print "$sec,$min,$hour,$mday,$mon,$year\n";
exit 0;

giver:
Hit ENTER or type command to continue
gmt: 0
gmt : 0
0,0,0,1,0,1970

#!/usr/bin/perl
use Time::Local;

$string = "1970";

$gmtTime [0] = 0;
$gmtTime [1] = 0;
$gmtTime [2] = 0;
$gmtTime [3] = 1;
$gmtTime [4] = 0;
$gmtTime [5] = $string;

#my @gmtTime = qw(0 0 0 1 0 $string);
$time_gm = timegm(@gmtTime);

print "gmt: $time_gm\n";

ja - jeg brugte en implicit array i mit eksempel ...