根据经纬度坐标计算实际距离

醒目

1. double hypot(double x, double y) {
   
2.       return sqrt(x * x + y * y);
   
3. }
   
4. 
   
5. double distance(double wd1, double jd1, double wd2, double jd2) {// 根据经纬度坐标计算实际距离
   
6.       double x, y, out;
   
7.       double PI = 3.1415926535898;
   
8.       double R = 6.371229 * 1e6;
   
9.        x = (jd2 - jd1) * PI * R * cos( ( (wd1 + wd2) / 2) * PI / 180) / 180;
   
10.        y = (wd2 - wd1) * PI * R / 180;
    
11.        out = hypot(x, y);
    
12.        return out;
    
13.   }

复制代码

请问上面程序转成perl的话是如何写？


http://blog.csdn.net/zgl_dm/archive/2007/11/23/1899598.aspx

枫影谁用了

[root@localhost ]# perl  1.pl
101.36658837807
[root@localhost ]# cat 1.pl
#!/bin/perl
#

my $JuLi= &distance(116.41794,39.99347,116.41787,39.99256);#$wd1, $jd1, $wd2, $jd2
print "$JuLi \n";

sub distance() {

my ($wd1, $jd1, $wd2, $jd2);
$wd1=$_[1];
$jd1=$_[0];
$wd2=$_[3];
$jd2=$_[2];

my $PI=3.1415926535898;
my $R = 6.371229 * 1e6;
#print "$R \n";

$x = ($jd2 - $jd1) * $PI * $R * cos( ( ($wd1 + $wd2) / 2) * $PI / 180) / 180;
$y = ($wd2 - $wd1) * $PI * $R / 180;

my $temp = sqrt($x * $x + $y * $y);

return $temp;


}

masylichu

省点力 嘿嘿
  my ($jd1,$wd1,$jd2,$wd2)=@_;