Perl中@_数组使用移位函数的问题

Question

希望在shift函数@_内部的数组上使用该函数，以便该函数可以接受可变数量的参数。根据 L Wall 等人在“Programming Perl”中的 p112，传递给子例程的参数被连接成一个平面数组，即@_。假设可以使用shift函数或foreach循环遍历此数组以确定传递的所有值中的最小值，即使某些参数本身就是数组并且这些值作为元素包含在这些数组中。

尝试了许多变体，并且 min2 和 min4 变体使用“for each”循环（最接近“Programming Perl”的 p113 上的循环）来遍历 @_ 数组。使用“shift”函数遍历@_ 数组的 min1 和 min3 变体确实正常工作。关于是什么导致基于“shift”功能的变体失败的任何想法？

下面给出了四个“min”子程序：

# Using perl5.14.2.exe 

use strict;

sub min {
    my $m=shift @_;
    my $b;
    while ($b = shift @_){
      if ($b < $m) {$m=$b};
    };
    return $m;
};


sub min_v2 {
    my $m = $_[0];
    foreach (@_){
      if ($_ < $m) {$m = $_};
    };
    return $m;
};


sub min_v3 {
    my $m=shift (@_);
    my $b;
    while ($b = shift (@_)){
      if ($b < $m) {$m=$b};
    };
    return $m;
};

sub min_v4 {
#   Similar to Programming Perl 2nd Ed by Larry Wall et al p113
    my $m=shift (@_);
    foreach $b (@_){
      if ($b < $m) {$m=$b};
    };
    return $m;
};

以下代码用于测试例程：

print 'min 10,50,5,1 = ' ,     min (10,50,5,1 ),     "\n" ;
print 'min ((10,50,5),1) = ' , min ((10,50,5),1) ,   "\n" ;
print 'min ((10,50,0),1) = ' , min ((10,50,0),1) ,   "\n" ;

print 'min (30,0) = ',         min (30,0),           "\n";
print 'min_v2 (30,0) = ',      min_v2 (30,0),        "\n";
print 'min_v3 (30,0) = ',      min_v3 (30,0),        "\n";
print 'min_v4 (30,0) = ',      min_v4 (30,0),        "\n";

my @a;
@a = (1,2,3,30);
print '@a = ' , "[ @a ]" , , "\n";
print 'min (@a,0) = '    , min (@a,0) , "\n";
print 'min_v2 (@a,0) = ' , min_v2 (@a,0) , "\n" ;
print 'min_v3 (@a,0) = ' , min_v3 (@a,0) , "\n" ;
print 'min_v4 (@a,0) = ' , min_v4 (@a,0) , "\n" ;

@a = (3,2,1,30);
print '@a = ' , "[ @a ]" , , "\n";
print 'min (@a,0) = '    , min (@a,0) , "\n";
print 'min_v2 (@a,0) = ' , min_v2 (@a,0) , "\n" ;
print 'min_v3 (@a,0) = ' , min_v3 (@a,0) , "\n" ;
print 'min_v4 (@a,0) = ' , min_v4 (@a,0) , "\n" ;

@a = (3,2,1,30);
print '@a = ' , "[ @a ]" , , "\n";
print 'min (@a,2) = '    , min (@a,2) , "\n";
print 'min_v2 (@a,2) = ' , min_v2 (@a,2) , "\n" ;
print 'min_v3 (@a,2) = ' , min_v3 (@a,2) , "\n" ;
print 'min_v4 (@a,2) = ' , min_v4 (@a,2) , "\n" ;

并生成了以下输出：

min 10,50,5,1 = 1

min ((10,50,5),1) = 1

min ((10,50,0),1) = 10

min (30,0) = 30

min_v2 (30,0) = 0

min_v3 (30,0) = 30

min_v4 (30,0) = 0

@a = [ 1 2 3 30 ]

min (@a,0) = 1

min_v2 (@a,0) = 0

min_v3 (@a,0) = 1

min_v4 (@a,0) = 0

@a = [ 3 2 1 30 ]

min (@a,0) = 1

min_v2 (@a,0) = 0

min_v3 (@a,0) = 1

min_v4 (@a,0) = 0

@a = [ 3 2 1 30 ]

min (@a,2) = 1

min_v2 (@a,2) = 1

min_v3 (@a,2) = 1

min_v4 (@a,2) = 1

Answer 1

问题与shift或无关@_。问题在于，当您0从列表中移动值时，while条件评估为 false，并且循环在您预期之前终止。考虑这个更简单的例子：

use warnings;
use strict;

my @nums = (50, 30, 0, 1);
my $m = 10;
my $b;
while ($b = shift @nums) {
    print "b=$b m=$m\n";
    if ($b < $m) {$m=$b}
}
print "m=$m\n";

输出：

b=50 m=10
b=30 m=10
m=10

数组 (50, 30) 中的 1st 2 个值的行为符合预期。一旦0从数组中移动 ，循环就结束。