有什么方法可以在perl中将函数声明为变量? [英] Is there any way to declare a function as a variable in perl?

问题描述

例如，伪代码如下.根据x的输入，如果x为func1，则必须调用func1().如果x为func2，则必须调用func2().有什么办法可以做到这一点.我不想使用if或switch case语句.还有其他方法可以根据用户输入进行函数调用吗?(类似于将函数视为变量?

For example, Pseudocode is given below. Based on the input given to x, if x is func1, then func1() must be called. If x is func2, then func2() must be called. Is there any way to do this. I do not want to use if or switch case statements. Is there any other approach to make function call based on user input? (something like treating a function as a variable?

sub func1()
{...}

sub func2() 
{...}

sub mainfunc()
{ 
   x = <STDIN>;
   x();
}

推荐答案

安全的方法是使用名称哈希来映射到子例程，这样恶意用户就无法调用任意子.像这样:

The safe way is to have a hash of names mapping to subroutines, so that arbitrary subs can't be called by a malicious user. Something like:

my %table = ("func1" => \&func1, "func2" => \&func2);
my $x = <STDIN>;
chomp $x;
if (exists $table{$x}) {
  $table{$x}->();
} else {
   # Handle error
}

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