小编Bob*_*les的帖子

我有以下C++(它还没有真正做任何事......)

#include "stdafx.h"
#include <iostream>

using namespace std;

class Ranker 
{
    int up, down;
  public:
    void set_ranks(int, int);
    int rank(int, int, int, double);
}

void Ranker::set_ranks(int a, int b)
{
    up = a;
    down = b;
}

int _tmain(int argc, _TCHAR* argv[])
{
    return 0;
}

当我运行它时,它会在MS V C++中显示以下错误消息

1>------ Build started: Project: rankclass, Configuration: Debug Win32 ------
1>  rankclass.cpp
1>c:\users\student\desktop\solomon w. c++\rankclass\rankclass\rankclass.cpp(17): error C2628: 'Ranker' followed by 'void' is illegal (did you forget a ';'?)
1>c:\users\student\desktop\solomon w. c++\rankclass\rankclass\rankclass.cpp(18): error …

我有以下代码:

Array.class_eval do
    def abs_sort
        new_array = self
        self.each do |x|
            new_array.push(x.abs)
        end
        return new_array.sort
    end
end

当我尝试运行代码时:

[1, 4, -2].abs_sort

没有任何反应,只是显示一个空白屏幕.为什么？

我有以下代码,它应该搜索数组并查看是否与第二个参数匹配.

def any(check: Set[Any], expr: Boolean): Boolean = {
  var checked = check.filter(_ => expr)
  if (checked == Set())
    return false
  else
    return true
}

它应该被称为这样: any(Set(3, 4, 5, 6), _ > 5)

但是当我打电话给它时:

error: missing parameter type for expanded function ((x$1) => x$1.$greater(5))

我对函数式语言和Scala的经验很少,所以,请给我一个彻底的解释,说明发生了什么以及如何解决它!

我想要做的是动态命名变量,如:

def instance(instance)
    @instance = instance #@instance isn't actually a variable called @instance, rather a variable called @whatever was passed as an argument
end

我怎样才能做到这一点？