Question

基于循环选择的条件。

if(valid)
for (std::multimap<int,int>::reverse_iterator rit=id_count.rbegin(); mcount<10 && rit!=id_count.rend();++rit)
else
for (std::multimap<int,int>::iterator rit=id_match.begin(); mcount<10 && rit!=id_match.end();++rit)

{
    //this is common for both for loop
}

如何在C ++中实现这一目标？

Answer 1

你别无选择，只能将公共部分放在一个函数中，大致如下：

void somefunction(...)
{
    //this is common for both for loops
}

if (valid)
{
  for (std::multimap<int,int>::reverse_iterator rit=id_count.rbegin(); mcount<10 && rit!=id_count.rend();++rit)
    somefunctiuon(...);
}
else
{
  for (std::multimap<int,int>::iterator rit=id_match.begin(); mcount<10 && rit!=id_match.end();++rit)
    somefunctiuon(...);
}

Answer 2

这可以说是最有用的例子，它的不值得合并循环逻辑，尽管它确实有效。这里提供了利息价值......

#include <iostream>
#include <map>

int main()
{
    std::multimap<int,int> id_count = { {1,2}, {9, -2}, {1,44}, {2,3}, {3,5}, {7,34} };

    for (int valid = 0; valid < 2; ++valid)
    {
        std::cout << "valid " << valid << '\n';
        int mcount = 0;
        for (std::multimap<int,int>::iterator it = valid ? id_count.rbegin().base()
                                                         : id_count.begin();
             mcount<10 && (valid ? it--!=id_count.begin() : it!=id_count.end());
             (valid ? it : ++it), ++mcount)
        {
            std::cout << "[mcount " << mcount << "] "
                << it->first << ',' << it->second << '\n';
        }
        std::cout << '\n';
    }
}

Answer 3

您可以创建模板功能：

#include <map>
#include <iostream>

template<typename I> void func(I begin, I end) {
    int mcount = 0;
    for (I it = begin; mcount < 10 && it != end; ++it) {
        ++mcount;
        std::cout << "[mcount " << mcount << "] "
            << it->first << ',' << it->second << '\n';
    }
}

int main() {
    std::multimap<int,int> id_count = { {1,2}, {9, -2}, {1,44}, {2,3}, {3,5}, {7,34} };
    for (int valid = 0; valid < 2; ++valid) {
        std::cout << "valid " << valid << '\n';
        if (valid) {
            func(id_count.rbegin(), id_count.rend());
        } else { 
            func(id_count.begin(), id_count.end());
        }
        std::cout << '\n';
    }
}

但是恕我直言这个解决方案有点复杂，所以请考虑其他方法（比如将循环体放在函数中）。

Answer 4

您可以尝试“#if valid”，例如：

#if 0 for(i=1;i<10;++i) #else for(i=2;i<9;++i) #endif { cout << i << endl; }

Answer 5

在C ++ 14中，您还可以选择使用通用lambdas：

auto common_code = [/* Capture state if needed */] ( auto& Iter )
{
    // Your common code
};

if ( valid )
    for ( std::multimap<int, int>::reverse_iterator rit = id_count.rbegin(); mcount < 10 && rit != id_count.rend(); ++rit )
        common_code( rit );
else
    for ( std::multimap<int, int>::iterator rit = id_match.begin(); mcount < 10 && rit != id_match.end(); ++rit )
        common_code( rit );

根据if条件选择for循环

5 个答案: