当代码流是这样的:

if(check())
{
  ...
  ...
  if(check())
  {
    ...
    ...
    if(check())
    {
      ...
      ...
    }
  }
}

我通常看到这种方法可以避免上述混乱的代码流:

do {
    if(!check()) break;
    ...
    ...
    if(!check()) break;
    ...
    ...
    if(!check()) break;
    ...
    ...
} while(0);

有什么更好的方法可以避免这种工作/黑客,从而使其成为更高级别(行业级别)的代码?

任何跳出盒子的建议都是欢迎的!


当前回答

TLDR: RAII、事务性代码(仅设置结果或在已经计算时返回内容)和异常。

长一点的回答:

在C语言中,这类代码的最佳实践是在代码中添加一个EXIT/CLEANUP/other标签,在该标签中进行本地资源的清理,并返回错误代码(如果有的话)。这是最佳实践,因为它将代码自然地划分为初始化、计算、提交和返回:

error_code_type c_to_refactor(result_type *r)
{
    error_code_type result = error_ok; //error_code_type/error_ok defd. elsewhere
    some_resource r1, r2; // , ...;
    if(error_ok != (result = computation1(&r1))) // Allocates local resources
        goto cleanup;
    if(error_ok != (result = computation2(&r2))) // Allocates local resources
        goto cleanup;
    // ...

    // Commit code: all operations succeeded
    *r = computed_value_n;
cleanup:
    free_resource1(r1);
    free_resource2(r2);
    return result;
}

在C语言中,在大多数代码库中,if(error_ok !=…goto代码通常隐藏在一些方便的宏(RET(computation_result), ENSURE_SUCCESS(computation_result, return_code)等)后面。

c++比C提供了额外的工具:

清理块功能可以作为RAII实现,这意味着您不再需要整个清理块,并允许客户端代码添加早期返回语句。 当无法继续时抛出抛出,转换所有if(error_ok !=…变成直接的通话。

等价的c++代码:

result_type cpp_code()
{
    raii_resource1 r1 = computation1();
    raii_resource2 r2 = computation2();
    // ...
    return computed_value_n;
}

这是最佳实践,因为:

It is explicit (that is, while error handling is not explicit, the main flow of the algorithm is) It is straightforward to write client code It is minimal It is simple It has no repetitive code constructs It uses no macros It doesn't use weird do { ... } while(0) constructs It is reusable with minimal effort (that is, if I want to copy the call to computation2(); to a different function, I don't have to make sure I add a do { ... } while(0) in the new code, nor #define a goto wrapper macro and a cleanup label, nor anything else).

其他回答

typedef bool (*Checker)();

Checker * checkers[]={
 &checker0,&checker1,.....,&checkerN,NULL
};

bool checker1(){
  if(condition){
    .....
    .....
    return true;
  }
  return false;
}

bool checker2(){
  if(condition){
    .....
    .....
    return true;
  }
  return false;
}

......

void doCheck(){
  Checker ** checker = checkers;
  while( *checker && (*checker)())
    checker++;
}

怎么样?

如何移动if语句到一个额外的函数,产生一个数值或enum结果?

int ConditionCode (void) {
   if (condition1)
      return 1;
   if (condition2)
      return 2;
   ...
   return 0;
}


void MyFunc (void) {
   switch (ConditionCode ()) {
      case 1:
         ...
         break;

      case 2:
         ...
         break;

      ...

      default:
         ...
         break;
   }
}

有些时候,使用goto实际上是正确的答案——至少对那些没有在宗教信仰中长大的人来说,“不管问题是什么,goto永远都不是答案”——这是其中的一个例子。

此代码使用了do{…}, (0);唯一的目的就是把一个傻瓜打扮成休息的样子。如果你打算使用goto,那就公开使用它。让代码更难阅读是没有意义的。

一种特殊的情况是当你有很多代码和相当复杂的条件时:

void func()
{
   setup of lots of stuff
   ...
   if (condition)
   {
      ... 
      ...
      if (!other condition)
      {
          ...
          if (another condition)
          {
              ... 
              if (yet another condition)
              {
                  ...
                  if (...)
                     ... 
              }
          }
      }
  .... 

  }
  finish up. 
}

没有如此复杂的逻辑,实际上可以更清楚地表明代码是正确的。

void func()
{
   setup of lots of stuff
   ...
   if (!condition)
   {
      goto finish;
   }
   ... 
   ...
   if (other condition)
   {
      goto finish;
   }
   ...
   if (!another condition)
   {
      goto finish;
   }
   ... 
   if (!yet another condition)
   {
      goto finish;
   }
   ... 
   .... 
   if (...)
         ...    // No need to use goto here. 
 finish:
   finish up. 
}

编辑:澄清一下,我绝不是建议使用goto作为通用解决方案。但在某些情况下,goto是比其他解决方案更好的解决方案。

例如,想象一下,我们正在收集一些数据,测试的不同条件是某种“这是正在收集的数据的结束”——这取决于某种“继续/结束”标记,这些标记根据您在数据流中的位置而变化。

现在,当我们完成之后,我们需要将数据保存到一个文件中。

是的,通常有其他解决方案可以提供合理的解决方案,但并不总是如此。

I'm adding an answer for the sake of completeness. A number of other answers pointed out that the large condition block could be split out into a separate function. But as was also pointed out a number of times is that this approach separates the conditional code from the original context. This is one reason that lambdas were added to the language in C++11. Using lambdas was suggested by others but no explicit sample was provided. I've put one in this answer. What strikes me is that it feels very similar to the do { } while(0) approach in many ways - and maybe that means it's still a goto in disguise....

earlier operations
...
[&]()->void {

    if (!check()) return;
    ...
    ...
    if (!check()) return;
    ...
    ...
    if (!check()) return;
    ...
    ...
}();
later operations

Try to extract the code into a separate function (or perhaps more than one). Then return from the function if the check fails. If it's too tightly coupled with the surrounding code to do that, and you can't find a way to reduce the coupling, look at the code after this block. Presumably, it cleans up some resources used by the function. Try to manage these resources using an RAII object; then replace each dodgy break with return (or throw, if that's more appropriate) and let the object's destructor clean up for you. If the program flow is (necessarily) so squiggly that you really need a goto, then use that rather than giving it a weird disguise. If you have coding rules that blindly forbid goto, and you really can't simplify the program flow, then you'll probably have to disguise it with your do hack.