c++不支持命名循环，就像Java和其他语言一样。您可以使用goto，或创建您使用的标志值。在每个循环结束时检查标志值。如果它被设置为true，那么您可以跳出该迭代。

虽然这个答案已经提出了，但我认为一个很好的方法是这样做:

for(unsigned int z = 0; z < z_max; z++)
{
    bool gotoMainLoop = false;
    for(unsigned int y = 0; y < y_max && !gotoMainLoop; y++)
    {
        for(unsigned int x = 0; x < x_max && !gotoMainLoop; x++)
        {
                          //do your stuff
                          if(condition)
                            gotoMainLoop = true;
        }
    }

}

只是使用lambdas添加一个显式的答案:

  for (int i = 0; i < n1; ++i) {
    [&] {
      for (int j = 0; j < n2; ++j) {
        for (int k = 0; k < n3; ++k) {
          return; // yay we're breaking out of 2 loops here
        }
      }
    }();
  }

当然，这种模式有一定的局限性，显然只适用于c++ 11，但我认为它非常有用。

我知道这是老帖子了。但我想提出一个更合理、更简单的答案。

for(unsigned int i=0; i < 50; i++)
    {
        for(unsigned int j=0; j < conditionj; j++)
        {
            for(unsigned int k=0; k< conditionk ; k++)
            {
                // If condition is true

                j= conditionj;
               break;
            }
        }
    }

用一个bool变量打破任意数量的循环，如下所示:

bool check = true;

for (unsigned int i = 0; i < 50; i++)
{
    for (unsigned int j = 0; j < 50; j++)
    {
        for (unsigned int k = 0; k < 50; k++)
        {
            //Some statement
            if (condition)
            {
                check = false;
                break;
            }
        }
        if (!check)
        {
            break;
        }
    }
    if (!check)
    {
        break;
    }
}

在这段代码中，我们中断了;所有的循环。

Breaking out of a for-loop is a little strange to me, since the semantics of a for-loop typically indicate that it will execute a specified number of times. However, it's not bad in all cases; if you're searching for something in a collection and want to break after you find it, it's useful. Breaking out of nested loops, however, isn't possible in C++; it is in other languages through the use of a labeled break. You can use a label and a goto, but that might give you heartburn at night..? Seems like the best option though.