在这样的情况下我应该避免"转到"吗？

Question

我正在创建一个"连接迭代器",即一个迭代器,它将遍历ints中的s int**.

它的构造函数需要:

• 一个数组T**,表示每个子数组的开头.
• 一个数组T**,表示每个子数组的结尾.

瞧,我遇到了一个goto似乎合适的情况.

但是我内心的某些东西尖叫着"不!" 所以我以为我会来这里问:

我应该尝试避免goto这种情况吗？(如果我这样做,它会提高可读性吗？)

#include <algorithm>

template<class T>
class lazy_concat_iterator
{
    // This code was meant to work for any valid input iterator
    // but for easier reading, I'll assume the type is: T**

    mutable T** m_endIt;              // points to an array of end-pointers
    mutable T** m_it;                 // points to an array of begin-pointers
    mutable bool m_started;   // have we started iterating?
    mutable T* m_sub;         // points somewhere in the current sub-array
    mutable T* m_subEnd;      // points to the end of the current sub-array

public:
    lazy_concat_iterator(T** begins, T** ends)
        : m_it(begins), m_endIt(ends), m_started(false) { }

    void ensure_started() const
    {
        if (!m_started)
        {
            m_started = true;

        INIT:
            m_sub = *m_it;
            m_subEnd = *m_endIt;

            if (m_sub == m_subEnd)  // End of this subarray?
            {
                ++m_it;
                ++m_endIt;
                goto INIT;  // try next one         <<< should I use goto here?
            }
        }
    }
};

你如何使用它:

#include <vector>
#include <cstring>

using namespace std;

int main(int argc, char* argv[])
{
    vector<char*> beginnings(argv, argv + argc);

    vector<char*> endings;
    for (int i = 0; i < argc; i++)
        endings.push_back(argv[i] + strlen(argv[i]));

    lazy_concat_iterator<char> it(&beginnings[0], &endings[0]);
    it.ensure_started();  // 'it' would call this internally, when dereferenced
}

Answer 1

是的,您可以而且应该避免goto,例如,此代码应该与您在INIT标签上执行的操作相同(这也适用于输入迭代器,这是一个"隐藏的要求",因为它不会取消引用,m_it并且m_endIt一旦条件有额外的时间与我以前的转型不同,我遇到了:

while ((m_subIt = *m_it) == (m_subEnd = *m_endIt))
{
    ++m_it;
    ++m_endIt;
}

以前的回答尝试:

即使是一个永远的循环也会比一个更清晰,更整洁goto.它突出了明显的"永不终止"的可能性甚至更好.

    for (;;)
    {
        m_sub = *m_it;
        m_subEnd = *m_endIt;

        if (m_sub != m_subEnd)
            break;

        ++m_it;
        ++m_endIt;
    }

虽然我不明白为什么你需要分配到m_subEnd并m_subIt在循环中.如果不这样做,你可以将其重写为while循环:

while (*m_it == *m_endIt)
{
    ++m_it;
    ++m_endIt;
}

m_subIt = *m_it;
m_subEnd = *m_endIt;

Answer 2

while (*m_it == *m_endIt)
{
    ++m_it;
    ++m_endIt;
}

m_sub = *m_it;
m_subEnd = *m_endIt;