我曾经使用过糟糕的编码标准，强迫你使用单一的退出路径，如果函数不是琐碎的，结果几乎总是非结构化的意大利面条——你最终会遇到许多中断和继续。

有时出于性能考虑，这是必要的(我不想获取不同的缓存线，就像继续一样;有时)。

如果你不使用RAII分配资源(内存、文件描述符、锁等)，那么多次返回很容易出错，而且肯定是重复的，因为释放需要手动执行多次，你必须仔细跟踪。

在这个例子中:

function()
{
    HRESULT error = S_OK;

    if(SUCCEEDED(Operation1()))
    {
        if(SUCCEEDED(Operation2()))
        {
            if(SUCCEEDED(Operation3()))
            {
                if(SUCCEEDED(Operation4()))
                {
                }
                else
                {
                    error = OPERATION4FAILED;
                }
            }
            else
            {
                error = OPERATION3FAILED;
            }
        }
        else
        {
            error = OPERATION2FAILED;
        }
    }
    else
    {
        error = OPERATION1FAILED;
    }

    return error;
}

我会把它写成:

function() {
    HRESULT error = OPERATION1FAILED;//assume failure
    if(SUCCEEDED(Operation1())) {

        error = OPERATION2FAILED;//assume failure
        if(SUCCEEDED(Operation3())) {

            error = OPERATION3FAILED;//assume failure
            if(SUCCEEDED(Operation3())) {

                error = OPERATION4FAILED; //assume failure
                if(SUCCEEDED(Operation4())) {

                    error = S_OK;
                }
            }
        }
    }
    return error;
}

这当然看起来更好。

这在手动资源释放的情况下尤其有用，因为在哪里和哪些释放是必要的是相当直接的。如下例所示:

function() {
    HRESULT error = OPERATION1FAILED;//assume failure
    if(SUCCEEDED(Operation1())) {

        //allocate resource for op2;
        char* const p2 = new char[1024];
        error = OPERATION2FAILED;//assume failure
        if(SUCCEEDED(Operation2(p2))) {

            //allocate resource for op3;
            char* const p3 = new char[1024];
            error = OPERATION3FAILED;//assume failure
            if(SUCCEEDED(Operation3(p3))) {

                error = OPERATION4FAILED; //assume failure
                if(SUCCEEDED(Operation4(p2,p3))) {

                    error = S_OK;
                }
            }
            //free resource for op3;
            delete [] p3;
        }
        //free resource for op2;
        delete [] p2;
    }
    return error;
}

如果在没有RAII(忘记异常问题!)的情况下使用多个出口编写这段代码，则必须多次写入删除。如果你用}else{then 这有点难看。

但是RAII使得多个出口资源问题变得毫无意义。

唯一重要的问题是“如何使代码更简单、更易读、更容易理解?”如果使用多个返回会更简单，那么就使用它们。

总是要求单一的返回类型是没有意义的。我认为这更像是一个标志，表明某些东西可能需要简化。有时有多个回报是必要的，但通常你可以通过至少尝试有一个退出点来简化事情。

是否有充分的理由说明在函数中只有一个return语句是一个更好的实践?

是的，有:

The single exit point gives an excellent place to assert your post-conditions. Being able to put a debugger breakpoint on the one return at the end of the function is often useful. Fewer returns means less complexity. Linear code is generally simpler to understand. If trying to simplify a function to a single return causes complexity, then that's incentive to refactor to smaller, more general, easier-to-understand functions. If you're in a language without destructors or if you don't use RAII, then a single return reduces the number of places you have to clean up. Some languages require a single exit point (e.g., Pascal and Eiffel).

这个问题通常被提出为多个返回或深度嵌套的if语句之间的错误二分法。几乎总有第三种解决方案，它是线性的(没有深度嵌套)，只有一个出口点。

更新:MISRA的指导方针显然也提倡单次退出。

需要澄清的是，我并不是说拥有多个回报总是错误的。但如果有其他等价的解决方案，有很多很好的理由选择单一回报的方案。

I always avoid multiple return statements. Even in small functions. Small functions can become larger, and tracking the multiple return paths makes it harder (to my small mind) to keep track of what is going on. A single return also makes debugging easier. I've seen people post that the only alternative to multiple return statements is a messy arrow of nested IF statements 10 levels deep. While I certain agree that such coding does occur, it isn't the only option. I wouldn't make the choice between a multiple return statements and a nest of IFs, I'd refactor it so you'd eliminate both. And that is how I code. The following code eliminates both issues and, in my mind, is very easy to read:

public string GetResult()
{
    string rv = null;
    bool okay = false;

    okay = PerformTest(1);

    if (okay)
    {
        okay = PerformTest(2);
    }

    if (okay)
    {
        okay = PerformTest(3);
    }

    if (okay)
    {
        okay = PerformTest(4);
    };

    if (okay)
    {
        okay = PerformTest(5);
    }

    if (okay)
    {
        rv = "All Tests Passed";
    }

    return rv;
}