控制到达非空函数的结尾 [英] Control reaches end of non-void function

问题描述

以下代码片段在编译时会生成一些警告消息:

The following code snippet generates some warning messages when compiling:

Cluster& Myclass::getCluster(const Point &p)
{
    foreach (Cluster c, *this)
        foreach (Point point, c)
            if (point == p)
                return c;
}

警告是:

1. 对局部变量c"的引用返回[默认启用]
2. 控制到达非空函数的结尾[使用-Wreturn-type时]

我知道如果条件失败，我不会返回值.但是，当我尝试 return 0 时，它给了我错误.

I know that I am not returning a value if the condition fails. However, when I try return 0 it gave me error.

我该如何解决这些问题?

How can I solve these issues?

推荐答案

如果您的函数可以合法地无法找到匹配的Cluster，那么您应该让它返回一个指针:

If your function can legitimately fail to find a matching Cluster, then you should have it return a pointer:

Cluster* Myclass::getCluster(const Point &p)
{
    foreach (Cluster c, *this)
        foreach (Point point, c)
            if (point == p)
                return &c;
    return 0; // or return nullptr; in C++11
}

但这还不行，因为 c 是一个局部变量.所以你把它作为一个参考，像这样:

But this doesn't work yet, because c is a local variable. So you make it a reference, like this:

Cluster* Myclass::getCluster(const Point &p)
{
    foreach (Cluster& c, *this)
        foreach (Point point, c)
            if (point == p)
                return &c;
    return 0; // or "return nullptr;" in C++11
}

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