类型' YAML :: Node&'的非const引用的初始化无效。来自类型' YAML :: Node'的左值

时间:2015-01-26 17:28:59

标签: c++ reference yaml-cpp

我正在编写一个函数,可以将值加载并保存到.yaml文件中(取决于toggle输入的内容;' S'用于保存和' L'用于加载)。出于某种原因,我在行中出现Invalid initialization of non-const reference of type 'YAML::Node&' from an rvalue of type 'YAML::Node'错误:

YAML::Node& child = parent[*currentToken];

我在网上找到的解决方案是通过添加const关键字使其保持不变。但是,正如我之前所说,我也需要能够节省价值。因此,我必须能够修改节点,并且不能使它成为const。我现在应该做些什么,我有点迷茫。我的代码接受.yaml文件和最后一个节点中第一个节点的迭代器。这是我的代码:

bool ReadParameter(YAML::Node& parent,
        boost::tokenizer< boost::char_separator< char > >::iterator& currentToken,
        const boost::tokenizer< boost::char_separator< char > >::iterator& lastToken, float& value,
        char toggle) {

    /*
     * When we reach the last token, the distance between the iterators is 1. At this point we must
     * check to see if the token node exists in the yaml file. If it does store the value associated
     * with the token key in the variable `value` to load or store the value back into the yaml file to save.
     */
    if (distance(currentToken, lastToken) == 1) {
        if (parent[*(currentToken)].IsScalar()) {
            if (toggle == 'L') { // Load
                value = parent[*(currentToken)].as< float >();
            }
            else if (toggle == 'S') { // Save
                parent[*(currentToken)] = value;
            }
            return true;
        } else {
            printf("Key %s does not give a scalar value.\n", (*currentToken).c_str());
            return false;
        }
    }

    // If the node is a map, get it's child. Else, our path does not exist.
    if (parent.IsMap()) {
        YAML::Node& child = parent[*currentToken];

        // If there is no child node, return false
        if (child.IsNull()) {
            printf("There are no more child nodes.\n");
            return false;
        }

        // Iterate to the next token.
        return ReadParameter(child, ++currentToken, lastToken, value, toggle);
    } else {
        printf("Too many parameters, the parameter path is incorrect.\n");
        return false;
    }

}

1 个答案:

答案 0 :(得分:1)

YAML::Node已经是参考语义类型。无需将YAML::Node的实例绑定到引用。它们被设计为按值传递,但行为类似于引用。例如:

YAML::Node a = 42;
YAML::Node b = a;
b = 43; // both a and b now contain the value 43

parent[*currentToken]按值返回YAML::Node,因此它不能绑定到非const左值引用,因为这意味着绑定到临时对象。

请删除&,这是设计使用的方式:

YAML::Node child = parent[*currentToken];

在第一行:

bool ReadParameter(YAML::Node parent,