C++ 类型不提供调用运算符
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type does not provide a call operator
提问by Julien Chien
I have this function, order, which returns vector<Node*>
我有这个函数order，它返回vector<Node*>
vector<Node*> order(vector<string> nodes, vector<pair<string, string>> dependencies) {
Graph graph = buildGraph(nodes, dependencies);
vector<Node*> order = buildOrder(graph.getNodes());
return order;
}
and I call it like this:
我这样称呼它：
vector<Node*> order2 = order(nodes, deps);
However, the compiler gives this error:
但是，编译器给出了这个错误：
error: type 'std::__1::vector<Node *, std::__1::allocator<Node *> >' does not provide a call operator
vector<Node*> order2 = order(nodes, deps);
^~~~~
1 error generated.
What is going wrong? 'std::__1::vector<Node *, std::__1::allocator<Node *> >'seems to suggest that there is a vector<Node*, <Node*>>or something going on, but I can't seem to figure this out.
出了什么问题？'std::__1::vector<Node *, std::__1::allocator<Node *> >'似乎暗示有vector<Node*, <Node*>>什么事情正在发生，但我似乎无法弄清楚这一点。
采纳答案by Ami Tavory
It's a bit hard to tell without your posting more complete code, but consider the following:
如果不发布更完整的代码，有点难以判断，但请考虑以下几点：
int order(int j, int k)
{
return 3;
}
int main(int argc, char *argv[])
{
char order;
// order(2, 3);
}
This code builds fine. However, uncommenting
此代码构建良好。但是，取消注释
// order(2, 3);
causes it to fail, as within main, orderis a character, not a function. From the error message, it looks like you might have some similar problem.
导致它失败，因为内main，order是一个字符，而不是功能。从错误消息来看，您可能遇到了一些类似的问题。
