英文:
C++ Lambda Local Variable Capturing Error
问题
我正在尝试使用lambda表达式使inorderTraversal函数可重用，但在尝试捕获printStandartDeviation函数的本地参数时出现以下错误：无法从'lambda at [...]/AVLTree.cpp:[...]'转换为'void (*)(AVLTree::TreeNode *)&'。
它会在我将本地变量设为静态或全局变量时修复，但实际上我不想这样做。
我也愿意接受替代方案，而不是使用lambda。
如果您能帮助我，我将不胜感激。
英文:
I am trying to use a lambda expression to make inorderTraversal function reusable, and I am having the following error while trying to capture the local parameters of printStandartDeviation function: No viable conversion from '(lambda at [...]/AVLTree.cpp:[...])' to 'void (*)(AVLTree::TreeNode *)'.
void AVLTree::printStandartDeviation() {
int totalWordCount = 0, totalWordFrequencies = 0;
inorderTraversal(root, [&totalWordCount, &totalWordFrequencies](TreeNode *tNode) {
totalWordCount++;
totalWordFrequencies += tNode->frequency;
});
}
void AVLTree::inorderTraversal(TreeNode *tNode, void (*visit)(TreeNode *)) {
if (tNode == nullptr) {
return;
} else {
inorderTraversal(tNode->left, visit);
visit(tNode);
inorderTraversal(tNode->right, visit);
}
}
It fixes if I make my local variables static or make them global, but I do not want to do either actually.
I am also open to alternative solutions as well instead of using lambda.
I would be grateful if you can help.
答案1
得分: 2
一个Lambda表达式不是一个函数指针。一个非捕获Lambda具有转换为函数指针的能力，但带有捕获的Lambda不具备这种能力。这在某种程度上类似于自由函数指针和成员函数指针之间的差异，仅有成员函数指针本身几乎没有用处。你需要一个对象来调用成员函数。在这里，你还需要数据，也就是捕获的变量，来调用Lambda表达式，它不仅仅是一个函数。
你不需要一个函数指针。只需将Lambda表达式原样传递，由于Lambda的类型没有名称，可以将这个方法设为模板以使其工作：
template <typename F>
void AVLTree::inorderTraversal(TreeNode *tNode, F&& visit) {
// 所有内容相同
}
英文:
A lambda is not a function pointer. A non-capturing lambda has a conversion to function pointer but one with captures not. It is somewhat similar to the difference between a pointer to free function and pointer to member function where the pointer to member function alone is of little use. You need an object to call a member function. Here you also need the data, the captured variables, to call the lambda, its not just a function.
You do not need a function pointer. Simply pass the lambda as it is, and because the lambdas type has no name, make the method a template to make this work:
template <typename F>
void AVLTree::inorderTraversal(TreeNode *tNode, F&& visit) {
// all the same
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论