将lambda传递到功能模板中 [英] Passing a lambda into a function template

问题描述

我正在学习C ++，并且正在尝试实现一个二进制搜索功能，该功能查找谓词所持有的第一个元素.函数的第一个参数是向量，第二个参数是对给定元素的谓词求值的函数.二进制搜索功能如下:

I'm learning C++, and I'm trying to implement a binary search function that finds the first element for which a predicate holds. The function's first argument is a vector and the second argument is a function that evaluates the predicate for a given element. The binary search function looks like this:

template <typename T> int binsearch(const std::vector<T> &ts, bool (*predicate)(T)) {
    ...
}

如果像这样使用，它将按预期工作:

This works as expected if used like this:

bool gte(int x) {
    return x >= 5;
}

int main(int argc, char** argv) {
    std::vector<int> a = {1, 2, 3};
    binsearch(a, gte);
    return 0;
}

但是，如果我使用lambda函数作为谓词，则会出现编译器错误:

But if I use a lambda function as a predicate, I get a compiler error:

search-for-a-range.cpp:20:5: error: no matching function for call to 'binsearch'
    binsearch(a, [](int e) -> bool { return e >= 5; });
    ^~~~~~~~~
search-for-a-range.cpp:6:27: note: candidate template ignored: could not match 'bool (*)(T)' against '(lambda at
      search-for-a-range.cpp:20:18)'
template <typename T> int binsearch(const std::vector<T> &ts,
                          ^
1 error generated.

以上错误是由

binsearch(a, [](int e) -> bool { return e >= 5; });

怎么了?为什么编译器不相信我的lambda具有正确的类型?

What's wrong? Why is the compiler not convinced that my lambda has the right type?

推荐答案

您的函数binsearch将函数指针作为参数. lambda 和函数指针是不同的类型:lambda可以视为实现operator()的结构的实例.

Your function binsearch takes a function pointer as argument. A lambda and a function pointer are different types: a lambda may be considered as an instance of a struct implementing operator().

请注意，无状态lambda(不捕获任何变量的lambda)可以隐式转换为函数指针.由于模板替换，此处的隐式转换不起作用:

Note that stateless lambdas (lambdas that don't capture any variable) are implicitly convertible to function pointer. Here the implicit conversion doesn't work because of template substitution:

#include <iostream>

template <typename T>
void call_predicate(const T& v, void (*predicate)(T)) {
    std::cout << "template" << std::endl;
    predicate(v);
}

void call_predicate(const int& v, void (*predicate)(int)) {
    std::cout << "overload" << std::endl;
    predicate(v);
}

void foo(double v) {
    std::cout << v << std::endl;
}

int main() {
    // compiles and calls template function
    call_predicate(42.0, foo);

    // compiles and calls overload with implicit conversion
    call_predicate(42, [](int v){std::cout << v << std::endl;});

    // doesn't compile because template substitution fails
    //call_predicate(42.0, [](double v){std::cout << v << std::endl;});

    // compiles and calls template function through explicit instantiation
    call_predicate<double>(42.0, [](double v){std::cout << v << std::endl;});
}

---

您应该使函数binsearch更通用，例如:

---

You should make your function binsearch more generic, something like:

template <typename T, typename Predicate>
T binsearch(const std::vector<T> &ts, Predicate p) {

    // usage

    for(auto& t : ts)
    {
        if(p(t)) return t;
    }

    // default value if p always returned false

    return T{};
}

从标准算法库中汲取灵感.

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