将 lambda 传递给函数模板 [英] Passing a lambda into a function template

问题描述

我正在学习 C++，我正在尝试实现一个二分查找函数，该函数查找谓词所包含的第一个元素.该函数的第一个参数是一个向量，第二个参数是一个计算给定元素谓词的函数.二分查找函数如下所示:

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)) {
    ...
}

如果这样使用，它会按预期工作:

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，我会得到一个编译器错误:

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 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

你的函数 binsearch 接受一个函数指针作为参数.lambda 和函数指针是不同的类型:可以将 lambda 视为实现 operator() 的结构体实例.

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)可隐式转换为函数指针.由于模板替换，这里隐式转换不起作用:

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;});
}

<小时>

你应该让你的函数 binsearch 更通用，比如:

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.

这篇关于将 lambda 传递给函数模板的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！