int main() { auto Square = [] (double a) { return a * a; }; auto Inc = [] (double a) { return a + 1; }; cout << combine<decltype(Square),decltype(Inc),int>(Square,Inc)(3) << endl; cout << combine<decltype(Inc),decltype(Square),double>(Inc,Square)(2.5) << endl; return 0; } 请利用模板函数补全代码使其输出169和10.75

以下是补全了的代码：

#include <iostream>
#include <functional>

using namespace std;

template <typename F, typename G, typename T>
auto combine(F f, G g) {
    return [=](T x) { return f(f(x) + g(x)); };
}

int main() {
    auto Square = [](double a) { return a * a; };
    auto Inc = [](double a) { return a + 1; };
    cout << combine<decltype(Square),decltype(Inc),int>(Square,Inc)(3) << endl;  // 输出169
    cout << combine<decltype(Inc),decltype(Square),double>(Inc,Square)(2.5) << endl;  // 输出10.75
    return 0;
}

在这个实现中，我们使用了C++11中的lambda表达式来定义函数Square和Inc。然后，我们使用combine函数模板将它们组合成一个新的函数。在调用combine函数时，我们需要显式指定函数Square和Inc的类型，以及新函数的输入类型（即T）。最后，我们分别调用新函数，输出结果169和10.75。

相关问题

#include <iostream> using namespace std; // 在此处补充你的代码 int main() { auto Square = [] (double a) { return a * a; }; auto Inc = [] (double a) { return a + 1; }; cout << combine<decltype(Square),decltype(Inc),int>(Square,Inc)(3) << endl; cout << combine<decltype(Inc),decltype(Square),double>(Inc,Square)(2.5) << endl; return 0; }

这段代码使用了 lambda 表达式和函数模板。其中，lambda 表达式是一个匿名函数，可以在代码中定义并直接使用，具有方便、简洁等特点。函数模板是一种泛型编程的方法，可以定义一种通用的函数，用于处理不同类型的数据。

在这段代码中，首先定义了两个 lambda 表达式 Square 和 Inc，分别用于计算一个数的平方和加一。接着，使用函数模板 combine，将这两个 lambda 表达式组合在一起，组合的方式由模板参数指定，返回一个新的 lambda 表达式。最后，调用这个新的 lambda 表达式，得到组合函数的结果。

具体实现见下方：

#include <iostream>
using namespace std;

template<typename T1, typename T2, typename T3>
auto combine(T1 f1, T2 f2) {
    return [=](T3 x) { return f1(f2(x)); };
}

int main() {
    auto Square = [](double a) { return a * a; };
    auto Inc = [](double a) { return a + 1; };
    cout << combine<decltype(Square),decltype(Inc),int>(Square,Inc)(3) << endl;
    cout << combine<decltype(Inc),decltype(Square),double>(Inc,Square)(2.5) << endl;
    return 0;
}

输出结果为：

16
7.25

Please use C language and pointers to define a function to compute the integral of any unary function.

Here is an example implementation of a function to compute the integral of any unary function using C language and pointers:

#include <stdio.h>

double integral(double (*f)(double), double a, double b, int n) {
    double h = (b - a) / n;  // width of each subinterval
    double sum = 0.0;  // initialize the sum
    int i;
    for (i = 0; i < n; i++) {
        double x1 = a + i * h;
        double x2 = a + (i + 1) * h;
        double y1 = (*f)(x1);
        double y2 = (*f)(x2);
        double area = (y1 + y2) * h / 2.0;
        sum += area;
    }
    return sum;
}

// example usage
double square(double x) {
    return x * x;
}

int main() {
    double a = 0.0;
    double b = 1.0;
    int n = 1000;
    double result = integral(&square, a, b, n);
    printf("Integral of x^2 from %g to %g with %d subintervals: %g\n", a, b, n, result);
    return 0;
}

In this example, the `integral` function takes four arguments: a pointer to the unary function to integrate (`f`), the lower and upper bounds of integration (`a` and `b`, respectively), and the number of subintervals to use in the approximation (`n`). The function then computes the width of each subinterval (`h`), initializes the sum to 0, and iterates over the subintervals, computing the area under the curve for each and adding it to the sum. The final result is the sum of all the subinterval areas, which is returned by the function.

To demonstrate the usage of this function, the `square` function is defined as an example of a unary function to integrate, and is passed as a pointer to the `integral` function along with the bounds of integration and number of subintervals. The result is printed to the console.