如何仅将选定的参数传递给函数？

Question

假设我有一个结构体 X，我想根据某些条件填写所有字段。我想区分要传递给函数的参数，并根据提供的参数创建然后返回该对象。我很好奇是否应该在 C++ 中使用可变参数模板？伪代码。

#include <iostream>

using namespace std;

struct X
{
    string a;
    string b;
    string c;
}

X generateTransaction(const string& e, const string& f, const string& g)
{
    X one;
    if (e)
        one.a = e;
    if (f)
        one.b = f;
    if (g)
        one.c = g;
    return one;
}

int main()
{
    generateTransaction(e = "first");
    generateTransaction(e = "abc", f = "second");
    generateTransaction(g = "third");
    generateTransaction(e = "test", g = "third");
    generateTransaction("m", "n", "o");
    return 0;
}

Answer 1

对于 C++20，您可以使用指定的初始值设定项：

X x1{.a = "first"};
X x2{.a = "abc", .b = "second"};
X x3{.c = "third"};
X x4{.a = "test", .c = "third"};
X x5{"m", "n", "o"};

演示

Answer 2

您可以传递参数 as std::optional，然后检查它是否有value。就像是：

X generateTransaction(const std::optional<std::string> &e, 
                      const std::optional<std::string> &f,
                      const std::optional<std::string> &g)
{
    X one;
    if (e.has_value())
    {
        one.a = e.value();
    }
    if (f.has_value())
    {    
        one.b = f.value();
    }
    if (g.has_value())
    {    
        one.c = g.value();
    }
    return one;
}

int main()
{
    generateTransaction("first", {}, {});
    generateTransaction("abc", "second", {});
    generateTransaction({}, {}, "third");
    generateTransaction("test", {}, "third");
    generateTransaction("m", "n", "o");
    return 0;
}