“简单”问题,是否可以在调用期望的函数时显式使用默认参数?类似的东西:
void function(int x, int y = 2, int z = 3)
{
// prints x, y and z
}
function(10, default, 13); // won't compile of course
// would return x = 10, y = 2 and z = 3
谢谢
答案 0 :(得分:3)
不是标准C ++,但您可以查看示例boost parameter library.
小例子:
#include <iostream>
#include <boost/parameter.hpp>
#include <boost/parameter/preprocessor.hpp>
BOOST_PARAMETER_NAME(x)
BOOST_PARAMETER_NAME(y)
BOOST_PARAMETER_NAME(z)
namespace tag { struct x; }
BOOST_PARAMETER_FUNCTION(
(void),
function,
tag,
(required (x, (int)))
(optional
(y, (int), 2)
(z, (int), 3)
)
)
{
std::cout << "Called with x = " << x << " y = "
<< y << " z = " << z << std::endl;
}
int main()
{
function(1, _z = 5);
function(1, _y = 8);
}
答案 1 :(得分:0)
不,目前在C ++中是不可能的。