可以传入参数std :: bind?
我试试:
主要代码:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <random>
#include <typeinfo>
#include <functional>
#include <type_traits>
// config random generator
std::default_random_engine generator;
const char *sequence = "rgbn";
const int sizeChar = strlen( sequence)-1;
std::uniform_int_distribution<int> distribution (0, sizeChar);
auto dice = std::bind ( distribution, generator);
std::cout << "type of dice : " << typeid(dice).name() << std::endl;
// i try that
std::bind< std::uniform_int_distribution<int>( std::default_random_engine) > dice2 = std::bind ( distribution, generator);
// or that
std::bind< std::uniform_int_distribution<int>, std::default_random_engine > dice3 = std::bind ( distribution, generator);
Map bibi;
//in a loop
randomInitializationGenerator(bibi, sequence, dice);
汽车的类型是:
St5_BindIFSt24uniform_int_distributionIiESt26linear_congruential_engineIjLj16807ELj0ELj2147483647EEEE
功能代码:
void randomInitializationGenerator( Map& aremplir, const char* sequence,
//std::bind< std::uniform_int_distribution<int>( std::default_random_engine) > dice)
std::function< std::uniform_int_distribution<int>( std::default_random_engine) > dice )
//std::function< std::uniform_int_distribution<int>, std::default_random_engine > dice )
{
const int* sizer = aremplir.size();
const unsigned int finRow = sizer[1];
const unsigned int finCol = sizer[0];
for(unsigned int y = 0; y < finCol; ++y)
for(unsigned int x = 0; x < finRow; ++x)
aremplir.set(y, x, sequence[ dice()]);
}
在randomInitializationGenerator
param中没有好的类型,因此如果我尝试使用dice()
,则无法auto dice = std::bind ( distribution, generator)
。
其他方式
std::bind< std::uniform_int_distribution<int>( std::default_random_engine) > dice2 = std::bind ( distribution, generator);
生成编译错误no the right type
error required from 'typename std::_Bind_helper<std::__or_<std::is_integral<typename std::decay<_Tp>::type>, std::is_enum<typename std::decay<_Tp>::type> >::value, _Func, _BoundArgs ...>::type std::bind(_Func&&, _BoundArgs&& ...) [with _Func = std::uniform_int_distribution<int>(std::linear_congruential_engine<unsigned int, 16807u, 0u, 2147483647u>); _BoundArgs = {}; typename std::_Bind_helper<std::__or_<std::is_integral<typename std::decay<_Tp>::type>, std::is_enum<typename std::decay<_Tp>::type> >::value, _Func, _BoundArgs ...>::type = std::_Bind<std::uniform_int_distribution<int> (*())(std::linear_congruential_engine<unsigned int, 16807u, 0u, 2147483647u>)>]'
最好的问候。
答案 0 :(得分:0)
没有&#34;一个绑定&#34 ;;它是一个功能,而不是一个对象。
如果您确实需要存储结果,请存储std::function
:
std::function<int()> dice3 = std::bind(distribution, generator);
请注意,dice3
没有参数,因为generator
已经绑定到distribution
,并且没有任何事情要做(毕竟,这是完整的一点! )。