我正在制作一个轮盘赌桌,想问一个问题,但也有帮助修复代码。
1,目前我正在使用rand()%37作为0-36号码的随机数生成器。是否有更好的发电机可以提供更随机的数字?
2,我不确定如何根据if语句检查数组中的值。 (查看下面的代码)
int red[18] = {1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36};
int main ()
{
srand(time(NULL));
r = rand() % 3;
cout << r << endl;
for(int i = 0; i<18; i++)
if (r==red[i])
{
cout << "hello" << endl;
break;
}
else
{
cout << "bye" << endl;
break;
}
return 0;
}
当我运行此代码时,即使r(随机数)在数组中为红色,它也会返回。
到woolstar:这是我实际代码的简化版本。我希望生成一个随机数(r),然后根据数组红色检查这个数字,如果r在数组中我希望它被认为是真的并且在这种情况下打印你好
E.g。 r = 14,是14的红色阵列?是的:打印你好
编辑:这是我的实际代码,当运行随机数并查找B和C时它会返回错误说:初始化正在跳过案例标签
// Random number generator
random_device rd;
mt19937 gen (rd());
uniform_int_distribution<>
dis (0, 36);
switch (play_method)
{
case 1: // Color Play
cout << "A - Play Red.\n";
cout << "B - Play Black.\n";
cout << "C - Play Green.\n";
cout << "Which color will you bet on?" << endl;
cin >> color;
switch (color)
{
case 'A': // Red Play
int rred = dis(gen);
auto
it = find(begin(red),end(red), rred);
if(it!=end(red))
{
win_amt = bet_amt*2;
total_amt = total_amt + (win_amt - bet_amt);
cout << "Congratulations you rolled a " << r << " and have won: $" << win_amt << "!" << endl;
cout << "You now have: $" << total_amt << " remaining." << endl;
}
else
{
total_amt = total_amt - bet_amt;
cout << "Unlucky, you rolled a " << r << " and have have lost $" << bet_amt << endl;
cout << "You have $ " << total_amt << " remaining." << endl;
} // Red play if/else end
break; // Break for case A: Red play
case 'B': // Black play
int rblack = dis(gen);
auto
it1 = find(begin(black),end(black), rblack);
if(it1!=end(black))
{
win_amt = bet_amt*2;
total_amt = total_amt + (win_amt - bet_amt);
cout << "Congratulations you rolled a " << r << " and have won: $" << win_amt << "!" << endl;
cout << "You now have: $" << total_amt << " remaining." << endl;
}
else
{
total_amt = total_amt - bet_amt;
cout << "Unlucky, you rolled a " << r << " and have have lost $" << bet_amt << endl;
cout << "You have $ " << total_amt << " remaining." << endl;
} // Black play if/else end
break; // Break for case B: Black play
case 'C':
int rgreen = dis(gen);
auto
it2 = find(begin(green),end(green), rgreen);
if(it1!=end(green))
{
win_amt = bet_amt*37;
total_amt = total_amt + (win_amt - bet_amt);
cout << "Congratulations you rolled a " << r << " and have won: $" << win_amt << "!" << endl;
cout << "You now have: $" << total_amt << " remaining." << endl;
}
else
{
total_amt = total_amt - bet_amt;
cout << "Unlucky, you rolled a " << r << " and have have lost $" << bet_amt << endl;
cout << "You have $ " << total_amt << " remaining." << endl;
} // Green play if/else end */
break;
} // Switch color end
break; // break for case 1: Color play
答案 0 :(得分:3)
1,目前我正在使用rand()%37作为0-36号码的随机数生成器。是否有更好的发电机可以提供更随机的数字?
绝对是的。请考虑使用C ++ 11随机数设施(如果没有C ++ 11,则使用Boost.Random)。
std::random_device rd;
std::mt19937 gen(rd());
std::uniform_int_distribution<> dis { 0, 36 };
//print random number between 0 and 36
std::cout << dis(gen);
请参阅以下参考资料:
http://en.cppreference.com/w/cpp/numeric/random
2,我不确定如何根据if语句检查数组中的值。 (查看下面的代码)
我会使用std::find
#include <random>
#include <iostream>
#include <algorithm>
int red[18] = {1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36};
int main () {
std::random_device rd;
std::mt19937 gen { rd() };
std::uniform_int_distribution<> dis { 0, 36 };
int r=dis(gen);
auto it=std::find(std::begin(red), std::end(red), r);
if(it!=std::end(red)) {
std::cout << "hello\n";
}
else {
std::cout << "hello\n";
}
return 0;
}
请参阅以下参考资料:
http://en.cppreference.com/w/cpp/algorithm/find
希望这足以让人放弃rand()
。
http://channel9.msdn.com/Events/GoingNative/2013/rand-Considered-Harmful
答案 1 :(得分:2)
由于我认为这是主要问题的一个问题,我会忽略(如果这就是这个评论)行
r = rand() % 3;
如果r为1,则打印hello。否则先执行else。这打印再见并打破循环。
for if永远不会超过一个循环迭代,因为if的两个子句中的break语句 - 如果你用这种方式编写循环就更容易看到(我强烈建议你养成更经常使用括号的习惯)
for(int i = 0; i<18; i++)
{
if (r==red[i])
{
cout << "hello" << endl;
}
else
{
cout << "bye" << endl;
}
break;
}
return 0;
基本上就是这个
for(int i = 0; i<18; i++)
{
// do something not affecting flow control
break;
}
return 0;
是
int i =0;
// do something not affecting flow control
我猜这不是你想要的,并建议一个标志可能是最好的方法,在循环完成后留下打印。