c++ - C++ 在循环中生成新随机数的问题/出现提示时
问题描述
我几乎完成了这个基本的骰子程序,但很难找出我对 srand 做错了什么。它应该为用户随机掷 2 个骰子,他们可以选择保留他们掷的 (K) 或再次掷 (R)。我希望程序在每次用户输入“R”时为掷骰子输出新的“随机”数字,当被问及是否要再次滚动时,但相同的数字会继续显示。
该程序的其余部分似乎运行良好——用户与计算机对战。当用户在要求保留或重新滚动时输入“R”时,如何为 roll1 和 roll2 生成新数字?
#include <stdlib.h> /// need this for srand() -- for random numbers
#include <time.h> /// need this for time() -- time
#include <iostream> /// need this for cout<< and cin>>
using namespace std; /// need this for cout<< and cin>>
int main()
{
int iseed = (int)time(0);
srand(iseed);
cout << "Beat the computer! \n";
int roll1 = 1+rand()%6; /// make a random number for die # 1 for user
int roll2 = 1+rand()%6; /// make a random number for die # 2 for user
int UserRoll = roll1 + roll2; /// totals the sum of die 1 and die 2 for the user
char keep;
do
{
cout << "You rolled a " << roll1 << " and a " << roll2 << " for a total of: " << UserRoll << "\n";
cout << "\n";
do
{
cout << "Would you like to keep this total, or roll again? \n";
cout << "\n";
cout << "Enter \"K\" for keep and \"R\" for roll again: \n";
cin >> keep;
if (keep != 'K' && keep != 'R')
{
cout << "That is not a valid choice. Please choose Y to keep your total or N to roll again. " << endl;
cout << "\n";
}
} while(keep != 'K' && keep != 'R');
if (keep == 'R')
{
cout << "You chose R--let's roll again. \n";
}
else
{
cout << "Great! Your total is " << UserRoll << "\n";
}
} while (keep == 'R');
int roll3 = 1+rand()%6; /// make a random number for die # 1 for computer
int roll4 = 1+rand()%6; /// make a random number for die # 2 for computer
int ComputerRoll = roll3 + roll4; /// totals the sum of die 1 and die 2 for the computer
cout << "The computer rolled a " << roll3 << " and a " << roll4 << " for a total of: " << ComputerRoll << "\n";
cout << "\n";
if (ComputerRoll < UserRoll)
{
cout << "Congratulations! You won! \n";
}
if (ComputerRoll > UserRoll)
{
cout << "Sorry. You lose. \n";
}
if (ComputerRoll == UserRoll)
{
cout << "It's a tie. \n";
}
return 0;
}
/*
SAMPLE RUNS:
------------
Beat the computer!
You rolled a 4 and a 6 for a total of: 10
Would you like to keep this total, or roll again?
Enter "K" for keep and "R" for roll again:
R
You chose R--let's roll again.
You rolled a 4 and a 6 for a total of: 10
Would you like to keep this total, or roll again?
Enter "K" for keep and "R" for roll again:
K
Great! Your total is 10
The computer rolled a 4 and a 6 for a total of: 10
It's a tie.
Process returned 0 (0x0) execution time : 8.763 s
Press any key to continue.
*/
解决方案
您只滚动一次并再次使用该滚动的结果。如果用户再次按 R。它应该再次滚动以更新新数字。为此,只需添加
void roll(){
roll1 = 1+rand()%6; /// make a random number for die # 1 for user
roll2 = 1+rand()%6; /// make a random number for die # 2 for user
UserRoll = roll1 + roll2; /// totals the sum of die 1 and die 2 for the user
}
记住:roll1,roll2 和 UserRoll 应该在 roll() 函数之前全局声明
就在 main() 方法之前,并在用户按下 R 以及之后调用它
cout << "Beat the computer! \n";
并将其从主要方法中删除。这就是你需要做的一切
推荐阅读
- spring - Spring Boot 2 - 将 Mono 转换为 rx.Observable?
- powershell - 使用 Cyberduck CLI for OneDrive 时找不到文件错误
- teamcity - 一个构建代理,用于多个 repo
- php - 用户批准我的 Facebook 应用后无法收到用户的电子邮件
- node.js - Angular Universal 应用程序无法向我的 Heroku 节点 API 发出发布请求
- python - 通过多列熊猫解析
- react-native - 我可以在单个 this.state 中使用两个不同的变量吗?
- sql - SQL Oracle - 条件旁边列上的值的where子句
- video - ffmpeg avcodec_receive_packet 返回 -11
- mongodb - Meteor 预填充集合