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练习了一些 C++,我从一本书中遇到了一些代码。对我来说,使用带有 break 接缝的 if 语句有点不必要:

if (!(cin >> dstep)) 中断;

这对我来说有点骇人听闻,只需使用没有 if 或 break 的 cin >> dstep 即可正常运行程序。有什么想法吗?

                    int main()
                {
                    using namespace VECTOR;

                    srand(time(0));     //seed random-number generator
                    double direction;
                    Vector step;        //creates default object
                    Vector result(0.0, 0.0);    //
                    unsigned long steps = 0;
                    double target;
                    double dstep;

                    cout << "Enter target distance (q to quit): ";

                    while (cin >> target)
                    {
                        cout << "Enter step length: ";
                        if (!(cin >> dstep))    //if NOT inputing into dstep THEN break/means if INPUTING is TRUE, keep going and don't break out of loop
                            break; 
                      //cin >> dstep  // why not just use this?
                        while (result.magval() < target)
                        {
                            direction = rand() % 360;
                            step.set(dstep, direction, 'p'); //sets the values dstep and direction based on the form; in this case 'p'
                            result = result + step;
                            steps++;
                        }
                        cout << "After " << steps << " steps, the subject "
                            "has the following location:\n";
                        cout << result << endl;
                        result.polar_mode();
                        cout << " or\n" << result << endl;
                        cout << "Average outward distance per step = "
                            << result.magval() / steps << endl;
                        steps = 0;
                        result.set(0.0, 0.0);
                        cout << "Enter target distance (q to quit): ";
                    }
                    cout << "Bye!\n";

                    cin.get();
                    return 0;
                }
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1 回答 1

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您必须启用异常或检查每个流输入操作,否则当输入格式错误时,您可能会陷入无限循环或得到错误结果。

于 2012-05-16T03:41:37.143 回答