Question

这是我的作业规范的最后一部分，我无法弄清楚：

使用默认情况来捕捉＆＃39;各种等级。不要将0到59之间的每个数字列为单独的案例！

这是我的代码，我的默认需要使用上面的说明考虑0到59之间的值。

感谢您提供任何帮助/建议。

double percentage;
int grade;

cout << "Enter your grade percentage " <<endl;
cin >> percentage;

if (percentage < 0 || percentage > 100 )
    cout << "Invalid entry " << endl;

grade = (int)round(percentage);

cout << fixed << setprecision(2);

switch(grade)

{
    case 100:
    case 99:
    case 98:
    case 97:
    case 96:
    case 95:
    case 94:
    case 93:
    case 92:
    case 91:
    case 90:
        cout << "Percentage: " << percentage << "% " << "Grade: A " << "Points: 4.00"; 
        break;
    case 89:
    case 88:
    case 87:
    case 86:
    case 85:
    case 84:
    case 83:
    case 82:
    case 81:
    case 80:
        cout << "Percentage: " << percentage << "% " << "Grade: B " << "Points: 3.00";
        break;
    case 79:
    case 78:
    case 77:
    case 76:
    case 75:
    case 74:
    case 73:
    case 72:
    case 71:
    case 70:
        cout << "Percentage: " << percentage << "% " << "Grade: C " << "Points: 2.00";
        break;
    case 69:
    case 68:
    case 67:
    case 66:
    case 65:
    case 64:
    case 63:
    case 62:
    case 61:
    case 60:
        cout << "Percentage: " << percentage << "% " << "Grade: D " << "Points: 1.00";
        break;
    default:
        cout << "Percentage: " << percentage << "% " << "Grade: F " << "Points: 0.00";
        break;
}
return 0;

}

Answer 1

与@ dvix的评论一致，如果您知道成绩将在0到100之间，则可以通过以下方式缩短代码：

switch( grade/10 ) {

    case 10:
    case 9:
        ... Do something ...
        break;
    case 8:
    case 7:
        ... Do something else...
        break
   ....

}

并在最后使用默认值，节省了几个案例。

话虽如此，if可能是正常处理这种情况的惯用方式。

切换语句默认范围？

1 个答案: