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我正在尝试创建一个程序，该程序将接受用户输入（标记）并将这些标记的平均值输出到字母等级。我相信我已经成功让程序成功计算平均值，但是它不会输出正确的字母等级？

有什么建议么？编辑：也许它在计算中，因为我的输出总是将百分比作为“F”级......

// Used iomanip to display averages for possible three digit or greater answers.
#include <iostream>
#include <iomanip>
using namespace std;

// Set up my program using only main()
int main()
{
    //Created the variable 'mark' so the user input can be stored and tested in the do-while condition.
    //Variable 'sum' and 'average' to calculate mark and then be tested against letter grades.
    //'counter' used to keep track of number of terms being averaged.
    double mark=0, sum=0, average=0;
    int counter=-1;

    // Do-while statement to test the loop. Gets input from user, and tests whether it is a a valid letter grade.
    // Marks below 0 or above 100 will test true and loop back re-promting the user for another mark.
    // If tested condition is false, then if statements then test the mark and output the letter grade.

    cout << "Please enter marks to calculate average, and enter -1 to stop and find letter equivalence. " << endl;

    do
    {
        if ( mark < 0 || mark > 100 )
        {
            cout << "Entered mark is invalid. Please try again, or enter -1 to stop and find letter equivalence. " << endl;
            break;
        }

        sum = sum + mark;
        counter++;
        cin >> mark;

    }
    while ( mark != -1 );

    average = sum / counter ;

    //What happens when above statement is false...
    if (mark >= 90)
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of A+ ";
        }
    else if (mark >=80)
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of A ";
        }
    else if (mark >=70)
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of B ";
        }
    else if (mark >=60)
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of C ";
        }
    else if (mark >=50)
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of D ";
        }
    else
        {
            cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of F ";
        }
return 0;
}
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3 回答 3

1

需要考虑的几件事：

  1. 你的成绩分配应该考虑average和不marks。如果您使用标记，那么最后输入的值（即-1退出 do while 循环）将用于评分，这将始终导致 F 级。

  2. 在循环后添加endl到所有cout语句。do-while输出流可能已被缓冲，这可能导致 cout 语句不会出现在您的监视器上。endl将刷新输出流。

例如。
cout << "The average of "<<setprecision(2)<<average<<"% is equivalent to a letter grade of A+ " << endl;

于 2011-10-28T03:07:40.020 回答
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仔细检查你的if陈述。我敢打赌，您总是看到正确的平均值相当于 F。

于 2011-10-28T03:08:16.980 回答
0

在循环结束时，mark总是-1. 所以所有的测试都失败了，你通过else子句，并输出一个 F。修复是测试average，而不是mark。

于 2011-10-28T03:19:24.567 回答