Question

我试图在同一行打印出标记和字母等级，但我无法在cout语句中调用void函数，有没有办法做到这一点？另外我理解我可以在cout语句之后的下一行调用它，但我需要它们在同一行上打印。

void printLetterGrade(float mark)
{
    float grade = mark;

    if (grade >= 90)
        cout << "Your Letter Grade is A+" << endl;

    else if ((grade >= 85) && (grade <= 89))
        cout << "Your Letter Grade is A" << endl;

    else 
        cout << "fail" << endl;

}

float calculateClassStats(float marks[], int length)
{
    const int arrayCount = length;
    for (int x = 0; x < arrayCount; x++){
        char letter = printLetterGrade(marks[x]);
        cout << marks[x] << letter << endl;
    }
    return 0;
}

Answer 1

您需要更改printLetterGrade，以便返回某些内容。例如。

const char *printLetterGrade(float mark)
{
   if ( ... )
      return "A+";
   if ( ... )
      return "A";
   ...
}

（当然，此时，您可能还希望将该功能称为其他内容）

Answer 2

在同一功能中打印

void printLetterGrade(float mark)
{
    float grade = mark;

    if (grade >= 90)
        cout <<mark<< "Your Letter Grade is A+" << endl;

    else if ((grade >= 85) && (grade <= 89))
        cout <<mark<< "Your Letter Grade is A" << endl;

    else 
        cout << "fail" << endl;

}

float calculateClassStats(float marks[], int length)
{
    const int arrayCount = length;
    for (int x = 0; x < arrayCount; x++){
        printLetterGrade(marks[x]);            
    }
    return 0;
}

Answer 3

如果您想在同一行打印某些内容，请不要打印endl：

void printLetterGrade(float mark)
{
    float grade = mark;

    if (grade >= 90)
        cout << "Your Letter Grade is A+" << endl;

    else if ((grade >= 85) && (grade <= 89))
        cout << "Your Letter Grade is A" << endl;

    else 
        cout << "fail" << endl;

}

float calculateClassStats(float marks[], int length)
{
    const int arrayCount = length;
    for(int x = 0; x < arrayCount; x++) {

        cout << marks[x] << " "; // don't print endl, just a space
        printLetterGrade(marks[x]); // goes on same line
    }
    return 0;
}

Answer 4

您无法GET / GET /favicon.ico cout函数的结果或将void转换为void上要使用的内容，因为没有任何内容被返回。< / p>

如果您需要的只是函数cout的副作用，并且副作用恰好发生在printLetterGrade序列中的某个点，您可以将其返回更改为一些微不足道的行为不打印，

例如，您可以将返回类型从cout更改为void或char *，并在您拥有string语句的任何地方返回空字符串""（如果有的话。

Answer 5

Click Element

只需将返回类型更改为float calculateClassStats(float marks[], int length) { const int arrayCount = length; for (int x = 0; x < arrayCount; x++) { cout << marks[x] << printLetterGrade(marks[x]) << endl; } return 0; }，然后将string更改为cout <<。

return

输出

string printLetterGrade(float mark)
{
    float grade = mark;

    if (grade >= 90)
        return " Your Letter Grade is A+";

    else if ((grade >= 85) && (grade <= 89))
        return " Your Letter Grade is A";

    else 
        return " Fail";

}

Answer 6

将您的cout语句修改为

 cout<<printLetterGrade(marks[x])<<marks[x];

同时删除char letter = printLetterGrade(marks[x]);行，似乎没必要。

编辑：修改后的代码将是：

void printLetterGrade(float mark)
{
float grade = mark;

if (grade >= 90)
    cout << "Your Letter Grade is A+" << endl;

else if ((grade >= 85) && (grade <= 89))
    cout << "Your Letter Grade is A" << endl;

else 
    cout << "fail" << endl;

}

float calculateClassStats(float marks[], int length)
{
const int arrayCount = length;
for (int x = 0; x < arrayCount; x++){
    cout<<marks[x];
    printLetterGrade(marks[x]); 
}
return 0;
}

如何在cout语句中调用void函数？

6 个答案: