为什么这一行给我一个c2120错误

Question

} while (mf_bingo(c) != 5);

这是我不断收到错误的那一行.我尝试在mf_bingo(c)之前放一个空格并取出(c)

编辑:这是完整的循环

do

    {
        char yesNo;
        int c, d;
        if (yesNo == 'y' || yesNo == 'Y')
        {
            cout << "Player 1 how many numbers are on your card" << endl;
            cin >> c;
            mf_bingo(c);
        }

        else if (yesNo == 'n' || yesNo == 'N')
        {
            cout << "Sorry player 1" << endl;
        }
        cout << "Player 2 do you see this number on your card." << endl;
        cin >> yesNo;
        if (yesNo == 'y' || yesNo == 'Y')
        {
            cout << "Player 2 how many numbers are on your card" << endl;
            cin >> c;
            mf_bingo(c);
        }

        else if (yesNo == 'n' || yesNo == 'N')
        {
            cout << "Sorry player 2" << endl;
        }
        cout << "Press 1 to continue generating random numbers." << endl;
        cin >> d;
        mf_numbers(d);
    } while (mf_bingo(c) != 5);

这是mf_numbers

void mf_numbers(int d)
{
    int xRan;
    srand(time(0));
    xRan = rand() % 50 + 1;
    cout << xRan << endl;

}

这是mf_bingo

void mf_bingo(int c)
{
    if (c == 0)
        cout << "You need to match get 5 more matching numbers to win." << endl;
    else if (c == 1)
    {
        cout << "You only need 4 more matching numbers to win" << endl;
    }
    else if (c == 2)
    {
        cout << "You only need 3 more numbers to win" << endl;
    }
    else if (c == 3)
    {
        cout << "You only need 2 more numbers to win" << endl;
    }
    else if (c == 4)
    {
        cout << "You only need 1 more number to win" << endl;
    }
    else if (c == 5)
    {
        cout << "You win!!!" << endl;
    }
}

Answer 1

问题是声明无效mf_bingo().你必须让这个函数返回一个int.

您可以使用以下代码重现此问题:

void mf_bingo(int c) {  // void doesn't return anything
}

int main() {
    do { } while(mf_bingo(c) != 5);
}

错误C2120解释说,当函数返回时void,即没有,您无法将其与整数进行比较.通常情况下,intellisence应该已经将mf_bingo强调为潜在错误,并提供更准确的描述.

您可以轻松纠正此问题:

int mf_bingo(int c) {  // here we return something 
    ...   // same logic as before 
    return c; 
}