Question

为什么编译器会在此代码的最后warning: Expression results unused语句中发出else？

PowerballLottery::WinningPossibility PowerballLottery::checkTicket(PowerballTicket ticket)
{
    int count = 0;
    if (ticket.getBall1() == getball1() || ticket.getBall1() == getball2() || ticket.getBall1() == getball3() || ticket.getBall1() == getball4() || ticket.getBall1() == getball5())
        count++;
    if (ticket.getBall2() == getball1() || ticket.getBall2() == getball2() || ticket.getBall2() == getball3() || ticket.getBall2() == getball4() || ticket.getBall2() == getball5())
        count++;
    if (ticket.getBall3() == getball1() || ticket.getBall3() == getball2() || ticket.getBall3() == getball3() || ticket.getBall3() == getball4() || ticket.getBall3() == getball5())
        count++;
    if (ticket.getBall4() == getball1() || ticket.getBall4() == getball2() || ticket.getBall4() == getball3() || ticket.getBall4() == getball4() || ticket.getBall4() == getball5())
        count++;
    if (ticket.getBall5() == getball1() || ticket.getBall5() == getball2() || ticket.getBall5() == getball3() || ticket.getBall5() == getball4() || ticket.getBall5() == getball5())
        count++;
    bool match = false;
    if (ticket.getPowerball() == getpowerball())
        match = true;
    if ((count == 0) && (match == false))
        return PowerballLottery::WinningPossibility::NOTWINNING;
    else if ((count == 0) && (match == true))
        return PowerballLottery::WinningPossibility::POWERBALL;
    else if ((count == 1) && (match == true))
        return PowerballLottery::WinningPossibility::ONEPLUSPOWERBALL;
    else if ((count == 2) && (match == true))
        return PowerballLottery::WinningPossibility::TWOPLUSPOWERBALL;
    else if ((count == 3) && (match == false))
        return PowerballLottery::WinningPossibility::THREE;
    else if ((count == 3) && (match == true))
        return PowerballLottery::WinningPossibility::THREEPLUSPOWERBALL;
    else if ((count == 4) && (match == false))
        return PowerballLottery::WinningPossibility::FOUR;
    else if ((count == 4) && (match == true))
        return PowerballLottery::WinningPossibility::FOURPLUSPOWERBALL;
    else if ((count == 5) && (match == false))
        return PowerballLottery::WinningPossibility::FIVE;
    else ((count == 5) && (match == true));
        return PowerballLottery::WinningPossibility::FIVEPLUSPOWERBALL;
}

Answer 1

实际上，代码是

if ((count == 0) && (match == false))
....
else if ((count == 5) && (match == false))
    return PowerballLottery::WinningPossibility::FIVE;
else
    ((count == 5) && (match == true)); //<- Expression results unused 

return PowerballLottery::WinningPossibility::FIVEPLUSPOWERBALL;

Answer 2

其他人已经指出了编译器警告你的问题。

现在让我们考虑一下它没有警告你的问题：你的代码要比它有任何好的理由复杂得多。

从本质上讲，用户选择和在抽奖中选择的球（除了力量球）形式集。我们关心的是这两组的交叉点的大小。 C ++标准库为我们提供了一些工具，使这个任务变得更加简单（而且我认为你设计PowerBallTicket类的方式比起帮助更多地阻碍了）：

std::set<int> lottery { 
    getball1(), 
    getball2(), 
    getball3(), 
    getball4(), 
    getball5()
};

std::set<int> tick { 
    ticket.getball1(), 
    ticket.getball2(), 
    ticket.getball3(), 
    ticket.getball4(), 
    ticket.getball5() 
};

std::vector<int> matches;

std::set_intersection(lottery.begin(), lottery.end(), 
                      tick.begin(), tick.end(),
                      std::back_inserter(matches));

这给了我们两者之间的匹配。从那里，我们有转换匹配数量（加上Powerball是否匹配）的问题，以选择返回哪个枚举值。一种简单的方法是使用数组：

WinningPossibility rets[][6] = {
    { NOTWINNING, ONE, TWO, THREE, FOUR, FIVE},
    { POWERBALL, ONEPLUS, TWOPLUS, THREEPLUS, FOURPLUS, FIVEPLUS }
};

有了这些，我们的回报看起来像这样：

return rets[ticket.getpowerball() == getpowerball()][matches.size()];

上面的代码指向了一个改进：代替所有那些getballX()成员函数，您应该只将数据存储在set中，然后直接在这些集合上运行。在这种情况下，看起来你有quasi-classes，其中“封装”失去了很多，并且获得了很少（可能没什么）。

Answer 3

正如@AlexD＆＃39; answer所述，这主要是您如何格式化代码的问题，并且出现了警告，因为

else ((count == 5) && (match == true));

该语句实际上并不用于存储在结果中的任何计算，或基于它做出分支决策。

return PowerballLottery::WinningPossibility::FIVEPLUSPOWERBALL;

语句无论如何都会被执行。

提醒一下着名的apple OpenSSL goto fail; bug：

if ((err = SSLHashSHA1.update(&hashCtx, &signedParams)) != 0)
goto fail;
goto fail;  /* MISTAKE! THIS LINE SHOULD NOT BE HERE */

Answer 4

您可以说else <statement>或else if (<expression>) <statement>;你有没有if的表达式，编译器正在尽力使用它。

警告：表达结果未使用

4 个答案: