Question

I m 试图在以下一栏中宣布变数:

int main(int argc, char *argv[]) {

    if (argv[3] == string("simple")) {
        Player & player = *get_Simple();
    } else if (argv[3] == string("counting")) {
        Player & player = *get_Counting();
    } else if (argv[3] == string("competitor")) {
        Player & player = *get_Competitor();
    }

    // More code
}

但是,在我试图汇编以下错误时,我会发现:

driver.cpp:38: error: unused variable ‘player’
driver.cpp:40: error: unused variable ‘player’
driver.cpp:42: error: unused variable ‘player’
driver.cpp:45: error: ‘player’ was not declared in this scope

任何想法?

Answer 1

你们的问题是,如果存在障碍,参与者就会分散在每一个领域。

您必须首先宣布发言中的变数。

但是,你可以援引这一点,因为你必须先设定参考权。

相反,你可能希望这样做:

int main(int argc, char *argv[]) {

    Player * pPlayer = NULL;
    if (argv[3] == string("simple")) {
        pPlayer = get_Simple();
    } else if (argv[3] == string("counting")) {
        pPlayer = get_Counting();
    } else if (argv[3] == string("competitor")) {
        pPlayer = get_Competitor();
    }

    //Then if you really want to...
    Player &player = *pPlayer;

}

Answer 2

另一些缔约方则提出了要点。但是,有条件的经营者也可以使用。

Player & player = argv[3] == string("simple") ? get_Simple()
                : argv[3] == string("counting") ? get_Counting() 
                : get_Competitor();

Answer 3

如果你在限定范围以<代码>{>}的范围之内出现静态变量,那么在范围结束时,变数将不复存在。

相反:

int main(int argc, char *argv[]) {

    // TODO: validate argc and argv here
    if (argc < 3) {
        printf("error: not enough arguments
");
        exit(1);
    }

    Player* player_ptr = NULL;
    if (argv[3] == string("simple")) {
        player_ptr = get_Simple();
    } else if (argv[3] == string("counting")) {
        player_ptr = get_Counting();
    } else if (argv[3] == string("competitor")) {
        player_ptr = get_Competitor();
    }

    if (!player_ptr) {
        printf("error: invalid argument %s
", argv[3]);
        exit(1);
    }

    Player& player = *player_ptr;

    // More code
}

Answer 4

您在三个不同领域分别宣布了三个不同的<条码>字句变量,错误的电文确切说它意味着什么。

您需要宣布在<代码>if之外的一个单一参与者变量,并分配结果。这是一种骗局,因为参与者是参考者,必须一度启动。

您可以将if -statement in a function (say GetPlayer(>) 退回标的,然后与*GetPlayer(签名)一起启动角色。

Answer 5

In

if (argv[3] == string("simple")) {
    Player & player = *get_Simple();
}

唯一变量存在于<代码>{}s之间。当你进入<代码>>><>>>>>时,该变量没有使用,将予以抛弃,从未使用过。

Answer 6

#include <map>

int main(int argc, char **argv)
{
    typedef std::map<std::string, Player*(*)()> lookup;
    lookup mapping;

    mapping["simple"] = get_Simple;
    mapping["counting"] = get_Counting;
    mapping["competitor"] = get_Competitor;

    lookup::const_iterator it = mapping.find(argv[3]);
    if (it == mapping.end())
    {
        std::cout << "illegal argument
";
    }
    else
    {
        Player& player = *it->second();
        // more code
    }
}

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