Question

我正在尝试将带有预定义数据的结构传递给函数，因此不必对多个结构使用多个函数，但是我可以对多个结构使用1个函数。

结构如下：

typedef struct {
    char* vehicleName;
    char* previewDir;
    char* previewName;
} vehicleSpawner;

vehicleSpawner vehicleSport[] = {{"adder","lgm_default","adder"},{"autarch","",""},{"banshee2","lsc_jan2016","banshee2"},{"bullet","lgm_default","bullet"}};
vehicleSpawner vehicleSportTwo[] = {{"adder","lgm_default","adder"},{"autarch","",""},{"banshee2","lsc_jan2016","banshee2"},{"bullet","lgm_default","bullet"}};
vehicleSpawner vehicleSportThree[] = {{"adder","lgm_default","adder"},{"autarch","",""},{"banshee2","lsc_jan2016","banshee2"},{"bullet","lgm_default","bullet"}};

现在要在结构中显示信息，我必须使用3个独立的函数，如下所示。

void vehicleSport() {
    menu.title("Sport");

    for (int i = 0; i < 3; i++) {
        char* name = UI::_GET_LABEL_TEXT(vehicleSport[i].vehicleName);
        Hash key = GAMEPLAY::GET_HASH_KEY(vehicleSport[i].vehicleName);

        if(!GAMEPLAY::ARE_STRINGS_EQUAL(name, "NULL")) menu.option(name).vehicleSpawn(key).vehiclePreview(vehicleSport[i].previewDir, vehicleSport[i].previewName);
    }
}

void vehicleSport2() {
    menu.title("Sport Two");

    for (int i = 0; i < 3; i++) {
        char* name = UI::_GET_LABEL_TEXT(vehicleSportTwo[i].vehicleName);
        Hash key = GAMEPLAY::GET_HASH_KEY(vehicleSportTwo[i].vehicleName);

        if(!GAMEPLAY::ARE_STRINGS_EQUAL(name, "NULL")) menu.option(name).vehicleSpawn(key).vehiclePreview(vehicleSportTwo[i].previewDir, vehicleSportTwo[i].previewName);
    }
}

void vehicleSport3() {
    menu.title("Sport Three");

    for (int i = 0; i < 3; i++) {
        char* name = UI::_GET_LABEL_TEXT(vehicleSportThree[i].vehicleName);
        Hash key = GAMEPLAY::GET_HASH_KEY(vehicleSportThree[i].vehicleName);

        if(!GAMEPLAY::ARE_STRINGS_EQUAL(name, "NULL")) menu.option(name).vehicleSpawn(key).vehiclePreview(vehicleSportThree[i].previewDir, vehicleSportThree[i].previewName);
    }
}

我要实现的目标是仅使用一个函数代替我所拥有的大量结构的函数，并每次重复相同的功能。< / p>

我尝试过的正在使用其他函数传递它，然后在“动态函数”中像这样访问它：

char* vehicleSpawnerTitle;
struct structurename;

void vehicleSpawnerItemMenu() {
    menu.banner(vehicleSpawnerTitle);

  for (int i = 0; i < sizeof(structurename)/sizeof(structurename[0]); i++) {
        Hash key = GAMEPLAY::GET_HASH_KEY(structurename[i].vehicleName);
        char* name = UI::_GET_LABEL_TEXT(VEHICLE::GET_DISPLAY_NAME_FROM_VEHICLE_MODEL(key));

        if(STREAMING::IS_MODEL_VALID(key) && STREAMING::IS_MODEL_A_VEHICLE(key) && !GAMEPLAY::ARE_STRINGS_EQUAL(name, "NULL"))  menu.option(name).vehicleSpawn(key).vehiclePreview(structurename[i].previewDir, structurename[i].previewName);
  }
}

void setVehicleSpawnerVar(char* title, struct structname) {
    vehicleSpawnerTitle = title;
    structurename = structname;
}

希望有人可以帮助我！

Answer 1

假设您要合并三个功能2 2018 6 1400 2 2017 1 2800 2 2018 3 8800 5 2018 2 1000 5 2018 1 950 3 2019 1 2800 2 2019 6 1400 2 2018 5 1550 3 2018 5 1550 3 2018 1 2800 1 2018 4 2300 5 2018 4 1300，class ConnectFour(): def __init__(self, width, height): self.width = width self.height = height self.board = [[0] * width] * height def dropLetter(self, letter, col): count = self.height - 1 while count > 0 and self.board[count][col] != 0: count -= 1 print self.board[count][col] self.board[count][col] = letter print self.board C = ConnectFour(4,4) C.dropLetter('X', 0) 和vehicleSport，您应该可以通过以下方式实现此目标：

vehicleSport2

并调用类似的函数

vehicleSport3

或者，如果您害怕数组的硬编码大小，请按以下方式调用它：

void vehicleSport(vehicleSpawner *spawner, size_t number_of_spawners_in_array) {
    menu.title("Sport >>>I ommitted the number here, you can add it if you wish easily<<<");

    for (size_t i = 0; i < number_of_spawners_in_array; i++) {
        char* name = UI::_GET_LABEL_TEXT(spawner[i].vehicleName);
        Hash key = GAMEPLAY::GET_HASH_KEY(spanwer[i].vehicleName);

        if(!GAMEPLAY::ARE_STRINGS_EQUAL(name, "NULL")) menu.option(name).vehicleSpawn(key).vehiclePreview(spawner[i].previewDir, spawner[i].previewName);
    }
}

此外，您可以完全省略长度，而很难-如果所有三个数组的大小始终相同，则将其编码到函数中。但我会避免这种情况，并始终明确地将其移交给我们。

但是请注意，由于初始示例未编译，因此上面的代码可能包含键入错误-您将必须自己添加适当的标头包含指令。此外，根据您的要求，这是一个 C解决方案-但您的代码显然是C ++。

希望这能回答您的问题，尽管我担心即使这样做，我仍可能不得不进一步解释为什么数组在函数中显示为指针，并且仍然可以为它们建立索引？

C将结构传递给函数

1 个答案: