Question

在尝试过滤MainMenu数组时

const byte menuLength = 10;

struct Menu {
    int id;
    char Description[16];
    int parentId;
};

Menu MainMenu[menuLength] = {
    { 1, "SYSTEM SETUP   ", -1 },
    { 2, "FUNCTIONS SETUP", -1 },
    { 3, "FIRMWARE VER.  ", -1 },

    //SYSTEM SETUP
    { 4, "< BACK         ", 1 },
    { 5, "MODEL SELECT   ", 1 },
    { 6, "RX SETUP       ", 1 },

    //FUNCTIONS SETUP
    { 7, "< BACK         ", 2 },
    { 8, "REVERSE        ", 2 },
    { 9, "ENDPOINTS      ", 2 },
};

与此一起

Menu GetSub(int parentId)
{
    int position = 0;
    Menu Filtered[menuLength];

    for (int i = 0; i < menuLength; i++)
    {
        if (parentId == MainMenu[i].parentId)
        {
            Filtered[position] = MainMenu[i];
            position++;
        }
    }
    return Filtered;
}

我遇到以下错误

“菜单”未命名类型
无法将'（Menu *）（＆Filtered）'从'Menu *'转换为'Menu

那么，我应该如何返回过滤后的数组？

Answer 1

首先，请在使用from ibapi.client import EClient from ibapi.wrapper import EWrapper from ibapi.contract import Contract def print_to_file(*args): with open('text.txt', 'a') as fh: fh.write(' '.join(map(str,args))) print = print_to_file class TestApp(EWrapper, EClient): def __init__(self): EClient.__init__(self, self) Layout = "{!s:1} {!s:2} {!s:3} {!s:4} {!s:5} {!s:6} {!s:7} {!s:8} {!s:8}" print(Layout.format("Ticker;", "Date;", "None;", "Time;", "Open;", "High;", "Low;", "Close;", "Volume ")) def historicalData(self, reqId, bar): print("AAPL", ";", bar.date.replace(' ', '; '), ";", bar.open, ";", bar.high, ";", bar.low, ";", bar.close, ";", bar.volume) def main(): app = TestApp() app.connect("127.0.0.1", 7497, 0) contract = Contract () contract.symbol = "AAPL" contract.secType = "STK" contract.exchange = "SMART" contract.currency = "USD" contract.primaryExchange = "NASDAQ" app.reqHistoricalData(0, contract, "", "1 D", "1 min", "TRADES", 0, 1, False, []) app.run() if __name__ == "__main__": main()时使用C++容器。不要使用可变大小的数组（VLA），有lots of条文章介绍了为什么使用它不好。改用std::vector和std::string。

C++

std :: copy_if

您可以使用std::copy_if过滤掉所需的菜单。

const byte menuLength = 10;

struct Menu {
    int id;
    std::string Description;
    int parentId;
};

std::vector<Menu> MainMenu = {
    { 1, "SYSTEM SETUP   ", -1 },
    { 2, "FUNCTIONS SETUP", -1 },
    { 3, "FIRMWARE VER.  ", -1 },

    //SYSTEM SETUP
    { 4, "< BACK         ", 1 },
    { 5, "MODEL SELECT   ", 1 },
    { 6, "RX SETUP       ", 1 },

    //FUNCTIONS SETUP
    { 7, "< BACK         ", 2 },
    { 8, "REVERSE        ", 2 },
    { 9, "ENDPOINTS      ", 2 },
};

LIVE DEMO

ranges :: view :: filter

有了Eric Niebler的range-v3库，这变得更加微不足道了。

std::vector<Menu> GetSub(const std::vector<Menu>& menu, int parentId)
{
  std::vector<Menu> sub;
  std::copy_if(menu.begin(), menu.end(), std::back_inserter(sub), [parentId](const Menu& m) {
    return m.parentId == parentId;
  });

  return sub;
}

Answer 2

您试图通过简单的Menu对象返回一个Menu对象数组，您需要更改函数的原型：

Menu GetSub(int parentId);

进入

Menu * GetSub(itn parentId);

此函数将为您返回指向Menu的数组的指针，但随后会出现另一个问题：您实际上并不知道数组中存储了多少个对象。可以通过在代码中添加简单的结构来解决：

struct MenuArray {
    Menu * ptr;
    int size;
}

然后像这样重新制作函数：

MenuArray GetSub(int parentId)
{
    int position = 0;
    Menu * Filtered = new Menu[menuLength];

    for (int i = 0; i < menuLength; i++)
    {
        if (parentId == MainMenu[i].parentId)
        {
            Filtered[position] = MainMenu[i];
            position++;
        }
    }
    return MenuArray{Filtered, position};
}

我们还可以简单地使用STL向量，这是您可以使用的最简单的解决方案。

using std::vector;
vector<Menu> GetSub(int parentId)
{
    vector<Menu> Filtered(menuLength);

    for (int i = 0; i < menuLength; i++)
        if (parentId == MainMenu[i].parentId)
        {
            Filtered.push_back(MainMenu[i]);
        }

    return Filtered;
}

如何在C ++中过滤数组？

2 个答案:

std :: copy_if

LIVE DEMO

ranges :: view :: filter