小编web*_*rc2的帖子

我写了这样的东西:

class Storage
{
public:
    Storage();
    QString key() const;
    int value() const;
    void add_item(QString&,int);
private:
    QMap<QString,int>* my_map_;
};

void Storage::add_item(QString& key,int value)//------HERE IS THE SLOT FOR ADDING
{
   *my_map_[key] = value;
}

当我试图通过以下方式添加项目时QMap:

class Dialog : public QDialog
{
    Q_OBJECT
public:
    Dialog(QWidget* = 0);
public slots:
    void add_item()
    {
        strg_->add_item(ui->lineEdit->text(),ui->spinBox->value());//---HERE I'M "PASSING" TWO OBJECTS: QString AND int
        ui->lineEdit->clear();
    }

private:
    Ui::Dialog* ui;
    Storage* strg_;
};  

我收到错误:

error: no matching function for call to 'Storage::add_item(QString, int)
note: candidates are: …

我想将一个现有的字符串转换为枚举(不作为枚举读取).我怎样才能做到这一点？

C#编译器抱怨包含以下代码new protected member declared in struct.问题是什么？

struct Foo {
    protected Object _bar;
}

我是Rails的新手,所以如果我的问题没有最有意义,我会道歉.

我有一个名为的类PaymentGatewayCipher看起来像:

require 'openssl'

# Encapsulates payment gateway encryption / decryption utility functions
class PaymentGatewayCipher
  class << self
    def encrypt(file, options = {})
      cipher = create_cipher
      cipher.encrypt(cipher_key)
      data = cipher.update(File.read(file))
      data << cipher.final

      if to_file = options[:to]
        # Write it out to a different file
        File.open(to_file, 'wb') do |f|
          f << data
        end
      end

      data
    end

    # Decrypts the given file
    def decrypt(file)
      cipher = create_cipher
      cipher.decrypt(cipher_key)
      encrypted_data = File.open(file, 'rb') {|io| io.read}
      data = cipher.update(encrypted_data)
      data << …

我使用 postgresql 作为我的后端数据库。

尝试扫描一个languagespoken文本数组字段

var user userprofile
row := core.db.QueryRow(
    "SELECT languagespoken FROM \"user\" WHERE id = $1",
    userId,
)

err := row.Scan(&user.Languages)

if err != nil {
    return user, err
}

我的结构看起来像这样

type userprofile struct {
    Languages []string `json:languages`
}

但出现错误

2014/06/30 15:27:17 PANIC: reflect.Set: **value of type []uint8 is not assignable to type []string**
/usr/lib/go/src/pkg/reflect/value.go:2198 (0x56c152)
    Value.assignTo: panic(context + ": value of type " + v.typ.String() + " is not assignable to type " + dst.String()) …

我注意到Vala不允许你const从非const变量初始化变量.为什么是这样？这是故意的设计决定还是错误/遗漏？分别考虑这些Vala和C示例; 在C程序编译并按预期运行时,Vala程序无法编译:

瓦拉

void main()
{
   const int constInt = 1;
   const int a = constInt;

   int plainInt = 0;
   const int b = plainInt;

   stdout.printf("A: %d\n", a);
   stdout.printf("B: %d\n", b);
}

// Compiler output:
//   test.vala:7.18-7.25: error: Value must be constant
//   const int b = plainInt;
//                 ^^^^^^^^

C

#include <stdio.h>

int main()
{
   const int constInt = 1;
   const int a = constInt;

   int plainInt = 0;
   const int b = plainInt;

   printf("A: …

我正试图在Genie中列出一个列表并且它似乎不起作用.编译代码:

[indent=2]
init
  var l = new list of string

产生这些错误:

someone@someone-UBook:~/Documents$ valac helloworld.gs helloworld.gs:2.10-2.24: error: The name `Gee' does not exist in the context of `main'
    var l = new list of int
            ^^^^^^^^^^^^^^^
helloworld.gs:2.8-2.24: error: var declaration not allowed with non-typed initializer
    var l = new list of int
          ^^^^^^^^^^^^^^^^^
Compilation failed: 2 error(s), 0 warning(s)

我已经安装了libgee2(通过sudo apt-get install libgee2),没有任何改变.有任何想法吗？

我试图覆盖基类中另一个方法使用的基类的方法; 但是,当派生类调用基类的using方法时,派生的used-method永远不会被执行,而是调用基类的used-method.这是一个例子:

#include <iostream>
using namespace std;
class Base {
public:
    Base() {}
    virtual ~Base() {}
    void printLeft() { cout << this->getLeft(); }
    int getLeft() { return 0; }
};

class Derived: public Base {
public:
    Derived() {}
    virtual ~Derived() {}
    int getLeft() { return 1; }
};
int main(int argc, char *argv[]) {
    Derived d = Derived();
    d.printLeft();
}

运行main()打印0,指示使用Base的getLeft()方法而不是派生对象的方法.

如何更改此代码,以便在从Derived实例Derived::getLeft()调用时调用？

我正在尝试使用Intent发送sdcard文件夹中包含的文件(.log文件).这是代码:

public void sendMail() {
            Intent intent = new Intent(Intent.ACTION_SEND_MULTIPLE);
            intent.setType("text/plain");
            intent.putExtra(Intent.EXTRA_EMAIL, new String[] {"name.surname@gmail.com"});
            intent.putExtra(Intent.EXTRA_SUBJECT, "Log files");
            intent.putExtra(Intent.EXTRA_TEXT, "body");
            //has to be an ArrayList
            ArrayList<Uri> uris = new ArrayList<Uri>();
            //convert from paths to Android friendly Parcelable Uri's
            if(Environment.MEDIA_MOUNTED.equals(Environment.getExternalStorageState())){
                File root = Environment.getExternalStorageDirectory();

                File logfolder = new File(root, "log");


                for (String file : logfolder.list()){
                    Uri u = Uri.parse(file);
                    uris.add(u);
                }
                intent.putParcelableArrayListExtra(Intent.EXTRA_STREAM, uris);
                startActivity(Intent.createChooser(intent, new String("Send mail...")));
            }

        }

我从菜单中选择Gmail.Gmail打开后,会正确显示包含收件人,主题,文本和附件的邮件.邮件被发送没有错误,但我收到状态栏通知,说"无法显示附件"!事实上,收件人正确收到电子邮件但没有附件.我无法弄清问题是什么.为什么附件没有被发现？请帮我!!

我发现字节godoctype byte byte非常混乱,不应该是？type byte uint8

byte是uint8的别名,在所有方面都等同于uint8.按照惯例,它用于区分字节值和8位无符号整数值.type complex128