小编gar*_*ese的帖子

在C#7中是否可以在字典中的foreach循环中使用解构？像这样的东西:

var dic = new Dictionary<string, int>{ ["Bob"] = 32, ["Alice"] = 17 };
foreach (var (name, age) in dic)
{
    Console.WriteLine($"{name} is {age} years old.");
}

它似乎不适用于Visual Studio 2017 RC4和.NET Framework 4.6.2:

错误CS1061:'KeyValuePair'不包含'Deconstruct'的定义,并且没有扩展方法'Deconstruct'可以找到接受类型'KeyValuePair'的第一个参数(你是否缺少using指令或汇编引用？)

When I resize my window, I need to resize my textures that are attached to my framebuffer. I tried calling glTexStorage2D again, with different size parameters. However that does not work.

How can I resize the textures attached to my framebuffer? (Including the depth attachment)

EDIT

Code I tried:

glBindTexture(m_target, m_name);
glTexStorage2D(m_target, 1, m_format, m_width, m_height);
glBindTexture(m_target, 0);

where m_name, m_target and m_format are saved from the original texture and m_width and m_height are the new dimensions.

EDIT2

Please tell …

struct Object {  
    Object() { cout << "constructor\n"; }
    Object(const Object &) { cout << "copy constructor\n"; }
    Object(Object &&) { cout << "move constructor\n"; }
};

int main() {
    vector<Object> v;
    v.reserve(10);
    v.emplace_back(Object{});
}

这给了我以下输出:

构造函数
移动构造函数

为什么？我认为emplace_back确实创建了Object,因此不必调用复制或移动构造函数.
从描述:

元素就地构造,即不执行复制或移动操作.

编辑:啊,好吧,似乎我从根本上误解了emplace_back().您不必将Object作为参数,因为它会自动为您创建.您只需将Object-constructor的参数提供给emplace_back().

所以,如果我有一个像这样的新构造函数:

Object(int) { cout << "int constructor\n"; }   

我会像这样调用emplace_back:

v.emplace_back(42);  

而不是这个:

v.emplace_back(Object(42));

现在有道理,非常感谢!

编辑2:我希望我能接受你所有的答案!:-P

我最近阅读了有关复制和交换的内容,现在我正在尝试在基类和派生类中实现ctors.我的base和派生类中都有四个构造函数,但是我不确定如何实现派生类的赋值运算符.

explicit Base(int i) : m_i{i} {}
Base(const Base & other) : m_i{other.m_i}
Base(Base && other) : Base(0) { swap(*this, other); }
Base & operator=(Base other) { swap(*this, other); return *this; }
friend void swap(Base & a, Base & b) noexcept {
    using std::swap;
    swap(a.m_i, b.m_i);
}

explicit Derived(int j) : Base(42), m_j(j) {}
Derived(const Derived & other) : Derived(other.m_j) {}
Derived(Derived && other) : Derived(other.m_j) { swap(*this, other); }
Derived & operator=(Derived other) { /*???*/ }
friend …

如何使用Clang 3.5在QtCreator 3.3中启用C++ 14支持？我添加了一个Clang工具包,我已添加CONFIG += c++14到我的项目文件中.但是当使用例如返回类型推导时,我得到以下错误:

错误:'auto'返回没有尾随返回类型; 推导的返回类型是C++ 1y扩展

是否可以将构造函数与固定参数和构造函数模板混合使用？

我的代码:

#include <iostream>

class Test {
    public:
        Test(std::string, int, float) {
            std::cout << "normal constructor!" << std::endl;
        }

        template<typename ... Tn>
        Test(Tn ... args) {
            std::cout << "template constructor!" << std::endl;
        }
};

int main() {
    Test t("Hello World!", 42, 0.07f);
    return 0;
}

这给了我"模板构造函数!".有没有办法,我的普通构造函数被调用？

我如何为qTreeView的qTreeWidget实现这个代码？

for (const auto & i : names) {
    QTreeWidgetItem * item = new QTreeWidgetItem(ui->treeWidget);
    item->setText(0, QString::fromStdString(i));
    ui->treeWidget->addTopLevelItem(item);
    const std::unordered_map<std::string, double> map = m_reader.getMapFromEntry(i);
    for (const auto & j : map) {
        QTreeWidgetItem * item2 = new QTreeWidgetItem();
        item2->setText(0,QString::fromStdString(j.first));
        item2->setText(1,QString::number(j.second));
        item->addChild(item2);
    }

}

我有一个模型和一个treeView,像这样:

m_model = new QStandardItemModel(m_reader.getAllNames().size(),2,this);
ui->treeView->setModel(m_model);

我尝试过这个,但只显示一列:

QStandardItem * parentItem = m_model->invisibleRootItem();
for (const auto & i : names) {
    QStandardItem * item = new QStandardItem(QString::fromStdString(i));
    parentItem->appendRow(item);
    const std::unordered_map<std::string, double> map = m_reader.getMapFromEntry(i);
    for (const auto & …

我有一个按钮样式.根据Button是否启用,我想更改背景.这就是它的样子:

<Style x:Key="MyButtonStyle" TargetType="Button">
    <Style.Triggers>
        <DataTrigger Binding="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType=Button}, Path=IsEnabled, PresentationTraceSources.TraceLevel=High}" Value="False">
            <Setter Property="Background" Value="Purple"/>
        </DataTrigger>
        <DataTrigger Binding="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType=Button}, Path=IsEnabled, PresentationTraceSources.TraceLevel=High}" Value="True">
            <Setter Property="Background" Value="Yellow"/>
        </DataTrigger>
    </Style.Triggers>
</Style>

这只是一个基本的例子.实际上我需要一个MultiDataTrigger,但它甚至不能使用常规的DataTrigger.我只看到一个灰色按钮.

这是跟踪:

System.Windows.Data警告:56:为绑定创建BindingExpression(hash = 31767240)(hash = 6303779)
System.Windows.Data警告:58:路径:'IsEnabled'System.Windows.Data
警告:60:BindingExpression(hash = 31767240):默认模式已解析为OneWay
System.Windows.Data警告:61:BindingExpression(hash = 31767240):解析为PropertyChanged的默认更新触发器
System.Windows.Data警告:62:BindingExpression(hash = 31767240):附加到系统. Windows.Controls.Button.NoTarget(hash = 24311680)
System.Windows.Data警告:66:BindingExpression(hash = 31767240):RelativeSource(FindAncestor)需要树上下文
System.Windows.Data警告:65:BindingExpression(hash = 31767240) :Resolve source deferred
System.Windows.Data警告:67:BindingExpression(hash = 31767240):解析源
System.Windows.Data警告:70:BindingExpression(hash = 31767240):找到数据上下文元素:(OK)
System.Windows.数据警告:73:查找类型Button的祖先:查询网格(哈希= 35 377238)
System.Windows.Data警告:73:查找类型Button的祖先:查询ContentPresenter(hash = 51189900)
System.Windows.Data警告:73:查找类型Button的祖先:查询边框(hash = 48541090)
System.Windows.数据警告:73:查找类型Button的祖先:查询StartStopControl(hash = 22721178)
System.Windows.Data警告:73:查找类型Button的祖先:查询网格(哈希= 32321338) …

我有一个类Input,它有默认的移动/复制构造函数.

Input(const Input &) = default;
Input(Input &&) = default;

但是,以下断言失败了.

static_assert(std::is_copy_constructible<Input>(), "Input is not copy-constructible");
static_assert(std::is_move_constructible<Input>(), "Input is not move-constructible");

这是为什么？

这是一个完整的例子:

#include <type_traits>

class A {
public:
    A(const A &) = default;
    static_assert(std::is_copy_constructible<A>(), "");
};

int main() {
    // your code goes here
    return 0;
}

这是这个问题的后续。

我已经像这样设置了复制构造函数：

glGenBuffers(1, &m_vertexBuffer);
glGenBuffers(1, &m_colorBuffer);
glGenBuffers(1, &m_indexBuffer);
glGenVertexArrays(1, &m_vertexArray);

glBindVertexArray(m_vertexArray);

GLint size = 0;

glBindBuffer(GL_COPY_READ_BUFFER, other.m_indexBuffer);
glGetBufferParameteriv(GL_COPY_READ_BUFFER, GL_BUFFER_SIZE, &size);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, m_indexBuffer);
glBufferData(GL_ELEMENT_ARRAY_BUFFER, size, nullptr, GL_STATIC_DRAW);
glCopyBufferSubData(other.m_indexBuffer, m_indexBuffer, 0, 0, size);

glBindBuffer(GL_COPY_READ_BUFFER, other.m_vertexBuffer);
glGetBufferParameteriv(GL_COPY_READ_BUFFER, GL_BUFFER_SIZE, &size);
glBindBuffer(GL_ARRAY_BUFFER, m_vertexBuffer);
glBufferData(GL_ARRAY_BUFFER, size, nullptr, GL_STATIC_DRAW);
glCopyBufferSubData(other.m_vertexBuffer, m_vertexBuffer, 0, 0, size);
glVertexAttribPointer(0, 3, GL_FLOAT, GL_FALSE, 0, (void *) 0);
glEnableVertexAttribArray(0);

glBindBuffer(GL_COPY_READ_BUFFER, other.m_colorBuffer);
glGetBufferParameteriv(GL_COPY_READ_BUFFER, GL_BUFFER_SIZE, &size);
glBindBuffer(GL_ARRAY_BUFFER, m_colorBuffer);
glBufferData(GL_ARRAY_BUFFER, size, nullptr, GL_STATIC_DRAW);
glCopyBufferSubData(other.m_colorBuffer, m_colorBuffer, 0, 0, size);
glVertexAttribPointer(1, 3, GL_FLOAT, GL_FALSE, 0, (void *) …