小编Pio*_*ach的帖子

哪种转换更好,有什么区别？

class Base
{};

class Derived : public Base, public std::enable_shared_from_this<Derived>
{};

int main(int argc, const char * argv[])
{
    std::shared_ptr<Base> ptr1 = std::dynamic_pointer_cast<Base>(std::shared_ptr<Derived>(new Derived())); // version 1
    std::shared_ptr<Base> ptr2 = std::shared_ptr<Derived>(new Derived()); // version 2
    return 0;
}

当第二个示例只调用基类中的析构函数时,为什么在使用std :: shared_ptr deallocation从基类和派生类调用析构函数？

class Base
{
public:
    ~Base()
    {
        std::cout << "Base destructor" << std::endl;
    }
};

class Derived : public Base
{
public:
    ~Derived()
    {
        std::cout << "Derived destructor" << std::endl;
    }
};

void virtual_destructor()
{
    {
        std::cout << "--------------------" << std::endl;
        std::shared_ptr<Base> sharedA(new Derived);
    }

    std::cout << "--------------------" << std::endl;
    Base * a = new Derived;
    delete a;
}

输出:

--------------------
Derived destructor
Base destructor
--------------------
Base destructor

在这两种情况下我都期待相同的行为.

我正在开发采用截图的静态库,从OpenGL应用程序中获取它们需要特殊处理.

当客户端应用程序链接到我的静态库时,它必须添加我的库使用的框架,例如采用OpenGL截图,即使客户端应用程序不使用OpenGL,它也必须链接OpenGLES.framework,这是不好的.我试图检查库是否客户端已与OpenGLES.framework链接并动态启用从OpenGL截取屏幕截图.

问题是当我尝试使用C函数时出现编译错误:

if(&glReadPixels != NULL) {
    glReadPixels(0, 0, size.width, size.height, GL_RGBA, GL_UNSIGNED_BYTE, buffer);
}

如您所见,我可以检查方法是否存在,但是如何调用它以不导致链接器错误？当我用我的库编译客户端时,我得到了这个:

Undefined symbols for architecture i386:
"_glReadPixels", referenced from:
  +[TakeScreenshotUtil takeOpenGLScreenshotWithContext:layerSize:] in libScr-iOS.a(TakeScreenshotUtil.o)

我正在尝试使用

__attribute__ ((weak)) 

但它不起作用(不会改变任何东西).

为什么这段代码会崩溃？

#include <iostream>
#include <functional>

int main(int argc, const char * argv[])
{
    std::function<void(int)> function = [](int)
    {
    };
    auto binding = std::bind(function, 10);

    std::function<void()> jobFunctor = binding; // crashes here with EXC_BAD_ACCESS

    return 0;
}

转换绑定结果时,构造函数中jobFunctor存在无限堆栈递归std::function.

我正在运行Mac OS X 10.8.5,我使用libc ++编译此代码与Xcode 5.0.2,编译器版本:

LO50F-04-198BX:$ clang++ --version
Apple LLVM version 5.0 (clang-500.2.79) (based on LLVM 3.3svn)
Target: x86_64-apple-darwin12.5.0
Thread model: posix

我正在尝试编写函数,func以便编译器可以推导出模板参数,它在我传入时有效std::function,但不能与lambdas一起使用:

template<typename TResult>
TResult func(std::function<TResult()> f)
{
    return TResult();
}

int main()
{
                                // Visual Studio 2013
    int result = func([]() {    // error: 'TResult func(std::function<TResult(void)>)' : could not deduce template argument for 'std::function<TResult(void)>' from 'main::<lambda_d9d7854806072a2cb711f56185602ccb>'
        return 100;
    });

    std::function<int()> testFunc = []() {
        return 100;
    };
    int result2 = func(testFunc); // this works

    return 0;
}

是否有可能推导出lambda的模板参数,以便该行编译？而不是写func<int>([](){ return 100; });我想写func([](){ return 100; });

我想编写一个将一个闭包作为参数的函数，该函数将另一个闭包作为参数，如下所示：

fn test<F: Fn(G), G: Fn()>(f: F) {
    let print = || {
        println!("hello");
    };
    f(print);
}

fn main() {
    test(|print| {
        print();
    })
}

但这给出了这个错误：

error[E0308]: mismatched types
 --> src/main.rs:5:7
  |
1 | fn test<F: Fn(G), G: Fn()>(f: F) {
  |                   - expected this type parameter
2 |     let print = || {
  |                 -- the found closure
...
5 |     f(print);
  |     - ^^^^^ expected type parameter `G`, found closure
  |     |
  |     arguments to this function are …

我使用CATiledLayer作为我的UIView的支持层,我把它放在UIScrollView中.在我的视图的init方法中,我正在创建绘制简单线条的CGPathRef对象.当我尝试在drawLayer:inContext中绘制此路径时,当我滚动/缩放时,它偶尔会与EXEC_BAD_ACCESS(很少)崩溃.

代码很简单,我只使用标准的CG*函数:

- (id)initWithFrame:(CGRect)frame
{
    self = [super initWithFrame:frame];
    if (self) {
        CATiledLayer *tiledLayer = (CATiledLayer *)[self layer];
        tiledLayer.levelsOfDetail = 10;
        tiledLayer.levelsOfDetailBias = 5;
        tiledLayer.tileSize = CGSizeMake(512.0, 512.0);

        CGMutablePathRef mutablePath = CGPathCreateMutable();
        CGPathMoveToPoint(mutablePath, nil, 0, 0);
        CGPathAddLineToPoint(mutablePath, nil, 700, 700);
        path = CGPathCreateCopy(mutablePath);
        CGPathRelease(mutablePath);
    }
    return self;
}

+ (Class) layerClass {
    return [CATiledLayer class];
}

- (void) drawRect:(CGRect)rect {
}

- (void) drawLayer:(CALayer *)layer inContext:(CGContextRef)ctx {
    CGContextSetRGBFillColor(ctx, 1, 1, 1, 1);
    CGContextFillRect(ctx, self.bounds);

    CGContextSetLineWidth(ctx, 5);

    CGContextAddPath(ctx, path);
    CGContextDrawPath(ctx, kCGPathStroke); …