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c++ - Lambda 捕获 shared_ptr 成员

转载 作者:可可西里 更新时间:2023-11-01 17:38:08 26 4
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我有一个类OpenGLRenderer它有一个类(class)成员mMemoryAllocator那是一个std::shared_ptr<MemoryAllocator> .我将内存分配器保留在 shared_ptr 中的原因是因为即使 shared_ptr<Texture>下面返回的时间超过了它的创建者 OpenGLRenderer , MemoryAllocator如果我按值捕获实例,实例仍然有效,因为它会增加引用计数:

std::shared_ptr<Texture> OpenGLRenderer::CreateTexture(TextureType textureType, const std::vector<uint8_t>& textureData, uint32_t textureWidth, uint32_t textureHeight, TextureFormat textureFormat)
{
return std::shared_ptr<Texture>(mMemoryAllocator->AllocateObject<Texture>(
textureData, textureWidth, textureHeight,
textureFormat, textureType, mLogger),
[=](Texture* texture) {
mMemoryAllocator
->DeallocateObject<Texture>(texture);
});
}

...但是,它不起作用。如果OpenGLRendererstd::shared_ptr<Texture> 之前超出范围, std::shared_ptr<MemoryAllocator>变得损坏,因此 lambda 表达式变得疯狂。我做错了什么?

最佳答案

这种情况下的问题是 lambda 不捕获对象的成员,而是捕获 this 指针。一个简单的解决方法是创建一个局部变量并绑定(bind)它:

std::shared_ptr<Texture> 
OpenGLRenderer::CreateTexture(TextureType textureType,
const std::vector<uint8_t>& textureData,
uint32_t textureWidth, uint32_t textureHeight,
TextureFormat textureFormat)

{
std::shared_ptr<AllocatorType> allocator = mMemoryAllocator;
return std::shared_ptr<Texture>(mMemoryAllocator->AllocateObject<Texture>(
textureData, textureWidth, textureHeight,
textureFormat, textureType, mLogger),
[=](Texture* texture) {
allocator
->DeallocateObject<Texture>(texture);
});
}

关于c++ - Lambda 捕获 shared_ptr 成员,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19528375/

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