Timothyoung97 / RenderingEngine

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Nvidia GPU Graphics Performance Architect Prep #4

Closed Timothyoung97 closed 4 months ago

Timothyoung97 commented 4 months ago
给定一个点,判断该点是否在三角形里面

To determine if a point is inside a triangle, you can follow these steps:

光栅化时,如果多个三角形共享一个顶点,如何制定一个合理的规则保证每个顶点只被光栅化一次;

One common approach is to use a data structure like an edge list or an active edge table (AET) along with a scanline algorithm. Here's a simplified explanation of how this can work:

The edge list ensures that edges are processed in a consistent order, while the active edge table manages which edges are currently active for each scanline.

为何硬件绘制时通常都以三角形为单位而不是其它多边形;
知道三角形三个顶点的颜色,光栅化时如何计算三角形内部其它点的颜色;
STL一定能提高效率吗?
Good
Bad
写一个函数,分配大小是32字节倍数的内存;
#include <cstdlib>
#include <iostream>

void* allocateMemory(size_t size) {
    // Calculate the nearest multiple of 32 bytes
    size_t alignedSize = ((size + 31) / 32) * 32;

    // Allocate memory using the standard library function malloc
    void* ptr = std::malloc(alignedSize);

    // Check if memory allocation was successful
    if (ptr == nullptr) {
        std::cerr << "Memory allocation failed." << std::endl;
        return nullptr;
    }

    return ptr;
}

int main() {
    // Example usage: allocate 50 bytes
    void* ptr = allocateMemory(50);

    // Check if allocation was successful
    if (ptr != nullptr) {
        std::cout << "Memory allocated successfully." << std::endl;
        // Don't forget to free the allocated memory when done using it
        std::free(ptr);
    }

    return 0;
}
写一个屏幕拷贝的函数,将屏幕上的一片区域拷贝到令外一个地方;
#include <iostream>
#include <vector>

// Represents a rectangular region on the screen
struct ScreenRegion {
    int x, y;       // Top-left corner coordinates
    int width, height;  // Width and height of the region
};

// Function to copy a screen region to another location
void copyScreenRegion(const std::vector<std::vector<char>>& screen, const ScreenRegion& srcRegion, 
                      std::vector<std::vector<char>>& destination, int destX, int destY) {
    // Ensure source and destination regions are valid
    if (srcRegion.x < 0 || srcRegion.y < 0 || srcRegion.x + srcRegion.width > screen[0].size() ||
        srcRegion.y + srcRegion.height > screen.size() || destX < 0 || destY < 0 ||
        destX + srcRegion.width > destination[0].size() || destY + srcRegion.height > destination.size()) {
        std::cerr << "Invalid screen regions or destination coordinates." << std::endl;
        return;
    }

    // Copy the screen region to the destination
    for (int y = 0; y < srcRegion.height; ++y) {
        for (int x = 0; x < srcRegion.width; ++x) {
            destination[destY + y][destX + x] = screen[srcRegion.y + y][srcRegion.x + x];
        }
    }
}

int main() {
    // Example usage
    // Define screen dimensions
    const int screenWidth = 80;
    const int screenHeight = 24;

    // Create the screen buffer
    std::vector<std::vector<char>> screen(screenHeight, std::vector<char>(screenWidth, '.'));

    // Draw some characters on the screen (just for demonstration)
    for (int y = 5; y < 10; ++y) {
        for (int x = 10; x < 20; ++x) {
            screen[y][x] = '*';
        }
    }

    // Define the source screen region to copy
    ScreenRegion srcRegion = {10, 5, 10, 5};

    // Define the destination screen buffer
    std::vector<std::vector<char>> destination(srcRegion.height, std::vector<char>(srcRegion.width, '.'));

    // Copy the source screen region to the destination at coordinates (30, 10)
    copyScreenRegion(screen, srcRegion, destination, 30, 10);

    // Print the destination buffer to verify the copy operation
    for (int y = 0; y < srcRegion.height; ++y) {
        for (int x = 0; x < srcRegion.width; ++x) {
            std::cout << destination[y][x];
        }
        std::cout << std::endl;
    }

    return 0;
}
Timothyoung97 commented 4 months ago

画一张GPU渲染流程图。 image

Timothyoung97 commented 4 months ago
Overview of graphics pipeline
API Questions
From Driver to GPU
DirectX11
Understanding DX11 Command Queues

DirectX 11 doesn't have explicit command queues like newer APIs such as DirectX 12 or Vulkan. Instead, command submission is handled implicitly by the device context.

However, it's essential to understand that DirectX 11 devices can still benefit from multithreaded command submission by using deferred contexts.

Timothyoung97 commented 4 months ago

Graphics performance team works on delivering an efficient and powerful graphics architecture every generation. The team studies graphics workloads and test out innovative HW/SW solutions on various platforms to address the inefficiencies in the current architecture. The work we do paves the path for real time rendering of some of the most complex and compute intensive visualization technique.

What you'll be doing:

What we need to see:

Ways to stand out from the crowd:

Good understanding of state-of-the-art rendering techniques and their usage of GPU
Timothyoung97 commented 4 months ago
给出stack结构,利用stack完成queue的操作。
Class stack{
Void push(data);
Void pop(&data);
Bool isempty;}

写出:

Class queue{

}

image image

一个链表,里面数字无序排列,要求给出代码,实现升序排序。

Void sort(*head){

}

put into vector, write a sort with a lambda function that using the member for comparison

Ascending:

std::ranges::sort(mMyClassVector, [](const MyClass &a, const MyClass &b)
{ 
    return a.mProperty < b.mProperty; 
});

Descending:

std::ranges::sort(mMyClassVector, [](const MyClass &a, const MyClass &b)
{ 
    return a.mProperty > b.mProperty; 
});
a=b*c+d; b,c,d均为unsigned 8bit,问a需要多大bit来存储。给出思考过程。

For unsigned 8-bit integers: The range of values is from 0 to 255 (2^8 - 1).

解释mipmap

You know this

数据发送端:100clock中工作 80clock,休息 20clock, 但这80是random分布的。(1bit/1clock. )

数据接收端:每10clock中前 8clock 工作,后2clock休息。

问:作为中间的一个缓冲器,其容量应该为多大?

一个格子图,大概如下:

b w b w b w b w b w b w b w b w b w b w b w b w b (1) 有多少个正方形? (2) 有多少个方形(包括长方形,正方形)? (3) 给你一个点,你如何判断它是黑色还是白色?写c代码。以左下角为原点。 注:b表示黑色,w表示白色。(上面所有小方格都是正方形:)。

Timothyoung97 commented 4 months ago

how to find if a number of a power of 2 in constant time?

// Function to check if x is power of 2
bool isPowerOfTwo(int n)
{
    if (n == 0)
        return false;

    return (ceil(log2(n)) == floor(log2(n)));
}
Timothyoung97 commented 4 months ago
什么时候用Virtual Destructor
class Base {
public:
    virtual ~Base() { } // Virtual destructor

    // Other virtual functions and non-virtual functions
};

class Derived : public Base {
public:
    ~Derived() {
        // Cleanup resources owned by Derived
    }

    // Other member functions
};

In this example, Base has a virtual destructor because it serves as a base class with polymorphic behavior. Derived inherits from Base and overrides the destructor to provide proper resource cleanup specific to Derived objects. When a Derived object is destroyed through a pointer to Base, the virtual destructor in Base ensures that the destructor of Derived is called.

Compiler对Virtual怎么处理
最后一道coding,用google doc做的,给16-bit RGB值,转成32-bit RGBX值
// Function to convert 16-bit RGB value to 32-bit RGBX value
uint32_t convertRGB16to32(uint16_t rgb16) {
    // Extract 5-bit components from 16-bit RGB value
    uint8_t r5 = (rgb16 >> 11) & 0x1F;  // 5 bits for red
    uint8_t g6 = (rgb16 >> 5) & 0x3F;   // 6 bits for green
    uint8_t b5 = rgb16 & 0x1F;          // 5 bits for blue

    // Expand 5-bit components to 8 bits
    uint8_t r8 = (r5 * 255) / 31;  // Scale to 8 bits
    uint8_t g8 = (g6 * 255) / 63;  // Scale to 8 bits
    uint8_t b8 = (b5 * 255) / 31;  // Scale to 8 bits

    // Create 32-bit RGBX value (0xFF for alpha channel)
    uint32_t rgbx32 = (uint32_t)r8 << 24 | (uint32_t)g8 << 16 | (uint32_t)b8 << 8 | 0xFF;

    return rgbx32;
}

解释点乘、叉乘用途 Z-buffer Z-fighting Z-buffer用什么数据类型,大小,取舍 Depth testing Stencil Buffer Deferred shading和forward shading 听没听说过TBDR (Tile-Based Deferred Rendering) 讲讲Ray Tracing

Timothyoung97 commented 4 months ago
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Timothyoung97 commented 4 months ago

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Timothyoung97 commented 4 months ago

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Timothyoung97 commented 4 months ago

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Timothyoung97 commented 4 months ago

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Timothyoung97 commented 4 months ago

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Timothyoung97 commented 4 months ago

Can you elaborate on some of the specific challenges the graphics performance team has encountered in previous projects, and how these challenges were addressed?

How does the team prioritize between optimizing existing architecture and introducing new features in each generation of GPU architecture?

Could you provide examples of real-world applications or industries where the advancements in GPU architecture directly impact performance or efficiency?

How does the team ensure compatibility and performance across different APIs such as D3D12, DX Machine Learning, DX, and Vulkan?

What methodologies or tools does the team employ to quantify and analyze the performance of existing and projected architectures?

Can you discuss any recent innovations or breakthroughs in real-time rendering techniques that the team has been investigating or implementing?

How does the team balance between theoretical performance gains and practical implementation feasibility when proposing ideas to improve GPU architecture?

What role does collaboration with other teams, such as software development or hardware engineering, play in the process of improving GPU architecture?

Could you walk me through the typical process of developing performance simulation models and infrastructure within the graphics performance team?

Can you provide insights into the approach the team takes in designing performance test plans and tests for new graphics units and architectural features?