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<html>
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<body>
<div id="maincontent" style="margin-top: 70px">
<h2>learn CUDA 2D</h2>
<p>
Last time we talked about 1D vector add operation at
<a href="https://malaysia-ai.org/learn-cuda"
>https://malaysia-ai.org/learn-cuda</a
>, it is straight forward, only use `x` coordinate.
</p>
<p>
Now let's go to 2D add operation, and actually CUDA supports up to 3D
for threads and blocks
<a
href="https://docs.nvidia.com/cuda/cuda-c-programming-guide/#thread-hierarchy"
>https://docs.nvidia.com/cuda/cuda-c-programming-guide/#thread-hierarchy</a
>, before this for 1D,
</p>
<pre>
```cuda
int N = 1000000;
int threadsPerBlock = 1024;
int blocksPerGrid = (N + threadsPerBlock - 1) / threadsPerBlock;
VecAdd<<<blocksPerGrid, threadsPerBlock>>>
```</pre
>
<p>Which is, view as,</p>
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<text x="350" y="40" text-anchor="middle">Block (1)</text>
<text x="550" y="40" text-anchor="middle">Block (2)</text>
<text x="400" y="280" text-anchor="middle">blockIdx.x</text>
<text x="90" y="90" text-anchor="middle" font-size="12">
Thread (0)
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Thread (1)
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<p style="font-size: 10px">Generated by Claude Sonnet 3.5</p>
<p>
All threads and blocks been aligned as one dimensional only, because we
only pass it as one dimensional! If you pass it as two dimensional, we
just expand Y dimension become size of one,
</p>
<pre>
```cuda
int N = 1000000;
dim3 threadsPerBlock(1024, 1); // (x, y)
dim3 blocksPerGrid((columns + threadsPerBlock.x - 1) / threadsPerBlock.x,
(rows + threadsPerBlock.y - 1) / threadsPerBlock.y);
VecAdd<<<blocksPerGrid, threadsPerBlock>>>
```</pre
>
<p>It will become,</p>
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<text x="150" y="40" text-anchor="middle">Block (0,0)</text>
<text x="350" y="40" text-anchor="middle">Block (1,0)</text>
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blockIdx.y
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Thread (0,0)
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Thread (1,0)
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<p style="font-size: 10px">Generated by Claude Sonnet 3.5</p>
<p>
Because we only initiate Y dimension as one, so the coordinate is always
0. In order to use multi-dimensional threads in CUDA, you must use
`dim3` data type, if we pass an integer N, behind the scence it passed
as `dim3(N, 1, 1)`.
</p>
<p>So, let us continue to write simple 2D vector sum operation,</p>
<pre>
```cuda
#include <stdio.h>
#include <cuda_runtime.h>
// Kernel definition
__global__ void VecAdd(int *a, int *b, int *c, int columns) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
int i = row * columns + col;
c[i] = a[i] + b[i];
}
int main()
{
int rows = 1000;
int columns = 1000;
size_t size = rows * columns * sizeof(int);
int *h_a, *h_b, *h_c;
int *d_a, *d_b, *d_c;
h_a = (int*)malloc(size);
h_b = (int*)malloc(size);
h_c = (int*)malloc(size);
for (int i = 0; i < rows * columns; i++) {
h_a[i] = i;
h_b[i] = i * 2;
}
cudaMalloc(&d_a, size);
cudaMalloc(&d_b, size);
cudaMalloc(&d_c, size);
cudaMemcpy(d_a, h_a, size, cudaMemcpyHostToDevice);
cudaMemcpy(d_b, h_b, size, cudaMemcpyHostToDevice);
// 16 * 16 = 256
dim3 threadsPerBlock(16, 16);
dim3 numBlocks((columns + threadsPerBlock.x - 1) / threadsPerBlock.x,
(rows + threadsPerBlock.y - 1) / threadsPerBlock.y);
VecAdd<<<numBlocks, threadsPerBlock>>>(d_a, d_b, d_c, columns);
cudaMemcpy(h_c, d_c, size, cudaMemcpyDeviceToHost);
for (int i = 0; i < rows * columns; i++) {
if (h_c[i] != h_a[i] + h_b[i]) {
printf("Error: %d + %d != %d\n", h_a[i], h_b[i], h_c[i]);
break;
}
}
free(h_a); free(h_b); free(h_c);
cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);
}
```</pre
>
<p>
As you can see, we defined `rows = 1000` and `columns = 1000`, but the
array definition is still one dimension `size = rows * columns`.
</p>
<p>
Because it is still one dimensional, in order to convert from 1D
coordinate to 2D coordinate, (1D // size of columns, 1D % size of
columns), if my index at 10040, so 2D is (10040 // 1000, 10040 % 1000)
== (10, 40)
</p>
<p>
If you have 2D coordinate and want to convert to 1D, 10 * columns + 40
== 10040.
</p>
<p>
So that is why we defined `int i = row * columns + col` in our kernel.
</p>
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blockIdx.y
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<p style="font-size: 10px">Generated by Claude Sonnet 3.5</p>
<p>
To know the actual row position, you must use `blockIdx.y * blockDim.y +
threadIdx.y`, and column position `blockIdx.x * blockDim.x +
threadIdx.x`, as we know, we defined the thread size is (16, 16), area
of 256, square of 16.
</p>
<p>
Current thread which `threadIdx.y` must add with current block
`blockIdx.y` multiply by size of thread, which is 16.
</p>
<p>
So if I am at the third row of thread, and this thread at the second row
of blocks, so the Y coordinate == 2 * 16 + 3 == 35.
</p>
<p>
If I am at nineth column of thread, and this thread at the fifth column
of blocks, so the X coordinate == 5 * 16 + 9 == 89.
</p>
<p>
To get 1D coordinate, 35 * 1000 + 89 == 35089, so that is why our
kernel,
</p>
<pre>
```cuda
__global__ void VecAdd(int *a, int *b, int *c, int columns) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
int i = row * columns + col;
c[i] = a[i] + b[i];
}
```</pre
>
<p>
Save the complete code above above (the long one) as `test.cu`, and then
you can continue to compile,
</p>
<pre>
```bash
nvcc test.cu -o test
./test
```</pre
>
<p>Let us try to use debugger,</p>
<pre>
```bash
nvcc -G -g -o test test.cu
cuda-gdb test
break VecAdd
run
```
```text
[New Thread 0x7ffb00204000 (LWP 561497)]
[New Thread 0x7ffafeef4000 (LWP 561498)]
[Detaching after fork from child process 561499]
[New Thread 0x7ffaf7fff000 (LWP 561506)]
[New Thread 0x7ffaf77fe000 (LWP 561507)]
[Switching focus to CUDA kernel 0, grid 1, block (0,0,0), thread (0,0,0), device 0, sm 0, warp 0, lane 0]
Thread 1 "test" hit Breakpoint 1, VecAdd<<<(63,63,1),(16,16,1)>>> (a=0x7ffacba00000, b=0x7ffacea00000, c=0x7ffacee00000, columns=1000) at test.cu:6
6 int row = blockIdx.y * blockDim.y + threadIdx.y;
```</pre
>
<p>No issue.</p>
</div>
</body>
</html>