GPU Programming

Fix PyCUDA float4 Kernel Args: 3 Crucial Steps 2025

Struggling with 'invalid argument' errors in PyCUDA? Learn the 3 crucial steps to correctly pass float4 and other vector types to your CUDA kernels in 2025.

D

Dr. Alex Carter

A computational scientist specializing in high-performance computing and GPU acceleration with Python.

7 min read15 views

You’re deep in the zone, accelerating your Python code with PyCUDA. Data is flying to the GPU, kernels are launching, and then... bam. You hit the wall:

pycuda._driver.LogicError: cuLaunchKernel failed: invalid argument

You frantically check your block sizes, grid dimensions, and data types. Everything looks right. But the error persists. If you're trying to pass a vector type like float4 to your kernel, you've likely stumbled upon one of PyCUDA's most common and frustrating hurdles.

The problem isn't your logic; it's the translation layer between Python's dynamic world and CUDA's rigid C-style memory layout. Fortunately, the fix is straightforward once you know the secret. In this post, we'll walk through the three crucial steps to correctly pass float4 arguments to your kernels, saving you hours of debugging.

Why Passing a Simple Tuple or Array Fails

At its core, CUDA is a C/C++ platform. A float4 in a CUDA kernel isn't just four float values; it's a specific C struct with a defined memory layout, typically 16 bytes (4 floats * 4 bytes/float). The fields are contiguous in memory: x, then y, then z, then w.

When you're in Python, you might instinctively create your vector like this:

# This will NOT work as a direct kernel argument
my_vector = (1.0, 2.0, 3.0, 4.0) # A Python tuple
# or
my_vector_np = np.array([1.0, 2.0, 3.0, 4.0], dtype=np.float32) # A standard NumPy array

PyCUDA is smart, but it's not a mind reader. It doesn't know how to take that Python tuple or standard NumPy array and map it to the 16-byte memory block that the CUDA kernel expects for a single float4 argument. It sees four separate numbers, not one single structured object. This mismatch is what triggers the invalid argument error.

To solve this, we need to build a bridge. We must create an object in Python that has the exact same memory representation as the C-level float4 struct.

The 3-Step Fix for float4 Arguments

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The solution lies in using NumPy's powerful structured data types (dtype). We'll explicitly define what a float4 looks like in memory, create our data using that definition, and then pass it to the kernel.

Step 1: Define the Struct in Python with NumPy

First, we need to tell NumPy how to build a float4. We do this by creating a custom dtype that mirrors the C struct. Each field (`x`, `y`, `z`, `w`) is defined as a 32-bit float (`np.float32`), which corresponds to a float in CUDA.

import numpy as np

# Define a structured dtype that mirrors the CUDA float4 struct
float4_dtype = np.dtype([
    ("x", np.float32),
    ("y", np.float32),
    ("z", np.float32),
    ("w", np.float32)
])

This float4_dtype object is our blueprint. It tells NumPy to allocate a contiguous block of memory and how to interpret the bytes within it: the first 4 bytes are 'x', the next 4 are 'y', and so on. The total size of an object with this dtype will be 16 bytes, a perfect match for CUDA's float4.

Step 2: Prepare Your Data with the Custom dtype

Now that we have our blueprint, we can create data with it. The key is to pass our custom float4_dtype to the np.array constructor.

If you need to pass a single float4 vector as a kernel argument, you create a one-element array and then extract that single element. This gives you a NumPy scalar object with the correct structured type.

# Create the data as a tuple inside a list
vector_data = [(10.0, 20.0, 30.0, 40.0)]

# Create a NumPy scalar object with our custom dtype
input_vector = np.array(vector_data, dtype=float4_dtype)[0]

print(f"Type of input_vector: {type(input_vector)}")
# Output: Type of input_vector: <class 'numpy.void'>

print(f"Value of input_vector: {input_vector}")
# Output: Value of input_vector: (10., 20., 30., 40.)

Notice the type is numpy.void. This is NumPy's way of representing a flexible, structured scalar type. Don't let the name confuse you; this is exactly what we want. This object is a single, 16-byte item that PyCUDA can pass by value to the kernel.

If you're passing an array of float4 vectors (e.g., using drv.In), you simply create the NumPy array without indexing the first element:

# For passing an array of vectors to GPU memory
input_vector_array = np.array([
    (1.0, 2.0, 3.0, 4.0),
    (5.0, 6.0, 7.0, 8.0)
], dtype=float4_dtype)

Step 3: Pass the Prepared Object to Your Kernel

With our data correctly structured, the final step is beautifully simple. You just pass the input_vector object directly to your kernel function. PyCUDA now understands exactly how to handle it.

Let's look at a complete, working example:

import pycuda.autoinit
import pycuda.driver as drv
from pycuda.compiler import SourceModule
import numpy as np

# --- Step 1: Define the float4 Struct ---
float4_dtype = np.dtype([("x", np.float32), ("y", np.float32), ("z", np.float32), ("w", np.float32)])

# --- Step 2: Prepare the Data ---
# Create a structured scalar object for our single float4 argument
input_vector = np.array([(10.0, 20.0, 30.0, 40.0)], dtype=float4_dtype)[0]

# Prepare an output buffer on the GPU
output_val = np.zeros(1, dtype=np.float32)
output_gpu = drv.mem_alloc(output_val.nbytes)

# --- CUDA Kernel Definition ---
# Note the kernel argument is `float4 my_vec`
mod = SourceModule("""
__global__ void test_float4_arg(float4 my_vec, float *result)
{
    // A simple test: sum the components of the float4
    *result = my_vec.x + my_vec.y + my_vec.z + my_vec.w;
}
""")

# Get the kernel function from the compiled module
kernel_func = mod.get_function("test_float4_arg")

print(f"Passing this structured object to kernel: {input_vector}")

# --- Step 3: Pass the structured object to the Kernel ---
# This now works perfectly!
kernel_func(input_vector, output_gpu, block=(1, 1, 1), grid=(1, 1, 1))

# Copy the result back from GPU to host
drv.memcpy_dtoh(output_val, output_gpu)

print(f"Result from GPU: {output_val[0]}") # Expected: 100.0

When you run this, it executes without error, and you get the correct result of 100.0. The invalid argument error is gone because we provided the kernel with an argument that has the exact memory layout it was compiled to expect.

Bonus Tip: What About float2 and float3?

This technique is not limited to float4. It works for any C-style struct, including other vector types. You just need to define the corresponding dtype.

# For float2 (8 bytes)
float2_dtype = np.dtype([("x", np.float32), ("y", np.float32)])

# For float3 (12 bytes, but see note below)
float3_dtype = np.dtype([("x", np.float32), ("y", np.float32), ("z", np.float32)])

An important note on float3: For performance reasons, CUDA's compiler often aligns float3 variables to 16 bytes, just like a float4, padding it with 4 bytes of unused space. This can cause major headaches if your host-side struct is only 12 bytes. The beauty of the np.dtype approach is that NumPy's alignment rules typically match the C compiler's rules, so it often handles this alignment padding for you automatically! You can always verify the byte size in Python with float3_dtype.itemsize to be sure.

Wrapping It Up

The dreaded cuLaunchKernel failed: invalid argument error when using vector types is almost always a memory layout mismatch between Python and CUDA. By using NumPy's structured dtypes, you can bridge this gap reliably.

Just remember the three crucial steps:

  1. Define the Struct: Create a np.dtype that mirrors the C struct in your kernel (e.g., [('x', np.float32), ...]).
  2. Prepare the Data: Use np.array(data, dtype=your_dtype) to create your data. Use [0] to get a scalar object for single arguments.
  3. Pass to Kernel: Pass the resulting NumPy structured object directly as a kernel argument.

Adopting this practice will make your PyCUDA code more robust, more readable, and save you from a world of debugging pain. Happy coding, and may your kernels always launch successfully!

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