Optimizing Memory Allocation: A Key to Efficient Performance
The Hidden Cost of Dynamic Memory Allocation
When working with containers like std::vector, it’s easy to overlook the impact of dynamic memory allocation on performance. While these containers automatically allocate memory as needed, they can sometimes hurt performance when dealing with small elements. This is because dynamic memory allocation can lead to memory fragmentation and slow down your program.
The Power of Small Object Optimization
One technique to mitigate this issue is small object optimization, also known as small string optimization or small buffer optimization. This approach involves using a union to store small objects or strings directly on the stack, rather than allocating memory on the heap. By doing so, you can significantly reduce memory allocation and improve performance.
A Closer Look at std::string
Modern implementations of std::string often employ small object optimization to handle short strings efficiently. They use a small buffer within the string class itself to store short strings, eliminating the need for dynamic memory allocation. However, this approach increases the size of the string class, even when the short buffer is not used.
A More Efficient Solution
A more memory-efficient solution is to use a union that can hold a short buffer when the string is in short mode and hold the necessary data members for a dynamically allocated buffer in long mode. This technique allows std::string to optimize memory allocation and improve performance.
Demonstrating Small Object Optimization
Let’s take a closer look at how std::string from libc++ behaves on a 64-bit system. By overloading the global operator new and operator delete, we can track dynamic memory allocations and see how std::string handles different string sizes.
“`cpp
auto allocated = sizet{0};
void* operator new(sizet size) {
void* p = std::malloc(size);
allocated += size;
return p;
}
void operator delete(void* p) noexcept {
return std::free(p);
}
int main() {
allocated = 0;
auto s = std::string{“”};
std::cout << “stack space = ” << sizeof(s)
<< “, heap space = ” << allocated
<< “, capacity = ” << s.capacity() << ‘\n’;
}
“`
The Results
When running this example, we can see that std::string occupies 24 bytes on the stack and has a capacity of 22 chars without using any heap memory. However, when we increase the string size to 23 characters, std::string is forced to use the heap to store the string, allocating 32 bytes and reporting a capacity of 31.
The Magic Behind Small Object Optimization
So, how does std::string manage to store strings of 22 characters in length without allocating any memory? The answer lies in the clever use of a union with two different layouts: one for short mode and one for long mode. By maximizing the use of the available 24 bytes, std::string can efficiently handle small strings and reduce memory allocation.