std::memcpy() or std::copy(): Which Offers Better Memory Copying Performance?

Performance Comparison of std::memcpy() and std::copy()

When it comes to memory copying, two common choices in C are std::memcpy() and std::copy(). However, the question arises: which one offers better performance?

std::memcpy() vs. std::copy()

Traditionally, std::memcpy() is considered faster as it operates on raw memory without any type information. Conversely, std::copy() retains type information, which can potentially lead to additional overhead.

Performance Benchmark

However, recent tests have shown that std::copy() may perform better in certain scenarios. A series of tests conducted on SHA-2 and MD5 implementations revealed that std::copy() consistently outperformed std::memcpy() with an average speed increase of 2.99% in the SHA-2 tests.

Explanation

This unexpected result can be attributed to several factors. First, modern compilers aggressively inline std::copy(), which eliminates the need for function calls and potential performance penalties. Second, std::copy() retains type information, which allows compilers to optimize memory access based on the data type being copied. Additionally, with link time optimization enabled, the performance of std::copy() further increased, indicating that the compiler was able to make more optimizations with this method.

Conclusion

Contrary to popular belief, std::copy() does not pose a significant performance penalty. In fact, it can even surpass std::memcpy() in efficiency, especially when dealing with large chunks of data. Therefore, for scenarios where memory copying is required, std::copy() is recommended for its convenience, versatility, and comparable or superior performance to std::memcpy().

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