函数第一次调用与第二次调用的性能

Question

考虑以下代码 (Julia)

bar(x) = for i = 1:9999 x+x*x-x+x end # Define the "bar" function
print("First try: "); @time bar(0.5)
print("Second try: "); @time bar(0.5)
bar(x) = for i = 1:9999 x+x*x-x+x end # Redefine the same "bar" function
print("Third try: "); @time bar(0.5)
print("Fourth try: "); @time bar(0.6)

输出为

First try: elapsed time: 0.002738996 seconds (88152 bytes allocated)
Second try: elapsed time: 3.827e-6 seconds (80 bytes allocated)
Third try: elapsed time: 0.002907554 seconds (88152 bytes allocated)
Fourth try: elapsed time: 2.395e-6 seconds (80 bytes allocated)

为什么第二次（和第四次）尝试比第一次（和第三次）尝试快得多（占用内存更少）？

Answer 1

据我所知，Julia 有一个即时编译器。所以第一个（和第三个）运行正在编译代码（为此需要分配），第二个（和第四个）运行只是运行编译之前编译的代码

Answer 2

只是为了扩展 Paul 的回答：加速的很大一部分来自 Julia 的类型推断和多重分派。假设您第一次使用浮点数评估函数：JIT（即时编译器）计算出参数的类型并编写适当的 LLVM 代码。如果您随后使用整数评估相同的函数，则会编译不同的 LLVM 代码。接下来调用该函数时，它将根据参数的类型分派不同的 LLVM 代码。这就是为什么在定义函数时编译通常没有意义的原因。

例如，您可以阅读更多相关内容 here（文档中有大量对多重分派的引用！）

考虑，例如：

bar(x) = for i = 1:9999 x+x*x-x+x end

print("First try with floats: "); @time bar(0.5)
print("Second try with floats: "); @time bar(0.5)

print("First try with integers: "); @time bar(1)
print("Second try with integers: "); @time bar(1)

给出：

First try with floats: elapsed time: 0.005570773 seconds (102440 bytes allocated)
Second try with floats: elapsed time: 5.762e-6 seconds (80 bytes allocated)    
First try with integers: elapsed time: 0.003584026 seconds (86896 bytes allocated)    
Second try with integers: elapsed time: 6.402e-6 seconds (80 bytes allocated)

函数第一次调用与第二次调用的性能

Performance of the first vs second call of a function

compilation

dynamic-compilation

julia