性能的函數

Question

考慮下面的代碼段的第一VS第二呼叫的（朱）

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) 

輸出是

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) 

爲什麼是第二（和第四）嘗試這樣比第一個（和第三個）嘗試更快（並佔用更少的內存）？

Answer 1

朱莉婭，據我所知，是一個即時編譯器。所以第一個（和第三個）運行是編譯代碼（爲此需要分配），第二個（和第四個）運行只是運行先前編譯的代碼

Answer 2

只是爲了擴展Paul的答案：加速來自Julia的類型推斷和多次調度。假設您第一次使用float評估函數：JIT（即時編譯器）計算出參數的類型並寫入適當的LLVM代碼。如果您隨後使用整數評估相同的函數，則會編譯不同的LLVM代碼。在您調用該函數的下列時間，它將根據參數的類型分派不同的LLVM代碼。這就是爲什麼在定義函數時編譯通常不合理的原因。

你可以閱讀更多關於這個here例如（有多個調度噸引用的文件中！）

考慮，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) 

這給：

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)