Cython：为什么size_t比int快？-Cython:whyissize_tfasterthanint?

9  

You'd likely have to do a line by line profiling to find out exactly, but one thing stands out to me from the produced C file: int version is checked for wraparound to negative numbers, size_t is assumed ok.

你可能不得不进行逐行分析以找出确切的结果，但是有一件事从我生成的C文件中脱颖而出：检查int版本是否为环绕负数，size_t假定为ok。

In the int loop: (t_3 is assigned from k, they're the same type)

在int循环中：（t_3从k分配，它们是相同的类型）

if (__pyx_t_3 <0) {
  __pyx_t_3 += __pyx_v_A.shape[0];
  if (unlikely(__pyx_t_3 <0)) __pyx_t_4 = 0;
} else if (unlikely(__pyx_t_3 >= __pyx_v_A.shape[0])) __pyx_t_4 = 0;

In the size_t loop:

在size_t循环中：

if (unlikely(__pyx_t_3 >= (size_t)__pyx_v_A.shape[0])) __pyx_t_4 = 0;

So no wraparound test is needed because size_t is unsigned and guaranteed not to wrap around when indexing items in memory. The rest is virtually the same.

因此，不需要进行环绕测试，因为size_t是无符号的，并且保证在索引内存中的项目时不会回滚。其余几乎是一样的。

Update: regarding your unsigned int results - what's your size of int and size_t? Any chance they're different size, causing the change? In my case the C code for uint and size_t is identical. (since size_t is unsigned and specifically unsigned int on this system)

更新：关于你的unsigned int结果 - 你的int和size_t的大小是多少？他们有什么不同的规模，导致变化？在我的例子中，uint和size_t的C代码是相同的。 （因为size_t是无符号的，在此系统上特别是unsigned int）

3  

On a 64 bit system there seem to be two reasons:

在64位系统上，似乎有两个原因：

1. Use an unsigned integer for the loop:

   对循环使用无符号整数：

   %%cython
   
   cimport numpy as cnp
   import numpy as np
   
   def sum_int_unsigned(cnp.int64_t[::1] A):
       cdef unsigned long s = 0
       cdef unsigned k
       for k in xrange(A.shape[0]):
           s += A[k]
       return s
   

2. Use a long instead of an int:

   使用long而不是int：

   %%cython
   
   cimport numpy as cnp
   import numpy as np
   
   def sum_int_unsigned_long(cnp.int64_t[::1] A):
       cdef unsigned long s = 0
       cdef unsigned long k
       for k in xrange(A.shape[0]):
           s += A[k]
       return s
   

Timings:

时序：

%timeit sum_int(a)
1000 loops, best of 3: 1.52 ms per loop

%timeit sum_size_t(a)
1000 loops, best of 3: 671 µs per loop

Using unsigned brings us half way:

使用unsigned将我们带到了一半：

%timeit sum_int_unsigned(a) 
1000 loops, best of 3: 1.09 ms per loop

Using long accounts for the rest:

使用长帐户来完成剩下的工作：

%timeit sum_int_unsigned_long(a)
1000 loops, best of 3: 648 µs per loop