这里有一些基本的Cython-在Cython中定义字符串数组的规范有效方法是什么具体来说,我想定义一个固定长度的char
常量数组。(请注意,我不想在这一点上引入NumPy。(
在C中,这将是:
/* cletters.c */
#include <stdio.h>
int main(void)
{
const char *headers[3] = {"to", "from", "sender"};
int i;
for (i = 0; i < 3; i++)
printf("%sn", headers[i]);
}
Cython:中的尝试
# cython: language_level=3
# letters.pyx
cpdef main():
cdef const char *headers[3] = {"to", "from", "sender"}
print(headers)
然而,这给出了:
(cy) $ python3 ./setup.py build_ext --inplace --quiet
cpdef main():
cdef const char *headers[3] = {"to", "from", "sender"}
^
------------------------------------------------------------
letters.pyx:5:32: Syntax error in C variable declaration
您需要两行:
%%cython
cpdef main():
cdef const char *headers[3]
headers[:] = ['to','from','sender`]
print(headers)
有些违反直觉的是,将unicode字符串(Python3!(分配给char*
。这是Cython的怪癖之一。另一方面,当只使用一个值初始化所有内容时,需要字节对象:
%%cython
cpdef main():
cdef const char *headers[3]
headers[:] = b'init_value` ## unicode-string 'init_value' doesn't work.
print(headers)
另一种选择是以下oneliner:
%%cython
cpdef main():
cdef const char **headers=['to','from','sender`]
print(headers[0], headers[1], headers[2])
这与上面的不完全相同,并导致以下C代码:
char const **__pyx_v_headers;
...
char const *__pyx_t_1[3];
...
__pyx_t_1[0] = ((char const *)"to");
__pyx_t_1[1] = ((char const *)"from");
__pyx_t_1[2] = ((char const *)"sender");
__pyx_v_headers = __pyx_t_1;
__pyx_v_headers
属于char **
类型,缺点是print(headers)
不再开箱即用。
对于python3 Unicode字符串,这是可能的-
cdef Py_UNICODE* x[2]
x = ["hello", "worlᏪd"]
或
cdef Py_UNICODE** x
x = ["hello", "worlᏪd"]