[英]Python ctypes: initializing c_char_p()
I wrote a simple C++ program to illustrate my problem: 我写了一个简单的C ++程序来说明我的问题:
extern "C"{
int test(int, char*);
}
int test(int i, char* var){
if (i == 1){
strcpy(var,"hi");
}
return 1;
}
I compile this into an so. 我把它编译成一个。 From python I call: 从python我打电话:
from ctypes import *
libso = CDLL("Debug/libctypesTest.so")
func = libso.test
func.res_type = c_int
for i in xrange(5):
charP = c_char_p('bye')
func(i,charP)
print charP.value
When I run this, my output is: 当我运行它时,我的输出是:
bye
hi
hi
hi
hi
I expected: 我期望:
bye
hi
bye
bye
bye
What am I missing? 我错过了什么?
Thanks. 谢谢。
The string which you initialized with the characters "bye"
, and whose address you keep taking and assigning to charP
, does not get re-initialized after the first time. 您使用字符"bye"
初始化的字符串,以及您继续使用并分配给charP
的地址,在第一次之后不会重新初始化。
Follow the advice here : 按照这里的建议:
You should be careful, however, not to pass them to functions expecting pointers to mutable memory. 但是,您应该小心,不要将它们传递给期望指向可变内存的函数。 If you need mutable memory blocks, ctypes has a create_string_buffer function which creates these in various ways. 如果你需要可变内存块,ctypes有一个create_string_buffer函数,它以各种方式创建它们。
A "pointer to mutable memory" is exactly what your C function expects, and so you should use the create_string_buffer
function to create that buffer, as the docs explain. “指向可变内存的指针”正是您的C函数所期望的,因此您应该使用create_string_buffer
函数来创建该缓冲区,正如文档所解释的那样。
I am guessing python is reusing the same buffer for all 5 passes. 我猜python正在为所有5次传递重用相同的缓冲区。 once you set it to "hi", you never set it back to "bye" You can do something like this: 一旦你把它设置为“hi”,你就不会把它设置回“再见”你可以这样做:
extern "C"{
int test(int, char*);
}
int test(int i, char* var){
if (i == 1){
strcpy(var,"hi");
} else {
strcpy(var, "bye");
}
return 1;
}
but be careful, strcpy
is just asking for a buffer overflow 但要小心, strcpy
只是要求缓冲区溢出
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