Python ctypes：初始化 c_char_p()
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1871375/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Python ctypes: initializing c_char_p()
提问by Mark
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. From python I call:
我把它编译成一个so。我从 python 调用：
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。
回答by Alex Martelli
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 初始化的字符串"bye"，以及您不断获取并分配给的地址charP，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. If you need mutable memory blocks, ctypes has a create_string_buffer function which creates these in various ways.
但是，您应该小心，不要将它们传递给需要指向可变内存的指针的函数。如果您需要可变内存块，ctypes 有一个 create_string_buffer 函数，它以各种方式创建这些。
A "pointer to mutable memory" is exactly what your C function expects, and so you should use the create_string_bufferfunction 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，因此您应该使用该create_string_buffer函数来创建该缓冲区，如文档所述。
回答by Evan Teran
I am guessing python is reusing the same buffer for all 5 passes. once you set it to "hi", you never set it back to "bye" You can do something like this:
我猜 python 正在为所有 5 次通过重用相同的缓冲区。一旦你将它设置为“hi”，你就永远不会将它设置回“bye”你可以做这样的事情：
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, strcpyis just asking for a buffer overflow
但要小心，strcpy只是要求缓冲区溢出
