当我从ctypes调用的c函数实例化Python对象时,我的嵌入式python3.3程序出现错误。在
设置好解释器后,我可以从c main成功地实例化python Int(以及自定义的c扩展类型):
#import <Python/Python.h>
#define LOGPY(x) \
{ fprintf(stderr, "%s: ", #x); PyObject_Print((PyObject*)(x), stderr, 0); fputc('\n', stderr); }
// c function to be called from python script via ctypes.
void instantiate() {
PyObject* instance = PyObject_CallObject((PyObject*)&PyLong_Type, NULL);
LOGPY(instance);
}
int main(int argc, char* argv[]) {
Py_Initialize();
instantiate(); // works fine
// run a script that calls instantiate() via ctypes.
FILE* scriptFile = fopen("emb.py", "r");
if (!scriptFile) {
fprintf(stderr, "ERROR: cannot open script file\n");
return 1;
}
PyRun_SimpleFileEx(scriptFile, scriptPath, 1); // close on completion
return 0;
}
然后我使用PyRun_SimpleFileEx运行一个python脚本。它看起来运行得很好,但是当它通过ctypes调用instantiate()时,程序在PyObject_CallObject中出现错误:
^{pr2}$lldb输出:
instance: 0
Process 52068 stopped
* thread #1: tid = 0x1c03, 0x000000010000d3f5 Python`PyObject_Call + 69, stop reason = EXC_BAD_ACCESS (code=1, address=0x18)
frame #0: 0x000000010000d3f5 Python`PyObject_Call + 69
Python`PyObject_Call + 69:
-> 0x10000d3f5: movl 24(%rax), %edx
0x10000d3f8: incl %edx
0x10000d3fa: movl %edx, 24(%rax)
0x10000d3fd: leaq 2069148(%rip), %rax ; _Py_CheckRecursionLimit
(lldb) bt
* thread #1: tid = 0x1c03, 0x000000010000d3f5 Python`PyObject_Call + 69, stop reason = EXC_BAD_ACCESS (code=1, address=0x18)
frame #0: 0x000000010000d3f5 Python`PyObject_Call + 69
frame #1: 0x00000001000d5197 Python`PyEval_CallObjectWithKeywords + 87
frame #2: 0x0000000201100d8e emb`instantiate + 30 at emb.c:9
为什么只从ctypes调用instantiate()失败?这个函数只有在调用python库时才会崩溃,所以可能某些解释器状态被ctypes FFI调用所破坏?在
多亏了阿明·里戈的提示。问题是通过ctypes.CDLL()创建在调用本机代码时释放GIL的函数。据我所知,这意味着为了让本机函数调用python代码,它需要首先使用pythoncapi获取GIL。在
更简单的选择是使用ctypes.PyDLL(),它不释放GIL(它还检查python错误标志)。文档说,“因此,这只对直接调用pythoncapi函数有用。”我的代码更间接,因为我有Python代码调用我自己的C函数,然后调用pythoncapi,但问题是一样的。在
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