我正在包装一些C库,我有一个函数,在某些情况下可能会导致分段错误。在这种情况下,我需要调用第二个函数,它将在那种情况下成功完成。
有人知道如何处理cython中的分段错误吗?
一个可能有帮助的简短示例(使用signal
):
示例.h(假设Cython扩展名为myext.pyx
)
// Header autogenerated by Cython when declaring "public api" functions
#include "../myext_api.h"
void handleFuncFailure(char *func1_name, char *func2_name);
void generateSegFault();
示例。c
#include <example.h>
#include <signal.h>
static char *func2name;
static void handler2(int sig)
{
// Catch exceptions
switch(sig)
{
case SIGSEGV:
fputs("Caught SIGSEGV: segfault !!n", stderr);
break;
}
int error;
// "call_a_cy_func()" is provided by "myext.pyx"
call_a_cy_func(func2name, &error);
exit(sig);
}
void handleFuncFailure(char *func1_name, char *func2_name) {
// Provided by autogenerated "myext_api.h" header
import_myext();
func2name = func2_name;
signal(SIGSEGV, handler2);
int error;
// "call_a_cy_func()" is provided by "myext.pyx"
call_a_cy_func(func1_name, &error);
}
void generateSegFault() {
*(int*) 0 = 0;
}
myext.pyx
# Helper function to call a function by its name
# "public api" enables to call this function from C side (See above)
cdef public api void call_a_cy_func(char* method, bint *error):
if (method in globals()):
error[0] = 0
globals()[method]();
else:
error[0] = 1
# Expose C functions
cdef extern from "src/example.h":
void handleFuncFailure(char *func1_name, char *func2_name)
void generateSegFault()
# The unreliable function
def func1():
print "hello1 ! Generating segfault..."
generateSegFault()
# The reliable function
def func2():
print "hello2 ! Running safe code..."
# To be called from the Cython extension inner code
cdef myHandleFuncFailure(f1, f2):
handleFuncFailure(f1, f2)
# To be called from Python source by importing "myext" module
def myHandleFuncFailure2():
myHandleFuncFailure("func1", "func2")
输出
你好!正在生成segfault。。。
捕获SIGSEGV:segfault!!
你好!正在运行安全代码。。。
我希望这能给你一些想法,至少。。。