c - cython 分段错误处理

Question

我正在包装一些 C 库，并且我有一个函数，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，在这种情况下它将成功完成。

有谁知道如何处理 cython 中的分段错误？

Answer 1

一个可能有帮助的简短示例（使用signal）：

example.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();

例子.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;
}

我的ext.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1！正在生成段错误...

抓到 SIGSEGV: segfault !!

你好2！运行安全代码...

我希望这能给一些想法，至少......