访问双结构指针，发生分段错误

Question

typedef struct
{
        int member;
} mystruct;

void myfunc(mystruct **data)
{
        mystruct *const *p;
        for(p = data; *p !=NULL; p++)
        {
                printf("hello\n");
        }
}

void main(int argc, char *argv[])
{
        myfunc(NULL);
}

tried with the above code am getting segmentation fault, its mainly something wrong in that for loop, how to just remove this segmentation fault in that for loop.... actually am learning this double pointer stuff, so i may be a little stupid in asking few question.... thanks in advance

Answer 1

The *p in the for loop dereferences the first pointer. But that pointer is NULL , as you call your function with myfunc(NULL); .

Answer 2

Whenever you dereference a nullpointer, you invoke undefined behaviour (eg a segmentation fault).

typedef struct
{
    int member;
} mystruct;

void myfunc(mystruct **data)
{
    mystruct *const *p;
    // this loop assumes data to be a valid pointer
    // to a NULL-terminated array!
    for(p = data; *p != NULL; p++)
    {
        printf("hello\n");
    }
}

int main()
{
    mystruct s;
    mystruct *arr[2];
    arr[0] = &s; // arr[0] points to s
    arr[1] = NULL; // null-terminator
    s.member = 13;
    myfunc(arr);
    // myfunc(NULL); // undefined behaviour

    return 0;
}

you can fix this by checking data in myfunc:

void myfunc(mystruct **data)
{
    mystruct *const *p;
    // first check data
    if(data != NULL)
    {
        // loop still assumes data to be a NULL-terminated array!
        for(p = data; *p != NULL; ++p)
        {
            printf("hello\n");
        }
    }
}
...
myfunc(NULL); // well defined as the pointer will be checked

if you just want to iterate over a range, consider the standard approach:

void myfunc_range(mystruct *begin, mystruct *end)
{
    mystruct const *it;
    for(it = begin; it != end; ++it)
    {
        printf("hello %d\n", it->member);
    }
}

int main()
{
    mystruct s;
    mystruct arr[2];
    s.member = 42;
    myfunc_range(&s, &s + 1); // iterate over a single element
    arr[0].member = 13;
    arr[1].member = 37;
    myfunc_range(arr, arr + sizeof(arr) / sizeof(*arr)); // iterate over the whole array
    myfunc_range(arr + i, arr + i + k); // iterate over elements arr[i..i+k-1]
    myfunc_range(NULL, NULL); // well defined as NULL == NULL (an empty range)
    // myfunc(arr, &s); // undefined behaviour as s is not part of the array arr

    return 0;
}

Answer 3

感谢您的所有答复，恰好现在在for循环的条件部分中访问NULL值时会导致分段错误。