用C语言包装alloca函数

Question

是否可以将C函数alloca包装成“另一个”？（当然只有宏）

类似于：

#define printStackFrameAddr(_fnc)   \
    printf("%-20s: %p \n", _fnc, __builtin_frame_address(0))


void _my_init(void *mem)
{

}

void *notSafeAlloca(size_t num)
{
    void *mem;

    printStackFrameAddr("notSafeAlloca");

    mem = alloca(num);

    _my_init(mem);

    return mem;
}

#define safeAlloca(_num)            \
    ({                              \
        void *ret = alloca(_num);   \
        _my_init(ret);              \
        printStackFrameAddr("safeAlloca");      \
    ret;})

int main(int argc, const char * argv[])
{
    printStackFrameAddr("main");


    {   /* <- creates no new stack frame ? */
        int *array = notSafeAlloca(sizeof(* array) * 4);

        printStackFrameAddr("main");

        printf("\t%-20s: %p\n", "array", array);
    }

    {   /* <- creates no new stack frame ? */
        int *array = safeAlloca(sizeof(* array) * 4);

        printStackFrameAddr("main");

        printf("\t%-20s: %p\n", "array", array);
    }

    return 0;
}

main                : 0x7fff5fbff8d0 
notSafeAlloca       : 0x7fff5fbff860 
main                : 0x7fff5fbff8d0 
        array               : 0x7fff5fbff820
safeAlloca          : 0x7fff5fbff8d0 
main                : 0x7fff5fbff8d0 
        array               : 0x7fff5fbff888

那么，上述做法安全吗？

Answer 1

我找到了解决问题的办法：）

void *_my_alloca(void *memory)
{
    // init memory & stuff

    return memory;
}

#define my_alloca(_size)   _my_alloca(alloca(_size))

int main(int argc, const char **argv)
{
    int *array = my_alloca(sizeof(* array) * 4);
}