将向量分配给结构的向量字段

Question

我不确定为什么会这样：

主.cpp：

int main(int argc, char * argv[]) {
    Pwm_Info tf_info;
    tf_info = get_pwm_info(library_data, motif_name);
}

根.cpp

        struct Pwm_Info {
    std::string motif_name;
    int width;
    std::vector < double > pwm;
    Pwm_Info(): motif_name(0),
    width(0),
    pwm(0) {}
}
TF_info;
Pwm_Info get_pwm_info(std::string library_data, std::string motif_name) {
    std::vector < double > double_list;
    strtk::parse(pwm_block, " nr", double_list);
    std::cout << double_list.size() << std::endl;
    Pwm_Info TF_info;
    TF_info.motif_name = motif_name;
    TF_info.width = n_width;
    std::vector < double > * pointer;
    std::vector < double > * pointer2;
    pointer = & TF_info.pwm;
    pointer2 = & double_list; * pointer = * pointer2;
    std::cout << TF_info.pwm[1] << std::endl;
    TF_info.pwm = double_list;
    return TF_info;
}

以前，我删除了指向TF_info和双列表的指针，只有以下行：

  TF_info.pwm = double_list;

。这导致了一个SegFault。创建指针（指针和指针 2）并将其相互分配有什么意义？

Answer 1

pointer = &TF_info.pwm;
pointer2 = &double_list;
*pointer = *pointer2;

和

TF_info.pwm = double_list;

完全等效！

---

唯一可能导致段错误的行是：

std::cout << TF_info.pwm[1] << std::endl;