about Nested Structures C++

Question

I want to learn c++, and here is my beginner question.

#include <iostream>
#include <string>

using std::cout;
using std::endl;

struct student{
    struct address{
        int no;
        std::string city;    //problem is here
    };

    char grade;
    int num;
    address *addr;
};

int main(){

    student st1;
    student *pSt1 = &st1;

    pSt1->grade = 'A';
    pSt1->num = 1234;

    pSt1->addr->no = 123;
    pSt1->addr->city = "Imladris";

    return 0;
}

It crashes, but if I change std::string city like this:

struct student{
    struct address{
        std::string city;   // here
        int no;
    };

    char grade;
    int num;
    address *addr;
};

It doesn't crash and returns 0... No errors nor warnings in either case. Maybe, there is no need to use a pointer here but as I said, I am learning c++. This is a sample code about pointers/structures.

I know how to fix it but I'd like to understand why it breaks. I change the declaration order of string city and it doesn't crash. Why?

And explain what "works" and "does not work" mean (compile error, program crash, unexpected results...) — SJuan76, Aug 26 '16 at 12:56
What does it mean that "it works or not" ?? show us compiler errors or sth like this — Krzysztof Bargieł, Aug 26 '16 at 12:57
Why are you using pointer here `address *addr;` and here `student *pSt1 = &st1;` actually? — Holt, Aug 26 '16 at 13:00
i mean first code crashes, but if i change its declaration order, it doesn't crash. Process returned 0. Also there is no error or warning in either case. — justaquestion, Aug 26 '16 at 13:08
@justaquestion Learn how to use your compiler. Any decent C++ compiler will give you a warning if you use the `Wall` option. — Nelfeal, Aug 26 '16 at 13:16
@justaquestion I second @Nelxiost on this, when programming in C or C++, *always* compile with `-Wall` and `-Wextra`. Warnings might seem annoying because "I can just ignore them" but they almost always point out that your code won't work as expected and tell you where to look. — sokkyoku, Aug 26 '16 at 13:41

score 1 · Accepted Answer · edited May 23 '17 at 10:29

1

You just came across the term Undefined behavior. The end result of your code is compiler and platform dependent. Dereferencing a pointer which value is not specified is (as mentioned here) considered to be one of these cases where you are not able to predict the outcome of an operation.

edited May 23 '17 at 10:29

Community

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answered Aug 26 '16 at 13:43

Adrian Jałoszewski

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score 0 · Answer 2 · edited May 23 '17 at 12:34

0

Keep your nested structure but add this to your member variable declaration:

address *addr = new address();

Or a better way is to use it as user @tobi303 mentioned: Don't declare it as a pointer, just an instance of address instead:

address addr;

But use it this way:

pSt1->addr.no = 123; // set
pSt1->addr.city = "Imladris"; // set

edited May 23 '17 at 12:34

Community

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answered Aug 26 '16 at 12:57

Khalil Khalaf

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would be worth mentioning that there is absolutely no need to make the member `addr` a pointer – 463035818_is_not_an_ai Aug 26 '16 at 13:00
@tobi303 it's mentioned in the question - "Maybe, there is no need to use pointer here but as i said, i learn c++". It's probably an exercise from a book. – Adrian Jałoszewski Aug 26 '16 at 13:49
1

@AdrianJałoszewski I think when OP says "maybe no need for pointer" he refers to `pSt1` which is also not needed. The fact that `address` does not need to be a pointer is maybe less obvious. If it is taken from a book, well.. also in books you can find nonsense sometimes – 463035818_is_not_an_ai Aug 26 '16 at 13:53
There are a very few need to use `new` in C++. This statement: `address *addr = new address();` is an invite to memory leak, manual memory management, and it is not an idiomatic way to deal with pointers with ownership semantics. – Amadeus Aug 26 '16 at 14:04

about Nested Structures C++

2 Answers2