Passing an object pointed to by an iterator by reference to a function C++

Question

I though that I understood iterators and addressing etc. but obviously not. See my below code below which is purely an example.

I need to be able to pass by pointer or reference each structure of mystructs to MyFunc(). The function should be able to update the actual structure that is passed, and not by copy or value.

I receive the compiler error :

error: cannot convert 'MY_STRUCT' to 'MY_STRUCT*' for argument '1' to 'void MyFunc(MY_STRUCT*)'

If I just pass the iterator address, this also doesn't work.

What is the correct way to do this. Thanks in advance.

typedef struct
{
   int var1;
   int var2;
   std::string name;

}MY_STRUCT;

std::list<MY_STRUCT> mystructs;

void MyFunc(MY_STRUCT*)
{
    // Do something
}


// populate the list with structs etc.. not included here
//.....


for (std::list<MY_STRUCT>::iterator it = mystructs.begin();it != mystructs.end(); ++it)
{
     MyFunc(*it);
}

Off topic but... `it != mystructs.begin()` should probably be `it != mystructs.end()`. — G.M., Mar 31 '20 at 14:42

score 2 · Answer 1 · answered Mar 31 '20 at 14:42

2

Passing by reference in C++ is done with:

void MyFunc(MY_STRUCT&)
{
    // Do something
}

So your call would be correct, what you currently want is to pass the pointer, which you can do with dereferencing the dereferenced iterator (by passing the address of the dereferenced object):

void MyFunc(MY_STRUCT*)
{
    // Do something
}


// populate the list with structs etc.. not included here
//.....
int main() {

    for (std::list<MY_STRUCT>::iterator it = mystructs.begin();it != mystructs.begin(); ++it)
    {
         MyFunc(&*it);
    }
}

answered Mar 31 '20 at 14:42

RoQuOTriX

2,871
14
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2

Or can I just have my function accept a MY_STRUCT& , then pass *it ... will I still get pass by reference then in this case? Thanks – Engineer999 Mar 31 '20 at 14:45
Yes you are right. If you declare the parameter as `PARAM&` it is "autoamtically" passed as reference in C++. In the function call you can't see, if is passed by value or reference directly, which I dislike a little bit in the C++ syntax – RoQuOTriX Mar 31 '20 at 14:47

score 0 · Answer 2 · answered Mar 31 '20 at 14:40

Your function requires a pointer, use & to get the address of something.

MyFunc(&*it);

*it returns a reference to the MY_STRUCT object, you need to use & to convert that reference to a pointer. This is normal, the fact that you are using iterators makes no difference at all.

The alternative (maybe better in C++) would be to convert your MyFunc function to take a reference instead of a pointer.

Passing an object pointed to by an iterator by reference to a function C++

2 Answers2