Vectors, structs and std::find

Question

Again me with vectors. I hope I'm not too annoying. I have a struct like this :

struct monster 
{
    DWORD id;
    int x;
    int y;
    int distance;
    int HP;
};

So I created a vector :

std::vector<monster> monsters;

But now I don't know how to search through the vector. I want to find an ID of the monster inside the vector.

DWORD monster = 0xFFFAAA;
it = std::find(bot.monsters.begin(), bot.monsters.end(), currentMonster);

But obviously it doesn't work. I want to iterate only through the .id element of the struct, and I don't know how to do that. Help is greatly appreciated. Thanks !

Great question, exactly what I needed. Not annoying at all:) — tomereli, Apr 05 '18 at 08:03

score 42 · Accepted Answer · edited Apr 23 '13 at 19:06

42

std::find_if:

it = std::find_if(bot.monsters.begin(), bot.monsters.end(), 
        boost::bind(&monster::id, _1) == currentMonster);

Or write your own function object if you don't have boost. Would look like this

struct find_id : std::unary_function<monster, bool> {
    DWORD id;
    find_id(DWORD id):id(id) { }
    bool operator()(monster const& m) const {
        return m.id == id;
    }
};

it = std::find_if(bot.monsters.begin(), bot.monsters.end(), 
         find_id(currentMonster));

edited Apr 23 '13 at 19:06

answered Feb 26 '09 at 10:41

Johannes Schaub - litb

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1

using boost, Perfect! – H'H Nov 14 '14 at 16:05
I wonder, if I have `{int, int}`, could I cast it to `long` and use normal find then? – Tomáš Zato Oct 20 '15 at 14:39
2

What is the meaning of "bot" in bot.monsters.begin()? – Semjon Mössinger Jan 07 '16 at 10:54
3

What is the datatype if "it"?? – nathan Oct 14 '16 at 17:35
1

good thing you gave an example without boost - we never use boost in embedded systems, it is huge. I just didn't understand why you inherited the functor object from std::unary_function - looks like it would work either way, won't it? appreciate your answer. – tomereli Apr 05 '18 at 07:50
1

@Hiyper - The type would be std::vector::iterator. Or you can use auto it = std::find_if... if you are using c++11 – tomereli Apr 05 '18 at 08:02
@tomereli IIRC it wouldn't work without it in C++03, but I think it will in C++11, which can figure out the return type itself using `decltype`. – Johannes Schaub - litb Apr 05 '18 at 23:24
1

@SemjonMössinger This was not explicitly stated, but `bot` should be a struct containing the following vector of structs: `std::vector monsters;` – Adam Erickson Jun 07 '18 at 01:05
unary_function is deprecated in C++11 and removed in C++17. – Darkproduct Oct 19 '18 at 17:01

score 26 · Answer 2 · answered Jan 08 '10 at 22:03

26

how about:

std::find_if(monsters.begin(), 
             monsters.end(), 
             [&cm = currentMonster]
             (const monster& m) -> bool { return cm == m; });

answered Jan 08 '10 at 22:03

2

Could someone who sees this walk me through what it's doing? Specifically the [&cm = currentMonster](const monster& m) -> bool { return cm == m; }); – 2kreate Nov 29 '14 at 19:57
7

This example uses a lambda function, which depends on C++11. `[&cm = currentMonster]` binds the variable `currentMonster` from the calling scope to a local reference in the lambda, called `cm`. Then `(const monster& m) -> bool` defines the signature of the lambda, taking one input parameter, `m`, and returning `bool`. The body of the lambda function is `{ return cm == m; }`, returning true if `cm` and `m` compare as equal. – Derek T. Jones Jun 25 '15 at 13:56
1

Finally I know how to properly bind local variables to lambdas. – Tomáš Zato Oct 20 '15 at 14:43
is the -> bool part necessary? it would compile either way, won't it? – tomereli Apr 05 '18 at 08:24
1

The `{ return cm == m; }` at the end requires an `operator==` defined in the struct (see @dirkgently answer), Without the `operator==`, change the return statement to `{ return cm.id == m.id; }` – Tzunghsing David Wong Mar 27 '19 at 01:18

Evgeny Lazin · Answer 3 · 2012-09-05T07:01:57.240

21

You need to write your own search predicate:

struct find_monster
{
    DWORD id;
    find_monster(DWORD id) : id(id) {}
    bool operator () ( const monster& m ) const
    {
        return m.id == id;
    }
};

it = std::find_if( monsters.begin(), monsters.end(), find_monster(monsterID));

edited Sep 05 '12 at 07:01

answered Feb 26 '09 at 10:40

Evgeny Lazin

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Nice answer, but there is a typo in the constructor. It should be ':' not ';' – Martin Sherburn Aug 24 '12 at 15:16
In addition to writing your own search predicate, you need to use `std::find_if` instead of `std::find`. – Christian Rau Aug 25 '12 at 16:34

score 9 · Answer 4 · answered Feb 26 '09 at 10:43

Take a look at the std::find template, the third parameter especially:

template<class InputIterator, class EqualityComparable>
InputIterator find(InputIterator first, InputIterator last,
               const EqualityComparable& value);

What is this EqualityComparable? Again from the documentation:

A type is EqualityComparable if objects of that type can be 
compared for equality using operator==, and if operator== is 
an equivalence relation.

Now, your type monster needs to define such an operator. If you don't the compiler generates one for you (as also the default ctor and the dtor) which does a memcmp sort of thing which doesn't work in your case. So, to use std::find first define a comparator function/functor that the algorithm can use to match your currentMonster i.e. something along the lines of:

 struct monster {
  // members
  bool operator==(const monster& l, const monster& r) const
  {
     return l.id == r.id;
  }
 };

Does this work? I did not have success because an operator definition inside a struct can only have 1 input — Kyle Heuton, Apr 15 '13 at 08:20
the concept is correct. this works for me: `bool operator==(const monster& r) const { return id == r.id; }` — Tzunghsing David Wong, Mar 27 '19 at 00:54

score 1 · Answer 5 · answered Jan 08 '10 at 22:07

1

or put the monsters in a map instead of a vector

or if they must be in a vector create an index map ie map of ID to vector index

answered Jan 08 '10 at 22:07

pm100

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score 0 · Answer 6 · answered Jan 07 '16 at 10:53

This is a complete sample based on the answer of Johannes Schaub (boost version).

#include <algorithm>
#include <boost/bind.hpp>

struct monster 
{
    DWORD id;
    int x;
    int y;
    int distance;
    int HP;
};

int main ()
{
    std::vector<monster> monsters;

    monster newMonster;
    newMonster.id    = 1;
    newMonster.x     = 10;
    monsters.push_back ( newMonster );

    newMonster.id    = 2;
    newMonster.x     = 20;
    monsters.push_back ( newMonster );

    newMonster.id    = 2;
    newMonster.x     = 30;
    monsters.push_back ( newMonster );

    DWORD monsterId = 2;

    std::vector< monster >::iterator it = std::find_if ( monsters.begin (), monsters.end (), 
        boost::bind ( &monster::id, _1 ) == monsterId );

    return 0;
}

score 0 · Answer 7 · answered Sep 18 '19 at 08:07

0

You can write a function as below:

monster* findMonster(DWORD currentMonster) {
    for (auto it = bot.monsters.begin(); it != bot.monsters.end(); it++) {
        if (it->id == currentMonster) {
            return &(*it);
        }
    }
    return NULL;
}

It returns a pointer to the stored node if it's found in the vector, otherwise returns NULL.

Please note that return it; won't work directly.

answered Sep 18 '19 at 08:07

Aditya Sharma

41
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How are you even compiling this? It fails on latest C++ and VS2019. – Mecanik Mar 05 '22 at 05:36

Vectors, structs and std::find

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