CODEWITHSUNDEEP
c++dictionary
Learning is a mess
Learning is a mess

Reputation: 8277

Update map in-place

I have a piece of code manipulating a map<int,int> object. Let's say that at some point I want to double all the entries, keys and values. Is a there a way to do it in place, without creating a new map, looping and entering updated entries?

My concern is saving memory space.

Upvotes: 2

Views: 211

Answers (2)

JuniorCompressor
JuniorCompressor

Reputation: 20015

You can traverse the map in reverse order of key for positive keys, in order to circumvent side effects, and create the new key/value pairs. You can do the equivalent for negative keys.

#include <cstdio>
#include <algorithm>
#include <map>

using namespace std;

int main() {
  map<int, int> m = {
    {10, 20},
    {-5, 23},
    {-10, 7},
    {20, 30},
    {15, 21},
    {18, 2},
  };

  for (auto it = m.begin(); it != m.end() && it->first < 0; it = m.erase(it)) {
    m[it->first * 2] = 2 * it->second;
  }

  for (auto it = m.rbegin(); it != m.rend() && it->first > 0; it++) {
    m[it->first * 2] = 2 * it->second;
    m.erase(----it.base());
  }

  for (auto &p: m) {
    printf("%d, %d\n", p.first, p.second);
  }
}

Output:

-20, 14
-10, 46
20, 40
30, 42
36, 4
40, 60

Upvotes: 1

Humam Helfawi
Humam Helfawi

Reputation: 20264

Doubling the values is possible like this:

std::map<int,int> my_map;
for(auto& item:my_map){
    item.second*=2; // double all values
}

However, it is not possible to double the Key since item is from the type std::pair<const int,int>. Notice the const for the Key.

Suggestion:

I think std::map is not best container for this case. Try this way:

std::vector<std::pair<int,int>> my_simi_map;
for(auto& my_simi_map){
    item.first*=2; // double all keys
    item.second*=2; // double all values
}

Edit:

My concern is saving memory space.

If it is just memory problem then you can pick an item from the map and insert a doubled version of it in the new map and directly delete it from the old map. In this case you will loose the size of just one element of the map not a whole another map.

Upvotes: 2

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