CODEWITHSUNDEEP
c++vectorstructerase-remove-idiom
djankooo
djankooo

Reputation: 21

How to erase value from vector of struct using erase-remove idiom?

How to erase all values from vector of struct, where struct value k equals to 0?

struct tabuRecord {
int x;
int y;
int k;
tabuRecord( int x, int y, int k)
    : x(x), y(y), k(k){}
};  

vector <tabuRecord> tabu;

v.insert(v.begin(), tabuRecord(1, 2, 3));
v.insert(v.begin(), tabuRecord(4, 5, 0));
v.insert(v.begin(), tabuRecord(7, 8, 9));
v.insert(v.begin(), tabuRecord(10, 11, 0));

I have tried to 

tabu.erase(std::remove(tabu.begin(), tabu.end(), tabu.k=0), tabu.end());

and

tabu.erase(std::remove(tabu.begin(), tabu.end(), tabuRecord.k=0), tabu.end());

Upvotes: 0

Views: 309

Answers (3)

Blaze
Blaze

Reputation: 16876

Try something like this:

tabu.erase(
    std::remove_if(tabu.begin(), tabu.end(), [valueToErase](const tabuRecord & t) {
    return (t.x==valueToErase.x) && (t.y == valueToErase.y) && (t.k == valueToErase.k);
}), tabu.end());

This uses a lambda that returns true if the three fields are equal, and it removes all values where this is the case.

Here's a full example:

#include <vector>
#include <algorithm>
#include <iostream>

int main(int argc, char **argv)
{
    tabuRecord valueToErase(1, 2, 3); // example value to remove

    tabu.push_back({ 1, 2, 3 });
    tabu.push_back({ 4, 5, 6 });
    tabu.push_back({ 1, 2, 3 });
    tabu.push_back({ 7, 8, 9 });

    tabu.erase(
        std::remove_if(tabu.begin(), tabu.end(), [valueToErase](const tabuRecord & t) {
        return (t.x==valueToErase.x) && (t.y == valueToErase.y) && (t.k == 
        valueToErase.k);
    }), tabu.end());

    for (tabuRecord t : tabu) {
        std::cout << "x: " << t.x << " y: " << t.y << " k: " << t.k << std::endl;
    } // print all entries to verify that the correct ones were removed

    return 0;
}

Also, there's an error in your construtor, you probably wanted this instead of setting all fields to the same value:

: x(x), y(y), k(k) {}

Upvotes: 3

super
super

Reputation: 12968

std::remove needs a tabuRecord to match against, so you need to do something like.

tabuRecord value_to_remove(1,2,3);
tabu.erase(std::remove(tabu.begin(), tabu.end(), value_to_remove), tabu.end());

If you want to remove only based on the k member, you need to use std::remove_if and pass an appropriate function to match for it.

auto match_func = [](const tabuRecord& obj) { return obj.k == 2; };
tabu.erase(std::remove_if(tabu.begin(), tabu.end(), match_func), tabu.end());

Upvotes: 4

Matthieu Brucher
Matthieu Brucher

Reputation: 22023

I guess what you want to do is to remove all objects that have k==0, so create a lambda for that:

tabu.erase(
    std::remove_if(tabu.begin(), tabu.end(),[](const tabuRecord& t){return t.k == 0;}),
    tabu.end());

std::remove cannot work because it's not one value that you want to remove, but all values with a specific pattern, which is what std::remove_if does.

Upvotes: 6

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