1

The two vectors the user enters will always be in alphabetical order, and the function merge_items places those values in one vector seeing which one comes before the other by using the < operator, the code initially gave a segmentation fault and at certain times, it doesn't show the last element. 

#include <iostream>
#include <string>
#include <vector>

using namespace std;

void merge_items(vector<string>& a1,vector<string>& b1,vector<string>& merged);

int main(){
vector<string> v1,v2;
string a,b;
int n1,n2;
cout << "How many values for v1? " << endl;
cin >> n1;
for(int i = 0;i < n1;i++){
    cin >> a;
    v1.push_back(a);
}
cout << "How many values for v2? " << endl;
cin >> n2;
for(int i = 0;i < n2;i++){
    cin >> b;
    v2.push_back(b);
}

vector<string> merge;

merge_items(v1, v2, merge);

for(int i = 0;i < merge.size();i++){
    cout << merge[i] << endl;
}

return 0;
}

void merge_items(vector<string>& a1,vector<string>& b1,vector<string>& merged){         int i1 = 0,i2 = 0;
string temp;
while(i1+i2 < (a1.size()-1+b1.size()-1)){
    if(a1[i1] < b1[i2]){
        temp = a1[i1];
        merged.push_back(temp);
        i1++;

    }else{
        temp = b1[i2];
        merged.push_back(temp);
        i2++;

    }
}
}
ApproachingDarknessFish
  • 14,133
  • 7
  • 40
  • 79
Afr0
  • 41
  • 1
  • 10
  • You can push back `a1[i1]` directly, there's no need to copy it into a local variable. I suppose this is an exercise, otherwise I would be tempted just to append both vectors to merged and then sort it. – Neil Kirk Mar 12 '15 at 00:57
  • possible duplicate of [Merging vectors](http://stackoverflow.com/questions/22894919/merging-vectors) – Christophe Mar 12 '15 at 01:05

1 Answers1

2

This is the appropriate way to merge:

std::merge(a1.begin(), a1.end(),
           b1.begin(), b1.end(),
           std::back_inserter(merged));

As far as what's wrong with your solution. A couple things.

First, once you reach the end of one of the two vectors, you need to stop comparing against that vector and just copy whatever elements are left from the other vector. So you need to compare i1 against a1.size(), separately from your comparison of i2 against b1.size(). With what you're doing now, when you reach the end of one vector, you continue comparing against out of bounds elements from that vector, which is undefined behavior, and likely the cause of your segmentation faults.

Second, you don't need to be subtracting 1 from the size of the vectors. The way you're doing it will leave you with a merged vector which has 2 fewer elements than the combined sizes of the source vectors.

Benjamin Lindley
  • 101,917
  • 9
  • 204
  • 274
  • Cool I didn't know about that merge function. – Neil Kirk Mar 12 '15 at 01:14
  • Can we use this function merge two source vector into one of source vector? If no, does STL has a function to do that?Thanks – Ron Tang Mar 12 '15 at 01:44
  • @RonTang: No, you cannot, and no, it does not. There is [`std::inplace_merge`](http://en.cppreference.com/w/cpp/algorithm/inplace_merge) though, which allows you to merge two adjacent sorted ranges in place. In other words, if you have, for example, a vector with the first half sorted, and the second half sorted, you can merge them so the whole vector is sorted. – Benjamin Lindley Mar 12 '15 at 01:50