合并排序中的C ++向量中的std :: bad_alloc错误

Question

I am trying to implement merge sort using vector in c++, i am getting following error - terminate called after throwing an instance of 'std::bad_alloc'. There were other solutions for bad_alloc but it didn't help.

problem is merging function-

#include<iostream>
#include<vector>
using namespace std;

void print(vector<int> v)
{
    for(int i=0; i<v.size(); i++)
        cout << v[i] << " ";
    cout << endl;
}

vector<int> merging(vector<int> left, vector<int> right)
{
    vector<int> result;
    while((int)left.size()>0 && (int)right.size()>0)
    {
        if((int)left.front()<=(int)right.front()){
            result.push_back(left.front());
            //left.erase(left.begin());
        }
        else{
            result.push_back(right.front());
            //left.erase(right.begin());
        }
    }
        while((int)left.size()>0){
            for(int j=0; j<(int)left.size(); j++)
                result.push_back(left[j]);
        }
        while((int)right.size()>0){
            for(int k=0; k<(int)right.size(); k++)
                result.push_back(right[k]);
        }
    cout << "check merging";
    return result;
}

int main()
{
    vector<int> a, b, result;
    a.push_back(38);
    a.push_back(27);
    b.push_back(43);
    b.push_back(3);
    cout << a.front() << endl;
    print(a);
    cout << endl;
    print(b);
    cout << endl;
    result = merging(a, b);
    print(result);
}

thank you

Answer 1

From the page on std::bad_alloc

Type of the exceptions thrown by the standard definitions of operator new and operator new[] when they fail to allocate the requested storage space.

You're running out of memory and an exception is being thrown when you can't allocate anymore.

Try taking a look at the exit conditions for your loops.

while((int)left.size()>0 && (int)right.size()>0)

This loop won't exit until both left and right are empty, however you're never changing their size (the erase calls are commented out).

The other two while loops have similar issues.

Answer 2

The first while loop never terminates. It keeps adding elements to result until the memory is exhausted.

There are other issues with your code: In the case of left.front() <= right.front you never remove the element you put in result from left, it is commented out. In that else, you do not remove from right either. The commented out code would remove from left , mixing an iterator from right into an erase from left , what is technically called "undefined behavior" meaning that your program is toast.

Also, why do you cast to int everywhere? that's so unnecessary

Even if you would erase from the front of left and right, you would use vectors for removing from the front, operations that are not supported efficiently by a vector. There is no need to do that, you can traverse left and right using iterators, checking instead of whether they are empty for whether the left iterator reached the end of left and whether the right iterator reached the end of right

Answer 3

You have an infinite loop.

while((int)left.size()>0 && (int)right.size()>0)
{
    if((int)left.front()<=(int)right.front()){
        result.push_back(left.front());
        //left.erase(left.begin());
    }
    else{
        result.push_back(right.front());
        //left.erase(right.begin());
    }
}

Is never going to end as you never remove any elements from left