Sorting a Vector of Custom Objects by overloading <

Question

I am trying to sort a vector of nodes. I followed the advice from this thread and overloaded my
struct's < operator. However I am not getting a sorted list after sort is called.

struct node
{

    int frequency ;
    char data;

    bool operator < (const node& n1) const
    {
        return (frequency < n1.frequency);
    }
};

I call sort by the following:

vector<node*> test
//fill test with nodes
sort(test.begin(),test.end());

Output:

Presort data is: 1,1,2,3,3,1,2,1,1
Postsort data is: 3,2,1,1,1,1,2,1,3

What do left and right refer to? They may need to be updated after the sort. — 1201ProgramAlarm
– 1201ProgramAlarm, Commented Oct 24, 2015 at 4:51
That was for other parts of the code not listed here. I removed them to reduce confusion — Matt
– Matt, Commented Oct 24, 2015 at 5:06

Sergey Kalinichenko · Accepted Answer · 2015-10-24 04:55:55Z

6

Since you are sorting a vector of pointers, but the operator applies to a struct, C++ ignores your operator < overload.

You can supply a custom comparer that calls your operator <, like this

std::sort(test.begin(), test.end(), [](const node* pa, const node* pb) {
    return (*pb) < (*pa);
});

or code the comparison straight into the lambda, and drop the unused overload of <, like this:

std::sort(test.begin(), test.end(), [](const node* pa, const node* pb) {
    return pb->frequency < pa->frequency;
});

answered Oct 24, 2015 at 4:55

Sergey Kalinichenko

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5 Comments

Ryan Over a year ago

Your first solution uses a pair of parenthesis for each pointer which is totally useless. Dereferencing a pointer does not require parenthesis.

Matt Over a year ago

I am attempting to get your solution to work, but as of yet, no luck

Sergey Kalinichenko Over a year ago

@Matt This requires C++11 or higher. VC does not have it; g++ use -std=c++11 option.

Matt Over a year ago

@dasblinkenlight I am using g++ with the -std=c++11 option

Matt Over a year ago

This works, it was user error on my part as to why it wasn't working for me. It's been a long night, thanks man.

doug · Accepted Answer · 2015-10-24 05:14:23Z

0

Easiest way to do this is to use lambdas:

sort(test.begin(),test.end(), [](const node &lhs, const node &rhs){return lhs->frequency < rhs->frequency;});

answered Oct 24, 2015 at 5:14

doug

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1 Comment

Matt Over a year ago

You need to change &lhs and &rhs to *lhs and *rhs. By doing those changes, I was able to get this to work immediately

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