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Home/ Questions/Q 9220883
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T03:27:07+00:00 2026-06-18T03:27:07+00:00

so I’m trying to copy over vectors of different lengths between MPI processes in

  • 0

so I’m trying to copy over vectors of different lengths between MPI processes in C++, namely taking vectors on all of the nodes and concatenating them into a new vector on node 0.

I have the following code, which does not return what I expected, driving me crazy, and causing trouble further down the line.

The code is this (abbreviated):

//previously summed all of numfrags to make _numFrag
//numfrags is a vector of the local sizes of _fragLoc
//_numFrag is the total of numfrags

MPI::COMM_WORLD.Barrier();
cout << _myid << "local numFrag = " << _fragLoc.size() << endl;
MPI::COMM_WORLD.Barrier();
for (unsigned i = 0; i < _fragLoc.size(); ++i)      cout << "fragloc(" << i << ") = " << _fragLoc[i] << endl;
MPI::COMM_WORLD.Barrier();

vector<int> outVector (_numFrag);
int displ[_numprocs]; 

if (_myid == 0) {
    double sum = 0;
    for (int i = 0; i < _numprocs; ++i) {
        displ[i] = sum;
        cout << _myid << " : " << i << " : " << sum << endl;
        sum += numfrags[i];
    }
}

MPI::COMM_WORLD.Barrier(); MPI::COMM_WORLD.Gatherv(&_fragLoc[0], numfrags[_myid], MPI::INT, &outVector[0], &numfrags[0], &displ[0], MPI::INT,0);

MPI::COMM_WORLD.Barrier();

if (_myid == 0) {
    cout << "X numFrag = " << _numFrag << endl;
    for (unsigned i = 0; i < _numFrag; ++i) cout << "outVector(" << i << ") = " << outVector[i] << endl;
}

Giving a simple example, I have a four-node run. Here are the variable inputs as pseudocode:

int _numprocs = 4;
vector<int> numfrags = {0,1,0,1};
vector<int> _fragLoc <node 0> = {};
vector<int> _fragLoc <node 1> = {12};
vector<int> _fragLoc <node 2> = {};
vector<int> _fragLoc <node 3> = {37};
int _numFrag = 2;

The output is:

2local numFrag = 0
3local numFrag = 1
0local numFrag = 0
1local numFrag = 1
fragloc(0) = 12
fragloc(0) = 37
0 : 0 : 0
0 : 1 : 0
0 : 2 : 1
0 : 3 : 1
0: after stage 2
X numFrag = 2
outVector(0) = 0
outVector(1) = 0

But I expected the individual fragLoc’s to be put together into outVector and this isn’t happening. Any advice? I’ll clean up the barriers when I’m done debugging.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T03:27:08+00:00Added an answer on June 18, 2026 at 3:27 am

    As far as I can tell, the code above works as expected.

    #include <iostream>
    #include <mpi.h>
    #include <vector>
    
    using namespace std;
    
    int main(int argc, char **argv) {
    
        MPI::Init(argc, argv);
        int _myid = MPI::COMM_WORLD.Get_rank();
        int _numprocs = MPI::COMM_WORLD.Get_size();
    
        vector<int> _fragLoc;
        switch(_myid) {
            case 0: break;
            case 1: _fragLoc.push_back(12); break;
            case 2: break;
            case 3: _fragLoc.push_back(37); break;
        }
    
        int locNumFrag = _fragLoc.size();
        cout << _myid << "local numFrag = " << locNumFrag << endl;
    
        MPI::COMM_WORLD.Barrier();  // for printing
    
        vector<int> numfrags(_numprocs);
        MPI::COMM_WORLD.Allgather(&locNumFrag, 1, MPI::INT, &numfrags[0], 1, MPI::INT);
    
        int _numFrag = 0;
        for (int i=0; i<_numprocs; i++) 
            _numFrag += numfrags[i];
    
        for (unsigned i = 0; i < _fragLoc.size(); ++i)
            cout << "fragloc(" << i << ") = " << _fragLoc[i] << endl;
    
        MPI::COMM_WORLD.Barrier(); // for printing
    
        vector<int> outVector (_numFrag);
        int displ[_numprocs]; 
    
        if (_myid == 0) {
            double sum = 0;
            for (int i = 0; i < _numprocs; ++i) {
                displ[i] = sum;
                cout << _myid << " : " << i << " : " << sum << endl;
                sum += numfrags[i];
            }
        }
    
        MPI::COMM_WORLD.Gatherv(&_fragLoc[0], numfrags[_myid], MPI::INT, &outVector[0], &numfrags[0], &displ[0], MPI::INT,0);
    
        if (_myid == 0) {
            cout << "X numFrag = " << _numFrag << endl;
            for (unsigned i = 0; i < _numFrag; ++i) cout << "outVector(" << i << ") = " << outVector[i] << endl;
        }
    
        MPI::Finalize();
        return 0;
    }
    

    Running gives

    $ mpirun -np 4 ./gatherv
    0local numFrag = 0
    1local numFrag = 1
    fragloc(0) = 12
    2local numFrag = 0
    3local numFrag = 1
    fragloc(0) = 37
    0 : 0 : 0
    0 : 1 : 0
    0 : 2 : 1
    0 : 3 : 1
    X numFrag = 2
    outVector(0) = 12
    outVector(1) = 37
    
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