title: Git Merge --no-ff makes copy of commits
body: <p>I had a curious git behaviour today, and I want to ask if this is normal. Now, this is the situation (new commits are added to the bottom):</p>
  
  <pre><code>br1  br2
    A
      \ 
       B
       C
       D
  </code></pre>
  
  <p>Now, when I make </p>
  
  <pre><code>git checkout br1
  git merge --no-ff br2
  </code></pre>
  
  <p>It becomes like this (please read up to down; newest commit is 'E')</p>
  
  <pre><code>br1  br2
    A
    | \ 
    |  B
    |  C
    |  D
    | /
    E
  </code></pre>
  
  <p>OK.. Now, weird thing is; now I call "git status"; it says I'm 4 commits ahead of the remote branch. How is this happening? Shouldn't I be only one commit ahead?</p>
  
  <p>And the peculiar thing is, when I check from the Stash (the Git Web UI, basically) it confirms this "4 commits" status and I see the same commits (B C and D) under "br1" and "br2" both... </p>
  
  <p>I assumed when I used "--no-ff" parameters, the commits at "br2" (B C D) won't be copied to the "br1" and only merge commit created. Am I mistaken here?</p>