From your other answers, I think I can help you further by showing the same numbers as above in binary form: 

  DB<5> printf "%04b", $mode ;
1000000111101101
  DB<6> printf "%04b",  07777;
111111111111
  DB<7>

[download]

So, 33261 is 1000000111101101 in binary representation, and 07777 is 111111111111 (or the equivalent 0000111111111111) in binary representation. If we now do manually a bitwise and (&) between these two numbers, we do this:

  1000000111101101 &
  0000111111111111
  ================
= 0000000111101101

[download]

We get a 1 in the result only at places where we have a 1 in both numbers above. This is in effect applying a mask to cancel out the four binary digits on the left in the original number. The same operation under the debugger: 

  DB<8> $c = $mode & 07777;

  DB<9> printf "%04b", $c;
111101101
  DB<10>

[download]

If I now print that same $c number in its octal representation: 

  DB<10> printf "%04o", $c;
0755
  DB<11>

[download]

Is it clearer in your mind now?

In reply to Re^3: What does ">>" do? (And other stat questions) by Laurent_R
in thread What does ">>" do? (And other stat questions) by three18ti

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