Performing AND operation on two decimal numbers (Octave code)

function  And_conversion(dec1,dec2)
i=1;
j=1;
dec_count=1;

for dec_count=1:1:2
  if dec_count==1
    dividend=dec1;
    i=i;
  end
  if dec_count==2
    dividend=dec2;
    i=j;
  end
  if dividend==0
    binary_value=0
  else
q_temp=dividend;
    while (q_temp>1)
      r_temp=mod(q_temp,2);
      binary_matrix(i,1)=q_temp;
      binary_matrix(i,2)=r_temp;
   
      q_temp=floor(dividend/2);
      dividend=q_temp;
      i=i+1;
    end
   binary_matrix(i,2)=1;
   a=binary_matrix(:,2);
   b=a';
   binary_value=flip(b);
   end
 
  if dec_count==1
    A_binary_value=binary_value
  end
  if dec_count==2
    B_binary_value=binary_value
  end
end

% bits padding
  A_count=size(A_binary_value);
  A_count=A_count(1,2);
  B_count=size(B_binary_value);
  B_count=B_count(1,2);
  count=max(A_count,B_count)
  for net_count=count: -1 : 1
    if A_count == 0
      A_binary_value(1,net_count)=0;
    else
      A_binary_value(1,net_count)=A_binary_value(1,A_count);
      A_count=A_count-1;
    end
    if  B_count == 0
     B_binary_value(1,net_count)=0;
    else
     B_binary_value(1,net_count)=B_binary_value(1,B_count);
     B_count=B_count-1;
    end
 
  end
    A= A_binary_value(1,:)
    B= B_binary_value(1,:)
 
% Zero padding end

% Anding the two values
  for And_count=count:-1:1
    if A(1,And_count)==B(1,And_count);
      And_result(1,And_count)=A(1,And_count);
    else
      And_result(1,And_count)=0;
    end
  end
    And_result=And_result(1,:)
   
end

Flaws of the above code:

1. It will only work correctly if smaller number is entered as dec1 and greater number as dec2.
2. This program contains many loops so speed of evaluation will be slow.