e 4 mb times 5 equals 20 mb then the temporary pointer is aligned with the current tlb size thus the u201c if u201d statement on line 4 resolves to true the algorithm continues by resolving the u201c if u201d statement on line 5 in this regard the end pointer is 21 mb and the temporary pointer is 20 mb the difference in the end pointer and the temporary pointer is 1 mb because 1 mb is not greater than or equal to two times the current tlb size i e 8 mb the u201c if u201d statement on line 5 resolves to false and the algorithm begins the u201c for_each u201d loop again with the next largest tlb size which is 1 mb in line 4 the algorithm resolves the u201c if u201d statement by determining alignment for the current tlb size which at this point is 1 mb with the temporary pointer which is the 20 mb address because 20 mb is a multiple of 1 mb i e 1 mb times 20 equals 20 mb then the temporary pointer is aligned with the current tlb size thus the u201c if u201d statement on line 4 resolves to true the algorithm continues by resolving the u201c if u201d statement on line 5 in this regard the end pointer is 21 mb and the temporary pointer is 20 mb the difference in the end pointer and the temporary pointer is 1 mb because 1 mb is not greater than or equal to two times the current.