onsider memory allocation for a process which has the follow

onsider memory allocation for a process which has the following segment table:

For each of the following logical addresses, determine the physical address or indicate if a segmentation fault occurs.

(a) 0, 2198

(b) 2, 2156

(c) 1, 3530

(d) 3, 2444

(e) 0, 2222

Starting Address Length (bytes) 2660 3752 2224 2996 248 422 198 604

Solution

a)2660+2198=4858 which is greater than the size 248 so segmentation fault occurs.i.e(2660+248=2908) and 4858>2908 which is a trap to operating system.

b)2224+2156=segmentation fault

c)3752+3530=segmentation fault\'

d)2996+2444=segmentation fault

e)2660+2222=segmentation fault