This problem considers a linked list that has been created in memory. We know the linked list consists of 2 nodes, we know one of the nodes resides at address 0x00007fffffffea40 and has the following declaration in C. struct lnode { struct lnode *prev; struct lnode *next; char first; char *word; int length; }; We decide to look at the node in gdb and see the following: (gdb) x/4xg 0x555555554774 0x555555554774: 0x6f72203136325343 0x6c65480021736b63 0x555555554784: 0x3b031b0100216f6c 0x000000060000003c (gdb) x/14xg $rsp 0x7fffffffea00: 0x00007fffffffeb58 0x00000001756e6547 0x7fffffffea10: 0x00007fffffffea40 0x0000000000000000 0x7fffffffea20: 0x0000000000000043 0x0000555555554774 0x7fffffffea30: 0x000000000000000c 0x000055555555473d 0x7fffffffea40: 0x0000000000000000 0x00007fffffffea10 0x7fffffffea50: 0x0000000000000048 0x0000555555554781 0x7fffffffea60: 0x0000000000000006 0x2248a3a88ddbb600 Fill-in the following blanks for the node starting at address 0x00007fffffffea40. - The field prev is stored at address Blank 1 (copy the value and paste it here including 0x) - The field word is stored at address Blank 2 (copy the value and paste it here including 0x) - The value stored in the field length is Blank 3 (number in decimal) - the fourth ASCII character referenced by word is Blank 4 (you can use an ASCII table) - The value stored in the field prev is Blank 5 (copy the value and paste it here including 0x)
This problem considers a linked list that has been created in memory.
We know the linked list consists of 2 nodes, we know one of the nodes resides at address 0x00007fffffffea40 and has the following declaration in C.
struct lnode {
struct lnode *prev;
struct lnode *next;
char first;
char *word;
int length;
};
We decide to look at the node in gdb and see the following:
(gdb) x/4xg 0x555555554774
0x555555554774: 0x6f72203136325343 0x6c65480021736b63
0x555555554784: 0x3b031b0100216f6c 0x000000060000003c
(gdb) x/14xg $rsp
0x7fffffffea00: 0x00007fffffffeb58 0x00000001756e6547
0x7fffffffea10: 0x00007fffffffea40 0x0000000000000000
0x7fffffffea20: 0x0000000000000043 0x0000555555554774
0x7fffffffea30: 0x000000000000000c 0x000055555555473d
0x7fffffffea40: 0x0000000000000000 0x00007fffffffea10
0x7fffffffea50: 0x0000000000000048 0x0000555555554781
0x7fffffffea60: 0x0000000000000006 0x2248a3a88ddbb600
Fill-in the following blanks for the node starting at address 0x00007fffffffea40.
- The field prev is stored at address Blank 1 (copy the value and paste it here including 0x)
- The field word is stored at address Blank 2 (copy the value and paste it here including 0x)
- The value stored in the field length is Blank 3 (number in decimal)
- the fourth ASCII character referenced by word is Blank 4 (you can use an ASCII table)
- The value stored in the field prev is Blank 5 (copy the value and paste it here including 0x)
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