main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
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Answer / susie
Answer :
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0
1
2
3
4
100 102 104 106 108
p
100
102
104
106
108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr
becomes 1002, if scaling factor for integer is 2 bytes. Now
ptr – p is value in ptr – starting location of array p,
(1002 – 1000) / (scaling factor) = 1, *ptr – a = value at
address pointed by ptr – starting value of array a, 1002 has
a value 102 so the value is (102 – 100)/(scaling factor) =
1, **ptr is the value stored in the location pointed by
the pointer of ptr = value pointed by value pointed by 1002
= value pointed by 102 = 1. Hence the output of the firs
printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in
ptr by scaling factor, so it becomes1004. Hence, the outputs
for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in
ptr by scaling factor, so it becomes1004. Hence, the outputs
for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the
value pointed by the value is incremented by the scaling
factor. So the value in array p at location 1006 changes
from 106 10 108,. Hence, the outputs for the fourth printf
are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4,
**ptr = 4.
| Is This Answer Correct ? | 26 Yes | 3 No |
Answer / rashmi
the logic given by u is wierd.......!!!!!!!!!!!!!!!!!!!
| Is This Answer Correct ? | 10 Yes | 8 No |
x=2 y=3 z=2 x++ + y++; printf("%d%d" x,y);
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