#include<stdio.h>
main()
{
int a[3];
int *I;
a[0]=100;a[1]=200;a[2]=300;
I=a;
Printf(“%d\n”, ++*I);
Printf(“%d\n”, *++I);
Printf(“%d\n”, (*I)--);
Printf(“%d\n”, *I);
}
what is the o/p
a. 101,200,200,199
b. 200,201,201,100
c. 101,200,199,199
d. 200,300
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Answer Posted / r.gopala krishnan (gk)
Explanation:
now 'I' variable pointing the base address of the
Array......
1.printf("%d\n",++*I); //I=a[0], bcos I is pointing the
base address. first Increamenting the value so,a[0]=100
become an ( a[0]=101 ).
2.printf("%d"\n,*++I); //This Increment will increment
the address not value. so, now ( I=a[1]=200 ).
3.printf("%d\n",*I--); // now I=a[1]=200 so value will
not change.
4.printf("%d\n",*I); //Now also I=a[1] pointing here only
but a[1]=199, bcos in previous printf after executing we
decrementing the value. a[1]=199......
ANS: a)101,200,200,199
| Is This Answer Correct ? | 12 Yes | 0 No |
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