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===Solution=== | ===Solution=== | ||
Assume start location of arr s 1000. | Assume start location of arr s 1000. | ||
| − | So, | + | So, the first two elements of the array will be stored in memory as follows: |
| − | 1000: 0 0 0 0 0 0 0 2 ( | + | 1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes) |
1004: 0 0 0 0 0 0 0 3 | 1004: 0 0 0 0 0 0 0 3 | ||
<syntaxhighlight lang="c"> int main() {
GATECSE int arr[3] = {2, 3, 4};
char *p;
p = (char*)arr;
printhex(p);
printf("%p %d ",p, *p);
p = p+1;
printhex(p);
printf("\n%p %d\n",p,*p);
return 0;
}
</syntaxhighlight>
(A) 2 3
(B) 2 0
(C) 1 0
(D) Garbage value
Assume start location of arr s 1000. So, the first two elements of the array will be stored in memory as follows:
1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes) 1004: 0 0 0 0 0 0 0 3
Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.
1001: 0 0 0 0 0 0 0 0
So, output will be 0.
<syntaxhighlight lang="c"> int main() {
int arr[3] = {2, 3, 4};
char *p;
p = (char*)arr;
printhex(p);
printf("%p %d ",p, *p);
p = p+1;
printhex(p);
printf("\n%p %d\n",p,*p);
return 0;
}
</syntaxhighlight>
(A) 2 3
(B) 2 0
(C) 1 0
(D) Garbage value
Assume start location of arr s 1000. So, the first two elements of the array will be stored in memory as follows:
1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes) 1004: 0 0 0 0 0 0 0 3
Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.
1001: 0 0 0 0 0 0 0 0
So, output will be 0.