C Programming - Memory Allocation - Discussion
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Read more:"Loneliness is the most terrible poverty."
- Mother Teresa
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Point out the correct statement will let you access the elements of the array using 'p' in the following program?
#include<stdio.h>
#include<stdlib.h>
int main()
{
int i, j;
int(*p)[3];
p = (int(*)[3])malloc(3*sizeof(*p));
return 0;
}
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| [A]. |
for(i=0; i<3; i++)
{
for(j=0; j<3; j++)
printf("%d", p[i+j]);
}
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for(i=0; i<3; i++)
printf("%d", p[i]);
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for(i=0; i<3; i++)
{
for(j=0; j<3; j++)
printf("%d", p[i][j]);
}
| | [D]. |
for(j=0; j<3; j++)
printf("%d", p[i][j]);
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Answer: Option E
Explanation:
No answer description available for this question.
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Bhavi said:
(Sun, Nov 14, 2010 01:21:58 PM)
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| Please give explanation to this program. |
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Gopi said:
(Fri, Jan 21, 2011 10:22:36 AM)
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Here, int(*p)[3]; //its like p[3]
p = (int(*)[3])malloc(3*sizeof(*p));
and again we are allocating more memory to it so like p[3][3] . |
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Teju said:
(Sat, Aug 20, 2011 11:32:19 AM)
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Does
p = (int(*)[3])malloc(3*sizeof(*p));
allocates 54 bytes of memory ? Anyone please explain. |
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Pragya said:
(Fri, Oct 7, 2011 04:05:00 AM)
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@ Teju
Yes. It allocates 54 bytes because
sizeof(*p)=6 bytes
So 3 * 6 = 18 bytes
and then int (*)[3] points to an array of 3 blocks each having 18 bytes.
So p = (18 + 18 + 18) = 54 bytes. |
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Tofik Kacchi said:
(Tue, Mar 20, 2012 03:03:17 PM)
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Here initially *p[3] will statically create an array of p[3].
Then again reallocating using malloc will create another array of size 3.
Therefore, its p[3][3].hence we require 2 nested for loop. |
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Saksham said:
(Fri, Jun 8, 2012 09:59:33 AM)
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| @pragya Type casting (int(*)[3]) will not increase the size. So it's just 18 bytes! |
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Bhargav said:
(Fri, Dec 7, 2012 01:16:35 AM)
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| Here type casting and it is array of pointers. So we have to access by using double pointer. So option C p[i][j] is similar to that. |
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