C Programming - Arrays - Discussion
Discussion Forum : Arrays - Find Output of Program (Q.No. 2)
2.
What will be the output of the program ?
#include<stdio.h>
int main()
{
static int a[2][2] = {1, 2, 3, 4};
int i, j;
static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
for(i=0; i<2; i++)
{
for(j=0; j<2; j++)
{
printf("%d, %d, %d, %d\n", *(*(p+i)+j), *(*(j+p)+i),
*(*(i+p)+j), *(*(p+j)+i));
}
}
return 0;
}
Discussion:
78 comments Page 7 of 8.
Shubham said:
1 decade ago
Can anyone please explain static int a[2][2] = {1, 2, 3, 4}; ?
Also static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
Also static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
Aabid khan said:
1 decade ago
I could not get the logic of this question please explain.
Trupti said:
1 decade ago
You are absolutely right @Paul.
If
static int *p[] = {a, a+1,a+2};
then o/p is:
1 1 1 1
2 3 2 3
3 2 3 2
4 4 4 4
If
static int *p[] = {a, a+1,a+2};
then o/p is:
1 1 1 1
2 3 2 3
3 2 3 2
4 4 4 4
Paul said:
1 decade ago
This is a terrible and confusing example for anyone!
It is a 2-dimensional array so (a + 1) would point to the first element of the second row, which is 3, and (a + 2) would point to the third row but in this case there is no third row.
Only because a, a+ 1 and a + 2 are cast to integer pointers in the line : static int *p[] = {(int*)a, (int*)a+1, (int*)a+2}, this means they increments by the size of an integer, therefore the next element in the array instead of the next row.
Anyone else agree?
It is a 2-dimensional array so (a + 1) would point to the first element of the second row, which is 3, and (a + 2) would point to the third row but in this case there is no third row.
Only because a, a+ 1 and a + 2 are cast to integer pointers in the line : static int *p[] = {(int*)a, (int*)a+1, (int*)a+2}, this means they increments by the size of an integer, therefore the next element in the array instead of the next row.
Anyone else agree?
Rashmi said:
1 decade ago
Can anyone tell me.
static int a[2][2] = {1, 2, 3, 4};
int i, j;
static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
What is meant by last line?
static int a[2][2] = {1, 2, 3, 4};
int i, j;
static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
What is meant by last line?
Vinod said:
1 decade ago
Substitute p=1 and i and j as 00,01,10,11 in all the arrays condition in printf.
So that for i=0 and j=0, p=1(every time) we get it as 1111.
Repeat the procedure for all. The answer is correct only.
So that for i=0 and j=0, p=1(every time) we get it as 1111.
Repeat the procedure for all. The answer is correct only.
Sourabh said:
1 decade ago
What may be the answer if we solve it by as two dimensional array?
Reena said:
1 decade ago
@Anushu.
The answer correct is:
1111
2222
2222
3333
The given answer is correctly.
The answer correct is:
1111
2222
2222
3333
The given answer is correctly.
Charan said:
1 decade ago
a[2][2] is an array stores all elements which is 2D (2-rows,2-columns).
column 0 1
0 row---1 2
1 row---3 4
int *a gives the base address of the first element.
int *a+1 gives the base address of the second element.
int *a+2 gives the base address of the third element.
All the base address are stored in the pointer *p[].
*p[0] gives 1 , *p[1] gives 2 ,*p[2] gives 3.
If we print *p then it will show the base address.
**p --- gives the value 1.
column 0 1
0 row---1 2
1 row---3 4
int *a gives the base address of the first element.
int *a+1 gives the base address of the second element.
int *a+2 gives the base address of the third element.
All the base address are stored in the pointer *p[].
*p[0] gives 1 , *p[1] gives 2 ,*p[2] gives 3.
If we print *p then it will show the base address.
**p --- gives the value 1.
Arun said:
1 decade ago
Why more operators are used in variable?
Does it show different meaning?
Does it show different meaning?
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