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 8 of 8.
Chand said:
1 decade ago
p[3]={a,a+1,a+2}
first (*(p+0)+0)=a[0]=1 etc
second *(*(p+0)+1)=a[1]=2 etc
third *(*(p+1)+0)=a[1]=2 etc
and last *(*(p+1)+1)=a[2]=3 etc
first (*(p+0)+0)=a[0]=1 etc
second *(*(p+0)+1)=a[1]=2 etc
third *(*(p+1)+0)=a[1]=2 etc
and last *(*(p+1)+1)=a[2]=3 etc
Vivek gupta said:
1 decade ago
I can not understand it please explain more.
Chakri said:
1 decade ago
Please explain clearly.
Saurav said:
1 decade ago
@sindhu:
Why p=0 in step 1?
Please explain it more clearly.
Why p=0 in step 1?
Please explain it more clearly.
Rahul aarthi said:
1 decade ago
i did't get u pls explain clearly
Sindhu said:
1 decade ago
step1:p=0;
step2:i=0,j=0
*(*(p+i)+j)=a[0]=1
*(*(j+p)+i)=a[0]=1
*(*(i+p)+j)=a[0]=1
*(*(p+j)+i)=a[0]=1
step2:i=0,j=1
*(*(p+i)+j)=a[1]=2
*(*(j+p)+i)=a[1]=2
*(*(i+p)+j)=a[1]=2
*(*(p+j)+i)=a[1]=2
step3:i=1,j=0
*(*(p+i)+j)=a[1]=2
*(*(j+p)+i)=a[1]=2
*(*(i+p)+j)=a[1]=2
*(*(p+j)+i)=a[1]=2
step4:i=1,j=1
*(*(p+i)+j)=a[2]=3
*(*(j+p)+i)=a[2]=3
*(*(i+p)+j)=a[2]=3
*(*(p+j)+i)=a[2]=3
step2:i=0,j=0
*(*(p+i)+j)=a[0]=1
*(*(j+p)+i)=a[0]=1
*(*(i+p)+j)=a[0]=1
*(*(p+j)+i)=a[0]=1
step2:i=0,j=1
*(*(p+i)+j)=a[1]=2
*(*(j+p)+i)=a[1]=2
*(*(i+p)+j)=a[1]=2
*(*(p+j)+i)=a[1]=2
step3:i=1,j=0
*(*(p+i)+j)=a[1]=2
*(*(j+p)+i)=a[1]=2
*(*(i+p)+j)=a[1]=2
*(*(p+j)+i)=a[1]=2
step4:i=1,j=1
*(*(p+i)+j)=a[2]=3
*(*(j+p)+i)=a[2]=3
*(*(i+p)+j)=a[2]=3
*(*(p+j)+i)=a[2]=3
Sitakanta Mishra said:
1 decade ago
Anyone please explain this question.
Aabid khan said:
1 decade ago
I could not get the logic of this question please explain.
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