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void main()
{
int s[4][2]={ {1234,56},{1212,33},{1434,80},{1312,78} };
int (*p)[2];
int i,j,*pint;
for(i=0;i<=3;i++)
{
p=&s[i];
pint=p;
printf("\n");
for(j=0;j<=1;j++)
printf("%d",*(pint+j));
}
} while running this program it shows a warning-suspicious
pointer conversion ie pint=p; my que is why should we assign
the value of p to pint again.why cant we use it directly as
*(p+j)..but if i use like tat the o/p is garbage value..
- 1 Answers
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Answer / test
p ia pointer to an array when p is incremented it will increment by the number of elements in the array...
for example
int (*p)[2]; //it is aponiter to an arry of 2 elements so when we do p+1 then it will be incremented by two*(sizeof(int))
so leading to the garbage value at the last loop..
can be properly analyzed by the below program
#include<stdio.h>
main()
{
int s[4][2]={ {1234,56},{1212,33},{1434,80},{1312,78} };
int (*p)[2];
int i,j,*pint;
for(i=0;i<=3;i++)
{
p=&s[i];
//pint=p;
printf("\n");
for(j=0;j<=1;j++)
printf("%p.....%d\n",*(p+j),**(p+j));
}
}
| Is This Answer Correct ? | 23 Yes | 2 No |
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