test Q

Question

main(){
static int n[3][3] = {2,4,3,6,8,5,3,5,1};
Int i,j;
for(i=2;i>=0;i--){
for (j=2;j>=0;j--){
printf(“%d ”,*(*(n+i)+j));
}
}

What will be the output of the program?

Comments

input in reverse order

1 5 3 5 8 6 3 4 2

#include <stdio.h>

int main(){
static int n[3][3]={2,4,3,6,8,5,3,5,1};
int i,j;
for(i=2;i>=0;i--){
for (j=2;j>=0;j--){
printf( "%d ",*(*(n+i)+j));

}
}
return 0;
}
 

run this here https://www.tutorialspoint.com/compile_c_online.php

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 problem is gate not provide any compiler :)

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??????

Answer 1

2	4	3	6	8	5	3	5	1
1000	1002	1004	1006	1008	1010	1012	1014	1016

• *(*(n+i)+j) = here *(n+i) means skipping 'i' rows which contains 3 elements of each size "c" bytes {here, i assumed c=2Bytes} and selecting row which is done by "*".
• after that, 'j' elements are skipping of size "c" bytes.

first iteration:

i=2, j=2

*(*(1000+2)+2) = *(*(1000+2*3*2)+2*2)  = *(*(1012)+4) =*(1016) = 1

similarly for i=2,j=1 

*(*(1000+2)+1) = *(*(1000+2*3*2)+1*2)  = *(*(1012)+2) =*(1014) = 5

final o/p: 1 5 3 5 8 6 3 4 2