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7 votes
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Consider the following C program

main()

{

int a[2][3][2]={{{1,2},{9,8},{3,7}},{{2,2},{1,4},{5,4}}}};

printf(“%d%d%d”, (a[1]-a[0]), (a[1][0]-a[0][0]), (a[1][0][0]-a[0][0][0]));

}

Which of the following represents the output of above program?

  1. $3\: 3\; 1$
  2. $3\; 6 \;1 $
  3. $6\;6\;1$
  4. $1\:1\:1$
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3 Answers

Best answer
14 votes
14 votes

Your answer is correct . It should be 361 only.
 

'

$a[1] - a[0] = \frac{112-100}{size of row} = \frac{12}{4} = 3$

$a[1][0] - a[0][0] = \frac{112-100}{size of element} = \frac{12}{2} = 6$

$a[1][0][0] - a[0][0][0] = 2-1 = 1$

Output : 3 6 1

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1 votes
1 votes
6,6,6 should be the correct answer.
0 votes
0 votes
ANS:-3,6,1

First of all a[1] is not equal to a[0][0][1]. a[0] is pointer of pointer which point to {1,2} and a[1] is point to {2,2}.

(PROOF:-If we print **a[o] then output is 1 and **a[1] then output is 2.)

so a[1]-a[0] is 3 beacause suppose address of {1,2} is 100 then address of {2,2} is 124 if int size consider 4. Then we add 3 in the adress of a[0] we get adress of a[1] because each { , } has size 8.so 100+8*3=124. so a[1]-a[0]=3.

simillarliy,a[1][0]-a[0][0]=6 because a[1][0] is adress of 2 and a[0][0] is adress of 1.suppose a[0][0] is 10 then a[1][0] is 34 each int has sze 4. 10+4*6=34

And a[1][0][0]-a[0][0][0]=1

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