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main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
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Answer Posted / susie
Answer :
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a
1-D array.
2
4
7
8
3
4
2
2
2
3
3
4
100 102 104 106 108 110 112 114 116 118
120 122
thus, for the first printf statement a, *a, **a give
address of first element . since the indirection ***a gives
the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension
thus points to value at 114, *a+1 increments in second
dimension thus points to 104, **a +1 increments the first
dimension thus points to 102 and ***a+1 first gets the value
at first location and then increments it by 1. Hence, the
output.
| Is This Answer Correct ? | 11 Yes | 17 No |
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3) Int Matrix of certain size was given, We had few valu= es in it like this. =97=97=97=97=97=97=97=97=97=97=97 1 = | 4 | | 5 | &= nbsp; | 45 =97=97=97=97=97=97=97=97=97=97=97 &n= bsp; | 3 | 3 | 5 | = | 4 =97=97=97=97=97=97=97=97=97=97=97 34 |&nbs= p; 3 | 3 | | 12 | &= nbsp; =97=97=97=97=97=97=97=97=97=97=97 3 | &nbs= p; | 3 | 4 | = | 3 =97=97=97=97=97=97=97=97=97=97=97 3 | = ; | | | = ; 3 | =97=97=97=97=97=97=97=97=97=97=97 &= nbsp; | | 4 | = ; | 4 | 3 We w= ere supposed to move back all the spaces in it at the end. Note: = If implemented this prog using recursion, would get higher preference.
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