Answer / rk

d) it will print 0.0000.

If we typecast the result to float as shown below then
expected output will be printed(i.e. 1.0000)
printf("%f",(float) i/(*p) ); // i is divided by pointer p

Answer / guest

c) Error becoz i/(*p) is 25/25 i.e 1 which is int & printed
as a float,

So abnormal program termination,

runs if (float) i/(*p) -----> Type Casting

Answer / km

the answer to this question is implementation dependent:
if tried in Turbo C++
a) Runtime error
abnormal termination
if tried in Unix using GNU C
non of the above
you get a junk result

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Find your day from your DOB?

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#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d..%d",*p,*q); }

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