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

main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

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