Main(){ int i, j, p; i = 25; j = 100; p = &i; // Address of i is assigned to pointer p printf("%f", i/(p) ); // i is divided by pointer p}

A) Compile error
B) 1.00000
C) Runtime error.
D) 0.00000
Explanation: Error because 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.

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Dec 10th, 2005
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Showing Answers 1 - 13 of 13 Answers

NAVNISH JAIN

Dec 15th, 2005

it is compiler dependent.

C_learner
Profile Answers by C_learner

Jan 18th, 2006

It will mostly give run time error and not compile time error....
Also use typecasting i.e. use (float)i/(*p); and you will get 1.0000.. as the answer

sravanan

May 15th, 2006

compiler error.because division is not possible usinsg pointers.

shweta

May 16th, 2006

hi,
this gives runtime error.
i tried to run program in vc with .c extension file.
it gives runtime error

noname2000
Profile Answers by noname2000

Mar 3rd, 2007

i try compile with turbo C++ and it print out a garbage value. No error.

Kalpitgdeewan
Profile Answers by Kalpitgdeewan

Jan 13th, 2008

it gives 0.00000 as the output when we run it :P

santhigkinterview
Profile Answers by santhigkinterview

Jan 16th, 2008

Run time Error – Floating point not loaded.
printf("%f", i/(*p) ); // i is divided by pointer p

by doing type casting (float)i/(*p) we can avoid this error.

sourabh_t
Profile Answers by sourabh_t

Aug 12th, 2008

C. Runtime Error

yousf
Profile Answers by yousf

Jun 7th, 2009

This is compiler dependent..

This will work on most popular compilers.. At least it will work on gcc!

abhimanipal
Profile Answers by abhimanipal

Jan 24th, 2010

In GCC this gives garbage value

The reason for this could be that when we write printf("%f", i/(*p) ); the printf statement expects a floating point value. However it is not sure what kind of value will result after the division. So it prints junk.

track92
Profile Answers by track92

Jun 15th, 2010

p is a pointer to the int variable i and *p prints the value stored in the address (of i) which it points. So answer will be 1 but data type is float so it will print 1.000000.

jbode
Profile Answers by jbode Questions by jbode

Jun 30th, 2010

The %f conversion specifier expects its corresponding argument to be of type double, but the result of i/(*p) is type int (since both operands have type int); therefore, the behavior is undefined (per 7.19.6.1.9); any result is possible, from printing 1.00000 to printing garbage to runtime errors.