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UGC NET CSE | December 2005 | Part 2 | Question: 11

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What is the output of the following $C$-program

main()
{
    printf("%d %d %d",size of (3.14f), size of (3.14), size of (3.141));
}
  1. 4 4 4
  2. 4 8 10
  3. 8 4 8
  4. 8 8 8
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6 Answers

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4 8 8

Float occupies 4 bytes and double occupies 8 bytes, and long double occupes 8 bytes

Here none of option matches....howz it possible
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answer is: D

Float occupies 8 bytes and double occupies 8 bytes, and long double occupes 8 bytes..

its also depends of compiler ..
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3.14f means floating point value---> so 4

3.14 is by default double hence----->8

3.141 is also by default double hence---->8

So 4 8 8 (using GNU GCC compiler)
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the answer is 4 8 8 (which is not there in options)

"By default,floating constant are stored as double precision numbers.To force the compiler to store a floating constant in floating format,put the letter f(or F) and in long double format put the letter l(or L) at the end "

now 3.14f is stored as float since f is specified at the end ...4 bytes(compiler dependent ofcourse)

3.14,3.141 are stored as double since  no letter at the end.....8 bytes,8 bytes

 

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