c-language basics

Question

int main()
{
       double i=4;
       printf("%d",i);
       return 0;
}

can anybody tell why does this program gives output as 0.

Answer 1

%d is for integers:

int main()
{

   int i=4;
   double f = 4;
   printf("%d",i); // prints 4
   printf("%0.f",f); // prints 4
   return 0;
}

Answer 2

Because "%d" specifies that you want to print an int, but i is a double. Try printf("%f\n"); instead (the \n specifies a new-line character).

Answer 3

Because i is a double and you tell printf to use it as if it were an int (%d).

Answer 4

Because the language allows you to screw up and you happily do it.

More specifically, '%d' is the formatting for an int and therefore printf("%d") consumes as many bytes from the arguments as an int takes. But a double is much larger, so printf only gets a bunch of zeros. Use '%lf'.

Answer 5

The simple answer to your question is, as others have said, that you're telling printf to print a integer number (for example a variable of the type int) whilst passing it a double-precision number (as your variable is of the type double), which is wrong.

Here's a snippet from the printf(3) linux programmer's manual explaining the %d and %f conversion specifiers:

   d, i   The int argument is converted to signed decimal  notation.   The
          precision,  if any, gives the minimum number of digits that must
          appear; if the converted value  requires  fewer  digits,  it  is
          padded  on  the  left  with  zeros.  The default precision is 1.
          When 0 is printed with an explicit precision 0,  the  output  is
          empty.

   f, F   The double argument is rounded and converted to decimal notation
          in the style [-]ddd.ddd, where the number of  digits  after  the
          decimal-point character is equal to the precision specification.
          If the precision is missing, it is taken as 6; if the  precision
          is  explicitly  zero,  no decimal-point character appears.  If a
          decimal point appears, at least one digit appears before it.

To make your current code work, you can do two things. The first alternative has already been suggested - substitute %d with %f.

The other thing you can do is to cast your double to an int, like this:

printf("%d", (int) i);

The more complex answer(addressing why printf acts like it does) was just answered briefly by Jon Purdy. For a more in-depth explanation, have a look at the wikipedia article relating to floating point arithmetic and double precision.

Answer 6

@jagan for the sub- question"What is Left most third byte. Why it is 00000001 ? Can somebody explain" 10000000001 is for 1025 in binary format.

Answer 7

Jon Purdy gave you a wonderful explanation of why you were seeing this particular result. However, bear in mind that the behavior is explicitly undefined by the language standard:

7.19.6.1.9: If a conversion specification is invalid, the behavior is undefined.²⁴⁸⁾ If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

(emphasis mine) where "undefined behavior" means

3.4.3.1: behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements

IOW, the compiler is under no obligation to produce a meaningful or correct result. Most importantly, you cannot rely on the result being repeatable. There's no guarantee that this program would output 0 on other platforms, or even on the same platform with different compiler settings (it probably will, but you don't want to rely on it).