Anomalous behavior of printf()? [duplicate]

Answer 1

Answer is simple. 9 is integer and 5 is also integer. when integer will be divided by integer then the result will be 0 and it will be escalated to 0.00000 because you have used "%f" in printf() If you want the correct result perform 9.0/5 Now you will for sure get the correct answer

Answer 2

9/5 is of type int. Passing an int argument to printf with "%f" has undefined behavior.

Try 9.0/5.0 if you want 1.800000, or (double)(9/5) if you want 1.000000

Answer 3

Why is this?

So 9 / 5 is obtained by dividing two integers -- and dividing two integers results in an integer value, 1. But, since it's an int, and the %f specified wants a float, this is undefined behavior.

Ps.: you probably want

9.0 / 5

or

9 / 5.0

or

9.0 / 5.0

Answer 4

That is because 9/5 is 1 and is type of integer. The fraction part is simply truncated. To achieve what you want, force the division to make floating point operations with 5.0/9 or 5/9.0 or 5.0/9.0 or (float) 5/9

When type of integer argument is bound with %f, then the behaviour is undefined.

This is stated in C99 standard Section 7.21.6.1 Paragraph 9. This paragraph is quoted below:

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

Answer 5

Re: Shouldn't be the output of program be 1.000000 ?

It would have been had your division resulted in a float value rather than an int. Since you tried to print an int with a %f format specification, the result was undefined behavior.

More details:

There are two problems, related though:

1. You are getting an integer result from your division, but trying format for a float value using %f

2. (Your question seems to imply you are aware of this) Integer division results in an integer which is why the %f format specifier isn't appropriate. When the format specifier and the argument don't match the behavior is undefined.

   To avoid an integer result, try:

   9.0 / 5 or 9 / 5.0 or even 9.0 / 5.0

   to get the result you were expecting and then the %f specifier will be appropriate and work as expected.

   Note that when you are doing integer division, the result will be another integer. If, however, at least one of the operands is a float, the other operand will be promoted to a float too and the division will result in a float value.

   With integer division you get truncation, which means the fractional part of the result is thrown away.

Answer 6

printf("%f",9/5);

This statement is undefined behavior because the argument has to be of type double and 9 / 5 is of type int.

Use 9 / 5.0 to have an argument of type double (and the correct floating division).