float variables in C [duplicate] - c

Question

This question already has answers here:
Using %f to print an integer variable
(6 answers)
Closed 3 years ago.
May be this is a simple question but I am not sure about how float variables are stored in memory and why it is behaving in this way, can someone please explain about the following behavior.
#include<stdio.h>
int main ()
{
int a = 9/5;
printf("%f\n", a);
return 0;
}
Output:
0.000000
I have looked at some information on how float variables are stored in memory, it has stuff about mantissa, exponent and sign. But I am not getting how to relate that here.

Answer 1

int a = 9/5;
performs integer division and ignores the remainder, so a is set to 1. Attempting to print that using %f gives undefined behavior, but by chance you got 0.000000 out of it.
Do
double a = 9./5.;
instead, or print with %d if integer division was the desired behavior. (float would also work, but a will be promoted to double when passed to printf, so there's no reason not to use double.)

Answer 2

It is an Undefined Behaviour.
You are using float format specifier (%f) to print an int (a). You should use %d to see correct output.

Answer 3

It is an undefined behaviour in C. Use %d format specifier instead of %f.
Does printf() depend on order of format specifiers? gives you detailed answer.

Answer 4

Here's a brief analysis of your code:
int a = 9/5; // 9/5 = 1.8, but since you are doing integer division and storing the value in an integer it will store 1.
printf("%f\n", a);//Using incorrect format specifiers with respect to datatypes, will cause undefined behavior
printf("%d\n",a);//This should print 1. And correct.
Or if you want the float:
instead of int use float :
float a=9.0f/5;//This will store 1.800000f
//float a=9/5 will store 1.000000 not, 1.8 because of integer divison
printf("%f\n",a); //This will print 1.800000
Also do read this: http://en.wikipedia.org/wiki/IEEE_754-2008 on how floating points work.
Clarification about integer division:
C99:
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.88) If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a
88) This is often called ‘‘truncation toward zero’’.

Answer 5

Just assuming that your division should result in a fraction in normal math won't magically make it a float.
int a = 9/5;
You need to use
float a = 9.0/5;
First you need the right data type i.e. float (or better yet double for higher precision) to store a float.
Secondly, if you are dividing two integers i.e. 9 and 5, it will simply follow integer division i.e. only store the integer part of division and discard the rest. To avoid that i have added .0 after 9. This would force compiler to implicitly covert into float and do the division.
Regarding your mentioning of why it is printing 0, it is already mentioned that trying %f on integer is undefined behavior. Technically, a float is 4 bytes containing 3 byte number and 1 byte exponent and these are combined to generate the resultant value.