What is problem in type declaration FLOAT?

[sauravamatyaw]

#include <stdio.h>
#include <conio.h>
void main ()
{
float a;
clrscr();
a=11/5;
printf ("%f",a);
getch ();
}

Output: 2.000000
Since "a" is defined as float it should have given "2.2" output but gives an integer value.
WHATS WRONG?

[JATINCSE]

use the value a is either in any format given 11.0/5.0 or 11.0/5 or 11/5.o so you will get the correct answer

[sauravamatyaw]

@JATINCSE cant i do it without putting ".0" behind ? is it compulsory?

[weaknessforcats]

You a mixing integer and floating point division:

a = 11/5;

11/5 is 2. Integers have no decimal portion.

Considering the variable is a float you would code:

a = 11.0f/5.0f;

If you code:

a = 11.0/5.0;

you will get a warning from the compiler. The 11.0/5.0 is a double and then you assign the double to a float, which is smaller, so you get a warning about possible data loss.

[donbock]

@sauravamatyaw:
The compiler first evaluates the expression on the right side of the equals sign and then assigns the computed value to the variable on the left side of the equals sign. The compiler doesn't notice the type of the left-side variable until the assignment stage; but by then it is too late -- the expression has already been evaluated using integer math.

[JATINCSE]

@sauravamatyaw:
i think so because it is depend upon the working of the compliler
first it evaluates the expression on right side as on right it is the ratio of two integers so compiler will evaluate in terms of int then assign its value to float