[✔️]Why output of this code is ''not equal'' ? Please explain! i don't understand this.!

Question

#include <stdio.h>
int main()
{
float f = 0.1;

if (f == 0.1)
printf ("equal");
else
printf ("not equal");

return 0;
}

Output :
not equal

Answer 1

It outputs "not equal" because your variable is a float but 0.1 is not. 0.1 is a double by default. To make it float add 'f' at the end. (e.g. 0.1f)

This code will output "equal", as required:

#include <stdio.h>
int main()
{
float f = 0.1;

if (f == 0.1f)
printf ("equal");
else
printf ("not equal");

return 0;
}

Answer 2

Here you decleared one float variable f and you assigning value 0.1 remember that this is double value not float becz by default float value treat as double type but here f type is float and you assigning double value to float variable f . If you type cast lower data type to higher like
Short to int ,int to float , float to double. but in case of double to float it means here u converting Higher data type to lower. Both data types have different different ranges so if u will try to assign value it will loose some precision and your data will be loose . If you avoid this kind of problems in your code in c++ you can use
static_cast<data type>

same thing happening in your case.
If u want to define float variable then you can write like this
float f=0.1f ;
if didn't get answer then try this one your doubts will be clear

0.1is a double value by default
when assigned to float it will loose some precision hence it will not compare equal to 0.1
printf("%.20f\n", a);
printf("%.20f\n",0.1);

Answer 3

Ooooooh..😅
Great to know.
And because of this reason i failed so many times in C programs.

Answer 4

#include <stdio.h>

int main() {

float a=9.6;
printf("%.20f\n", a);
printf("%.20f\n",9.6);

return 0;
}

Answer 5

Aleksei Radchenkov ,
😮 Ohh.
But in cpp by default any decimal number is double, right ?
So is it also in c ?

Answer 6

I don't use C too much, but I believe it's the same as C++

Answer 7

Okay.
Anyways, thanks for explaining. 😊

Answer 8

#...._._...._<........> Yeah it will give same in c ,cpp

Answer 9

A.S. ,
Yeah this statement is working in C -->float f = 0.1f;

Thank you so much for this help.
and explaining. 😊

Answer 10

#...._._...._<........> are you sure

Answer 11

#...._._...._<........> Just try this:

printf("%0.12f\n%0.12f", (double) 0.1f, 0.1);

Edit: It should be obvious that this should be included inside the main function.

Answer 12

Saifullah ,
Thank you. This code is working. 😊

Answer 13

Calvin Thomas ,
But still i don't understand this. 😅
And its not working on sololearn's compiler. 🙃

Answer 14

User_dead ;
Calvin Thomas

Sorry both of you.
Your given code is right and working.
🙌

that was my mistake, i wrote my code in a wrong way.😬

Answer 15

Just Practice on it then you will be become a good programmer.
I am also beginner bit I trying my best for the best.

Answer 16

#...._._...._<........> Try this statement your doubts will be more clear
float a=9.6;
printf("%.20f\n", a);
printf("%.20f\n",9.6);

Answer 17

Whenever you have to deal with floating points number use double data type becz it give more accurate value than float

Answer 18

A.S. ,
I use this statement [ printf ("%0.20f \n",9.6); ] but this line give me error.

Answer 19

A.S. ,
Yeah ! Iam sure.
Now i just this.
These two line gives error.

Answer 20

#...._._...._<........> 🙃😆 Check the code which i have pasted