Float data type

Question

I am initializing a variable a of type float with value 1.2.

If I print it on screen, it give me 1.2, but when I check it with if/else, it does not run the if statement whose condition is if(a == 1.2). Instead it runs the else part.

Can someone please help, I am still learning the basics.

Here is my code:

#include<iostream>
#include<iomanip>
using namespace std;

int main ()
{
    float a = 1.2;
    cout<<fixed<<setprecision(1)<<a<<endl;
    
    if (a == 1.2)
        cout<<"\n a is 1.2";
    else
        cout<<"\n a is not 1.2";
    return 0;
}

Try if (a == 1.2f) instead of if (a == 1.2) That is use a constant of the type float instead of the type double. — Vlad from Moscow, Nov 02 '22 at 18:05
`1.2` is a `double`, not a `float`. When you do `a == 1.2` it converts the value held by `a` to a `double` for the comparison. However a float can't store as precise a representation of `1.2` as double so the value you get when you convert back to `double` is slightly different from the original `1.2` double value. — François Andrieux, Nov 02 '22 at 18:05
Read this in its entirety https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html — possum, Nov 02 '22 at 18:07
Another good read: [Is floating point math broken?](https://stackoverflow.com/q/588004/3282436) — 0x5453, Nov 02 '22 at 18:10
Does this answer your question? [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — Richard Critten, Nov 02 '22 at 18:23
[Why are floating point numbers inaccurate?](https://stackoverflow.com/questions/21895756/) — Remy Lebeau, Nov 02 '22 at 19:18

score 0 · Answer 1 · answered Nov 02 '22 at 19:28

0

The it-statement is comparing a float to a double. 1.2 is a double and not a float.

Maybe this will give you some clarity: Why comparing double and float leads to unexpected result?

Hope this helps!

answered Nov 02 '22 at 19:28

Iwan de Jong

51
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Float data type

1 Answers1