Remove the last digit from float

Question

0.00/5 (No votes)

See more:

I have a float value like 67,800003 and the last digit can also vary. How can I remove the last digit as in this case digit 3?

Thanks
M.H

Posted 19-Dec-12 4:20am

merh

Add a Solution

Comments

Albert Holguin 19-Dec-12 10:49am

Remove it or replace it with a certain known number (like 0)? This is important because the outcome is very different, if you remove the number as in solution 1, you'll end up with a completely different number (i.e. it will be different by a factor of ten). If you replace the number with a zero, it will only be slightly different (i.e. the number in your example will still be 67.8K).

lewax00 19-Dec-12 11:07am

In all honesty, you probably can't...not from the float itself anyways. Floats cannot represent every number, in many cases it is only an approximation. You can change how it's displayed however (like Stefan_Lang's answer below), or find a library for scientific/mathematical purposes that allows greater/arbitrary precision (I don't know of any for C/C++, so I can't help you there).

Aswin Waiba 20-Dec-12 9:23am

Will your variable only have whole numbers or would you expect real numbers too. Because the code would be different in these cases.

4 solutions

Solution 3

C++

bool removeLastDigit(float& number)
{
 int32_t tempNum = static_cast<int32_t>(number);
 tempNum = tempNum % 10;
 if( 0 == tempNum )
    return false; //No digit to remove
 number = number - tempNum;
 return true;
}

Won't work for real number though. :)

Posted 20-Dec-12 3:28am

Aswin Waiba

Updated 20-Dec-12 3:32am

v2

Solution 4

As other already stated, float is just an approximation - you can not represent every number accurately. The more digits you use in the integer part, the less precision you have in the fraction part of a float. When I display floating point numbers on user interfaces I usually use "%g" instead of "%f" as "%g" does rounding on the fraction part of the float and looks much nicer than "%.3f" or something similar.

Posted 20-Dec-12 8:02am

pasztorpisti

Solution 5

hi dear friend
i think this code can help you:

float f=45.723;        //main digit
float df=0;   int i1=0,t=0,t1=0;
t=f;
df=f-t;
t1=df*1000000;
while(!t1%10)
   t1/=10;

t1/=10;      //remove last digit
df=t1/(float)1000000;
f=t+df;
//f is your number

Posted 20-Dec-12 10:19am

Mahdi Nejadsahebi

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Stefan_Lang · Accepted Answer · 2012-12-19T04:59:00

That looks like a precision error to me: if you assign 67+4/5 to a float variable and print that, you'll get pretty much the same! The reason is that floats and doubles are internally converted to and stored as binary numbers which cannot accurately store a number like 4/5 anymore than in decimal you cannot accurately store 1/3 (at best you can write 0.333... , with as much digits as your storage allows).

Depending on which method you use for printing this variable, you can limit the output to less digits:

C++

// using C-style printf
// no more than 4 digits after the decimal point, no more than 8 characters total:
printf("%8.4f\n", var); // prints "67.8000"

// using C++ style cout
// no more than 8 meaningful digits (no trailing 0s)
cout << setprecision(8) << var << endl; // prints "67.8"
// exactly 6 digits
cout <<fixed <<setprecision(6) << var << endl; // prints "67.8000"

See setprecision for more info on using cout manipulators.