Printf Question

[Mandragon03]

I am looking for a way to take a floating point number and get rid of
any extraneous 0's at the end. For instance if I have

myFloat = 9999;

printf("%f", myFloat);

I get 9999.0000. I want to get rid of these 0's.

Here is one catch. I also need to be able to do this:

myFloat = .531;

printf("%f", myFloat);

I get .5310000

I need to get .531.

I realize I can do this

printf("%2f", myFloat);

which will limit the mantissa to two places, but this will not work
for what i am doing because I can not lose that much precision on
floats like .05369.

Thank you for you time,

Mandragon03

 

[Victor Bazarov]

I am looking for a way to take a floating point number and get rid of
any extraneous 0's at the end. For instance if I have

myFloat = 9999;

printf("%f", myFloat);

I get 9999.0000. I want to get rid of these 0's.

Here is one catch. I also need to be able to do this:

myFloat = .531;

printf("%f", myFloat);

I get .5310000

I need to get .531.

I realize I can do this

printf("%2f", myFloat);

which will limit the mantissa to two places, but this will not work
for what i am doing because I can not lose that much precision on
floats like .05369.

The problem actually is rather simple. The default is 7 decimal
digits in the result, which means that 0.5310000 and 0.530999951
will give you the same output. The zeros are there to indicate to
what digit the output was _rounded_ to.

If you really need to lose the trailing zeros, you will have to
trim them yourself. Convert your number to a string and drop all
chars from it until the first non-zero:

std:stringstream s;
s << myFloat;
std::string str = s.str();
while (str.size() > 1 && *str.rbegin() == '0')
str.erase(str.size() - 1);

V

 

[Jack Klein]

The problem actually is rather simple. The default is 7 decimal

Correction, 6 decimal places.

--
Jack Klein
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[Mandragon03]

The problem actually is rather simple. The default is 7 decimal
digits in the result, which means that 0.5310000 and 0.530999951
will give you the same output. The zeros are there to indicate to
what digit the output was _rounded_ to.

If you really need to lose the trailing zeros, you will have to
trim them yourself. Convert your number to a string and drop all
chars from it until the first non-zero:

std:stringstream s;
s << myFloat;
std::string str = s.str();
while (str.size() > 1 && *str.rbegin() == '0')
str.erase(str.size() - 1);

V

I figured as much. Thank you for taking the time to answer my
question. You guys are great =)

 

[Juha Nieminen]

I am looking for a way to take a floating point number and get rid of
any extraneous 0's at the end. For instance if I have

myFloat = 9999;

printf("%f", myFloat);

I get 9999.0000. I want to get rid of these 0's.

Have you tried "%g" instead of "%f"?