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AlGorism
April 21st, 2011, 03:02 PM
Hi everybody. I have a question, again.

I have a float variable, num.
After many operations, from debug screen, I know the value of num is 24.339999972362185... and so on.
When I use printf function like this,

printf("%f",num);
program prints 24.340000
But, I must print all the fraction part.
What can I do to do it?
Thanks.

Telengard C64
April 21st, 2011, 03:14 PM
I think you can get more precision by using a double.

April 21st, 2011, 03:22 PM
Are you asking if you can print a certain number of decimal places? You can use format specifiers (http://www.codingunit.com/printf-format-specifiers-format-conversions-and-formatted-output) like so:

printf("%.2f", num);
Which would print 24.34

Arndt
April 21st, 2011, 03:58 PM
Hi everybody. I have a question, again.

I have a float variable, num.
After many operations, from debug screen, I know the value of num is 24.339999972362185... and so on.
When I use printf function like this,

printf("%f",num);
program prints 24.340000
But, I must print all the fraction part.
What can I do to do it?
Thanks.

As someone said, you can use %.15f, for example. But I think you are getting all the digits that are really there. Look:

int main()
{
float f1 = 24.34;
float f2 = 24.339999972362185;

printf("%.20g %.20g\n", f1, f2);
if (f1 == f2)
printf("f1 and f2 are equal\n");
}

\$ ./floattest
24.340000152587890625 24.340000152587890625
f1 and f2 are equal

As was also suggested, if you use double instead, you get more precision.

Telengard C64
April 21st, 2011, 09:31 PM
As was also suggested, if you use double instead, you get more precision.

I should explain myself. If you need so many digits after the decimal point then I presume they are important to you. Storing your non-integer into a float is probably a mistake.

As was already pointed out by more patient respondents, you can set the desired precision with your format specifier.

foo\$ cat a.c
#include <stdio.h>

int main(void)
{
float num;
num = 24.339999972362185;
printf("%20.20f\n", num);
return 0;
}
foo\$ gcc -Wall a.c
foo\$ ./a.out
24.34000015258789062500
foo\$ # ^ GARBAGE!

If those numbers after the decimal are really important to you then use a double or some other type with more precision.

foo\$ cat a.c
#include <stdio.h>

int main(void)
{
double num;
num = 24.339999972362185;
printf("%20.20f\n", num);
return 0;
}
foo\$ gcc -Wall a.c
foo\$ ./a.out
24.33999997236218604257
foo\$ # ^ MUCH MORE PRECISE

HTH

stchman
April 21st, 2011, 11:15 PM
There are 2 problems:

1. Your format specifier is wrong.
2. You need to use double over float for more precision.

#include <stdio.h>

int main( void )
{
float num1 = 24.339999972362185;
double num2 = 24.339999972362185;

printf( "num1 = %f\n", num1 );
printf( "num2 = %f\n", num2 );

printf( "num1 (format specifier) = %20.15f\n", num1 );
printf( "num2 (format specifier) = %20.15f\n", num2 );

return 0;
}

Output:

num1 = 24.340000
num2 = 24.340000
num1 (format specifier) = 24.3400001525878906
num2 (format specifier) = 24.3399999723621860

As you can see a double will give you far greater precision over a float and the proper format specifier will give you more precise output.