C program; trouble with doubles

[Math Is Hard]

Hi - what do I need to do to take a double value as input and display it? Thanks.

//This prints garbage:

#include <stdio.h>
int main(void)
{
double mydouble;

printf("Enter a number\n");
scanf("%f",&mydouble);
printf("you typed %f\n", mydouble);

return 0;
}

[plover]

scanf("%f",&mydouble);

%f reads a float, for a double you want %lf

However, %f works to print a double with printf.

printf and scanf codes are often faux amis.

[AKG]

If plover's idea doesn't help, use "%g" instead of "%f".

[Math Is Hard]

plover and AKG - thanks so much for the suggestions. I am going to give this a try tonight.
I had not heard of "%g". I better go look that up!

[plover]

According to Harbison & Steele %f %g and %e should all act identically (and C99 adds %a). So:

%f %g %e %a read a float
%lf %lg %le %la read a double
%Lf %Lg %Le %La read a long double

[AKG]

Plover is right, ignore what I was saying. My book says %g is used to printf a double.

[Math Is Hard]

//This works better
#include <stdio.h>
int main(void)
{
double mydouble;

printf("Enter a number\n");
scanf("%lf",&mydouble);
printf("you typed %f\n", mydouble);

return 0;
}

[Math Is Hard]

//This works pretty good too
#include <stdio.h>
int main(void)
{
double mydouble;

printf("Enter a number\n");
scanf("%lg",&mydouble);
printf("you typed %g\n", mydouble);

return 0;
}

[Math Is Hard]

Playing around with it, I entered the number 23.4567891 into each program.
Without using any conversion mods,
The first program with "%lf" returned 23.456789
The second program with "%lg" returned 23.4568
I am not exactly sure of the difference since I don't have any reference here.

[AKG]

There are too many variables in your experiment. Try having two programs, one that scans with "%lf" and one with "%lg" but have them both print out with "%f." Then do the same where they both print out with "%g." Because, as you have it, it's hard to tell if it's the fact that you're scanning wiht "%lg" or printing with "%g" that's shrinking your output.

[Math Is Hard]

you're absolutely right. I'll give that a try.

[plover]

For printf (again from Harbison and Steele):
%f or %lf prints a float or a double (use %Lf for long double) to a minimum precision
%e or %le (respectively %Le) prints in scientific notation, again to a minimum precision
%g or %lg (respectively %Lg) decides whether %f or %e fits the situation better, uses that, but then strips off trailing zeros

%e %g use a lower case 'e' before the exponent, %E %G use upper case (%F is not a standard code).

%a (%la %La %A etc) is a C99 addition for printing in hexadecimal floating point, but doesn't seem to be implemented in the libc I'm using (cygwin's).

Try this:

#include <stdio.h>
int main(void)
{
double mydouble;

printf("Enter a number\n");
scanf("%lg",&mydouble);
printf("you typed (%%f) %f\n", mydouble);
printf("you typed (%%g) %g\n", mydouble);
printf("you typed (%%e) %e\n", mydouble);

return 0;
}

For more details on setting the precision and other formatting gewgaws, check out the floating point conversions (http://www.gnu.org/software/libc/manual/html_node/Floating-Point-Conversions.html#Floating-Point%20Conversions) page for glibc.

[plover]

The default for %f is six digits after the decimal point. The default for %g is six significant figures. This explains the results for 23.4567891.

Try these inputs: .000123456789 and .0000123456789

[Math Is Hard]

This is going to be a pretty dumb question, but I am not very clear on what is meant by "significant figures". Can you help?

oh, I tried your samples, plover. Thanks for that piece of code demostrating the different displays. %f displays 0.000123 and %g displays 0.0001234567
This was very helpful! :smile:

[plover]

Check the display for %g, it should be: 0.000123457 (no 6 - the 6 was rounded up). If it's not, something very strange is going on.

Here's a tutorial on significant figures (http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/) I found. If you don't like this one, googling turns up plenty more.

Now that I think about it though, "significant figures" is not the correct term to describe what is going on. Apparently, the correct term in this case is "significant digits", which is the phrase that is actually used on the glibc page linked to above, and in my reference book. It had not registered with me earlier that there were two distinct terms here. However the description on the glibc page is a bit confusing.

Here's my version of what's going on (in the default case):

Call the value to be converted v. If |v|>=10^7 or |v|<10^-3, use scientific notation, else use decimal notation.
If more than 6 digits remain to the right of any leading zeros, round the number so that 6 such digits remain.
Strip off trailing zeros if they are to the right of the decimal point.

Thus:
23.00 -> 23
23.00001 -> 23
2300.001 -> 2300
23.67676 -> 23.6768
0.001234566 -> .00123457


You might also add the line:

printf("you typed (%%#g) %#g\n", mydouble);

to the program. The '#' causes trailing zeros not to be stripped.

Here's a version of the program for playing with the "precision" specification. (And to puzzle over the notation... :wink: ) If you want to fiddle with this some more, you might try constructing the statements so that the "field width" can be set instead of (or in addition to) the "precision".

#include <stdio.h>
int main(void)
{
double mydouble;
int precision;

printf("Enter a number\n");
scanf("%lg",&mydouble);
printf("Enter a precision\n");
scanf("%d",&precision);
printf("\n");

printf("With default precision:\n");
printf(" %%f converts your number to: %f\n", mydouble);
printf(" %%e converts your number to: %e\n", mydouble);
printf(" %%g converts your number to: %g\n", mydouble);
printf(" %%#g converts your number to: %#g\n", mydouble);
printf("\n");

printf("With precision %d:\n", precision);
printf(" %%.%df converts your number to: %.*f\n", precision, precision, mydouble);
printf(" %%.%de converts your number to: %.*e\n", precision, precision, mydouble);
printf(" %%.%dg converts your number to: %.*g\n", precision, precision, mydouble);
printf(" %%.%d#g converts your number to: %.*#g\n", precision, precision, mydouble);

return 0;
}

[Math Is Hard]

plover, you are a wealth of information! Thanks so much!