I have an array of floats where data are stored with varying decimal points so some are 123.40000, 123.45000, 123.45600...now if i want to print these values in the string without the 0s in the end in printf() so that they are 123.4, 123.45, 123.456, without those 0s in the end. Is this possible? If so, how?
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4 Answers
Use the %g formatter:
printf( "%g", 123.4000 );
prints
123.4
Trailing zeros are removed, but unfortunately so is the trailing decimal point if the fractional part is zero. I don't know if there is actually any way of doing what you want directly using printf() - I think something like this is probably your best bet:
#include <stdio.h>
#include <math.h>
void print( FILE * f, double d ) {
if ( d - floor(d) == 0.0 ) {
fprintf( f, "%g.", d );
}
else {
fprintf( f, "%g", d );
}
}
int main() {
print( stdout, 12.0 );
print( stdout, 12.300 );
}
2 Comments
himself
Isn't it a bad idea to compare doubles? "if (d-floor(d) == 0.0)"?
Pascal Cuoq
@himself No, it's fine when you know what you are doing. “Never compare floating-point numbers for
==” is just an approximation for the unsophisticated.I don't know how hacky this is but:
float f = 124.000;
if (f == (int) f) {
printf("%.1f\n", f); /* .1 can be changed */
} else {
printf("%g\n", f);
}
Returns 124.0.
float f = 124.123000;
if (f == (int) f) {
printf("%.1f\n", f); /* .1 can be changed */
} else {
printf("%g\n", f);
}
Returns 124.123.
Comments
Use %g --
Print a double in either normal or exponential notation, whichever is more appropriate for its magnitude. 'g' uses lower-case letters, 'G' uses upper-case letters. This type differs slightly from fixed-point notation in that insignificant zeroes to the right of the decimal point are not included. Also, the decimal point is not included on whole numbers.
1 Comment
sfactor
just a small query, what if i want to remove the trailing zeros but want to keep the decimal in whole numbers, like 124.00000 is 123.
Print to a (large enough) buffer. Print the buffer ... and if the there's no '.' in the buffer print a dot.
char buf[100];
sprintf(buf, "%g", val);
printf("%s", buf);
if (strchr(buf, '.') == NULL) putchar('.');
edit
The Standard specifies the # flag:
# The result is converted to an ``alternative form''. [...] For a, A, e, E, f, F, g, and G conversions, the result of converting a floating-point number always contains a decimal-point character, even if no digits follow it. [...] For g and G conversions, trailing zeros are not removed from the result. [...]
... but you get the trailing zeros :(
Comments
