How do I format decimals in C? - c

Question

This is my print statement:
printf("%d %f\n",kPower, raisePower);
This is my output:
-4 0.000100
-3 0.001000
-2 0.010000
-1 0.100000
0 1.000000
1 10.000000
2 100.000000
3 1000.000000
4 10000.000000
I want it to be printed like this:
UPDATE
So I made my positive values line up:
-4 0.0
-3 0.0
-2 0.0
-1 0.1
0 1.0
1 10.0
2 100.0
3 1000.0
4 10000.0
This is my new code so far:
printf("%d %10.1f\n",kPower, raisePower);
I don't know, should I make a for loop to print each one (positive results vs negative result) in a different format?

Answer 1

#include <stdio.h>
char *get_number_formatted(double f)
{
static char buf[128]; // this function is not thread-safe
int i, j;
i = snprintf(buf, 128, "%20.10f", f) - 2;
for (j = i - 8; i > j; --i)
if (buf[i] != '0')
break;
buf[i + 1] = '\0';
return buf;
}
int main(void)
{
int i;
for (i = -4; i < 5; ++i)
printf("%5d %s\n", i, get_number_formatted(pow(10.0, i)));
return 0;
}
http://ideone.com/KBiSu0
Output:
-4 0.0001
-3 0.001
-2 0.01
-1 0.1
0 1.0
1 10.0
2 100.0
3 1000.0
4 10000.0
printf() cannot print a variating length of decimal digits, so basically what I did was print the formatted number into a buffer and then cut the exceeding zeros.

Answer 2

Try calculating the powers first using pow() from math.h and then:
You can use %10f to precede the number with blanks in the example total of 10 spaces:
printf ("Preceding with blanks: %10f \n", 10000.01);
Source: cplusplus.com

Answer 3

Basicly you can use variable length to perform this:
printf("%d %.*lf", kPower, -kPower, raisePower);
Advantage over other methods is that this method does not need any extra buffer(s)

Answer 4

With a little help of modf, you can use %g to skip the trailing zeroes and \b to skip the leading zero:
#include <stdio.h>
#include <math.h>
int main(void)
{
int i, iarr[] = {-4, -3, -2, -1, 0, 1, 2, 3, 4};
double darr[] = {0.0001, 0.001, 0.01, 0.1, 1., 10., 100., 1000., 10000.};
double intpart, fractpart;
for (i = 0; i < 9; i++) {
fractpart = modf(darr[i], &intpart);
if (fractpart == 0.0)
printf("%10d%10d.0\n", iarr[i], (int)intpart);
else
printf("%10d%10d\b%g\n", iarr[i], (int)intpart, fractpart);
}
return 0;
}
Output:
-4 0.0001
-3 0.001
-2 0.01
-1 0.1
0 1.0
1 10.0
2 100.0
3 1000.0
4 10000.0

Answer 5

Try this example code
float y[7]={0.000100f,0.0010f,0.0100,0.1000f,1.0f,10.000f,100.00f};
int a[7]={-4,-3,-2,-1,0,1,2};
for(int i=0;i<7;i++)
printf("%2d%20f\n",a[i],y[i]);
Output will like that.

Answer 6

You can use sprintf and then trim the zeros. This is the same idea as #Havenard's answer, but writing spaces over the zeros instead of cutting the string.
And my C-style is somewhat different FWIW. My style is that I don't want to count or do any arithmetic in my head; that's what the C optimizer is for :).
#include <math.h>
#include <stdio.h>
#include <string.h>
int main() {
int kPower;
for(kPower=-4; kPower<5; kPower++){
enum { bufsize = 2+5+10+1+4+1+1 };
char buf[bufsize];
int j,n,i;
double raisePower = pow(10,kPower);
//printf("%2d %10.4f\n",kPower,raisePower);
snprintf(buf,bufsize,"%2d %10.4f\n",kPower,raisePower);
j=strchr(buf,'.')-buf;
j+=1;
n=strchr(buf+j,'\n')-buf;
for (i=n-1; i>j; i--)
if (buf[i]=='0')
buf[i]=' ';
else
break;
printf("%s",buf);
}
return 0;
}
Output:
-4 0.0001
-3 0.001
-2 0.01
-1 0.1
0 1.0
1 10.0
2 100.0
3 1000.0
4 10000.0

Answer 7

use printf like this:
printf( "%5d %10.1f\n",kPower, raisePower);
this will result in kPower being printed,
right justified,
in 5 columns
- sign in the right place
this will result in raisePower being printed with:
10 columns
leading 0s replaced by spaces
except 1 digit (could be 0) to the left of the decimal point
1 (rounded) digit to the right of the decimal point
- signs being printed at the proper location
decimal point being aligned