Any decent way of printing out floats and doubles with commas? - c

Question

I'm working on a program that regards with currency. Ive been finding a solution to display money values decently like this:
9,999.99 USD
Remember when assigning a certain variable with a value (money), you musn't insert commas.
I.e.:
double money=9999.99;
And when accessing it;
printf("%.2l USD",money);
Which will output:
9999.99 USD
This is not what I want, especially on bigger amounts exceeding the hundredth, thousandth, millionth, or even billionth place value.
Now I can't find any solution than printing out the desired output directly on the printf.
printf("9,999.99");
Which is undesirable with many variables.
Can anyone help me out?

Answer 1

Please take a look and printf manual page taking note of the following bit:
*"For some numeric conversions a radix character ("decimal point") or thousands' grouping character is used. The actual character used depends on the LC_NUMERIC part of the locale. The POSIX locale uses '.' as radix character, and does not have a grouping character. Thus,
printf("%'.2f", 1234567.89);
results in "1234567.89" in the POSIX locale, in "1234567,89" in the nl_NL locale, and in "1.234.567,89" in the da_DK locale."*
This can be changed by the function setlocale

Answer 2

There is a function, strfmon which might be able to help you

Answer 3

First, don't use floating-point types to represent money because normally floating-point types are binary and as such cannot represent all decimal fractions (cents) exactly, further these types are prone to rounding errors. Use integers instead and count cents instead of dollars.
#include <stdio.h>
#include <limits.h>
unsigned long long ConstructMoney(unsigned long long dollars, unsigned cents)
{
return dollars * 100 + cents;
}
void PrintWithCommas(unsigned long long n)
{
char s[sizeof n * CHAR_BIT + 1];
char *p = s + sizeof s;
unsigned count = 0;
*--p = '\0';
do
{
*--p = '0' + n % 10;
n /= 10;
if (++count == 3 && n)
{
*--p = ',';
count = 0;
}
} while (n);
printf("%s", p);
}
void PrintMoney(unsigned long long n)
{
PrintWithCommas(n / 100);
putchar('.');
n %= 100;
putchar('0' + n / 10);
putchar('0' + n % 10);
}
int main(void)
{
PrintMoney(ConstructMoney(0, 0)); puts("");
PrintMoney(ConstructMoney(0, 1)); puts("");
PrintMoney(ConstructMoney(1, 0)); puts("");
PrintMoney(ConstructMoney(1, 23)); puts("");
PrintMoney(ConstructMoney(12, 34)); puts("");
PrintMoney(ConstructMoney(123, 45)); puts("");
PrintMoney(ConstructMoney(1234, 56)); puts("");
PrintMoney(ConstructMoney(12345, 67)); puts("");
PrintMoney(ConstructMoney(123456, 78)); puts("");
PrintMoney(ConstructMoney(1234567, 89)); puts("");
return 0;
}
Output (ideone):
0.00
0.01
1.00
1.23
12.34
123.45
1,234.56
12,345.67
123,456.78
1,234,567.89

Answer 4

If you're using the standard library, there's no way to do this -- you have to write some code that does it by hand.
I would recommend multiplying the value by 100, casting to integer, and printing the digits with separators as needed -- it's much easier to handle individual digits on an integer.
The following code, for instance, will fill a char * buffer with the string representation of the value you have:
void formatString (double number, char * buffer) {
if (number < 0) {
*buffer = '-';
formatString(number, buffer + 1);
return;
}
unsigned long long num = (unsigned long long) (number * 100);
unsigned long long x; // temporary storage for counting the digits
unsigned char digits;
for (x = num / 1000, digits = 1; x; digits ++, x /= 10);
// counts the digits, also ensures that there's at least one digit
unsigned char pos; // digit position
for (pos = 1, x = 100; pos < digits; pos ++, x *= 10);
// reuses x as a value for extracting the digit in the needed position;
char * current = buffer;
for (pos = digits; pos; pos --) {
*(current ++) = 48 + (num / x);
// remember 48 + digit gives the ASCII for the digit
if (((pos % 3) == 1) && (pos > 1)) *(current ++) = ',';
num %= x;
x /= 10;
}
*(current ++) = '.';
*(current ++) = 48 + num / 10;
*(current ++) = 48 + num % 10;
*current = 0;
}