Hello
So the function below can print any int using putchar. But I can't figure out how to print floats using putchar. Any ideas on this?
Regards!
Code:void putDigit(int n){ int pow = 1; char lastChar = ' '; if (n < 0){ putchar('-'); lastChar = (char)((int)'0' + (n % 10)); n /= 10; n *= (-1); } while (pow * 10 <= n) pow *= 10; while (pow != 0){ int d = n / pow; putchar((char)((int)'0' + d)); n = n - d * pow; pow /= 10; } if (lastChar != ' '){ putchar(lastChar); } }
Dear Friend,
You can't get the floating value correctly. Because in the putDigit function you gave the argument type as int. So if we pass the floating point value , it will convert the float into int. So it will take the number before "." only.
If we change the argument type as float also, it will give the error. Because we can't perform the % operation for the floating point value.
Thank you,
floating value is not possible
you can print the integer values using your function. you cannot print the floating values.
because.
[%] modules operations are only allowed for integer. see in your code the following line
The negative number's is not possible your code.Code:lastChar = (char)((int)'0' + (n % 10));
================================================== ========================
you can handling the '.' dot in your code.
see the following code in this code I am handling [.] floating values
Code:double atof(char s[]) { double val, power; int i, sign; for(i=0;isspace(s[i]); i++)// skip white space ; sign = (s[i] == '-') ? -1 : 1; if (s[i] == '+' || s[i] == '-') i++; for (val =0.0; isdigit(s[i]);i++) val = 10.0 * val + (s[i]-'0'); if (s[i] == '.') i++; for(power = 1.0; isdigit(s[i]); i++) { val = 10.0 * val + (s[i]-'0'); power *=10.0; //printf("val:%d",val); } return sign * val/ power; }
Last edited by ungalnanban; 02-18-2010 at 12:54 AM. Reason: some spell mistakes
Here's one way (not necessarily the best, but very intuitive), which I will describe to you in an example and let you do the coding.
Say our number is 6543.21
I take the log10 of the number rounded down, which is 3.
I divide the number by 10^3 and round down, which is 6.
I print out the 6.
subtract 6x10^3 from the number. Now the number is 543.21
I take the log of the number rounded down: 2
and so on.
keep doing this until you have printed out the number of digits that you want.
I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo
Thanks so much for clearing that up, Murugaperumal.Originally Posted by murugaperumal
The first step is to study and understand the floating-point specs for your particular machine (typically IEEE 754). Short of that, you can always fall back on ssprintf or similar, of course, tho I'm assuming that the purpose of this exercise is to do everything from scratch, correct?
what do you mean when you say "I divide the number by 10^3 and round down, which is 6." you divide 3 by 1000 ?
6543.21/1000 = 6.54321
I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo
I'm having some trouble making this work. Any one want to give this a try?
Code:void putcf( float f ) { char buf[ BUFSIZ ] = {0}; size_t x; for( x = 0, sprintf( buf, "%f", f ); x < strlen( buf ); x++ ) putchar( buf[ x ] ); }
Quzah.
Hope is the first step on the road to disappointment.
Here is what I coded it and it prints everything upto the decimal point.
But how can I print the dot and everything after that now?Code:void putFloat(double n){ double x; int r, second; x = log10(n); r = n/1000; putDigit(r); second = n - (6*1000); putDigit(second); double x; int r, second; x = log10(n); r = n/1000; putDigit(r); second = n - (6*1000); putDigit(second); }
No offense, but you've really taken a wrong turn on this, imo.
Take what you have up above this post, and toss it - and the farther the the better. Look carefully at what Quzah has posted above. Despite the "little surprise" he's placed in it, that is how it should be done.
It's clear, direct, concise.
What more could you ask for?
