How can I round a float value (such as 37.777779) to two decimal places (37.78) in C?
If you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:
#include <math.h>
float val = 37.777779;
float rounded_down = floorf(val * 100) / 100; /* Result: 37.77 */
float nearest = roundf(val * 100) / 100; /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100; /* Result: 37.78 */
Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.
As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.
For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.
Using %.2f in printf. It only print 2 decimal points.
Example:
printf("%.2f", 37.777779);
Output:
37.77
Assuming you're talking about round the value for printing, then Andrew Coleson and AraK's answer are correct:
printf("%.2f", 37.777779);
But note that if you're aiming to round the number to exactly 37.78 for internal use (eg to compare against another value), then this isn't a good idea, due to the way floating point numbers work: you usually don't want to do equality comparisons for floating point, instead use a target value +/- a sigma value. Or encode the number as a string with a known precision, and compare that.
See the link in Greg Hewgill's answer to a related question, which also covers why you shouldn't use floating point for financial calculations.
How about this:
float value = 37.777779;
float rounded = ((int)(value * 100 + .5) / 100.0);
printf("%.2f", 37.777779);
If you want to write to C-string:
char number[24]; // dummy size, you should take care of the size!
sprintf(number, "%.2f", 37.777779);
Always use the printf family of functions for this. Even if you want to get the value as a float, you're best off using snprintf to get the rounded value as a string and then parsing it back with atof:
#include <math.h>
#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>
double dround(double val, int dp) {
int charsNeeded = 1 + snprintf(NULL, 0, "%.*f", dp, val);
char *buffer = malloc(charsNeeded);
snprintf(buffer, charsNeeded, "%.*f", dp, val);
double result = atof(buffer);
free(buffer);
return result;
}
I say this because the approach shown by the currently top-voted answer and several others here -
multiplying by 100, rounding to the nearest integer, and then dividing by 100 again - is flawed in two ways:
For some values, it will round in the wrong direction because the multiplication by 100 changes the decimal digit determining the rounding direction from a 4 to a 5 or vice versa, due to the imprecision of floating point numbers
For some values, multiplying and then dividing by 100 doesn't round-trip, meaning that even if no rounding takes place the end result will be wrong
To illustrate the first kind of error - the rounding direction sometimes being wrong - try running this program:
int main(void) {
// This number is EXACTLY representable as a double
double x = 0.01499999999999999944488848768742172978818416595458984375;
printf("x: %.50f\n", x);
double res1 = dround(x, 2);
double res2 = round(100 * x) / 100;
printf("Rounded with snprintf: %.50f\n", res1);
printf("Rounded with round, then divided: %.50f\n", res2);
}
You'll see this output:
x: 0.01499999999999999944488848768742172978818416595459
Rounded with snprintf: 0.01000000000000000020816681711721685132943093776703
Rounded with round, then divided: 0.02000000000000000041633363423443370265886187553406
Note that the value we started with was less than 0.015, and so the mathematically correct answer when rounding it to 2 decimal places is 0.01. Of course, 0.01 is not exactly representable as a double, but we expect our result to be the double nearest to 0.01. Using snprintf gives us that result, but using round(100 * x) / 100 gives us 0.02, which is wrong. Why? Because 100 * x gives us exactly 1.5 as the result. Multiplying by 100 thus changes the correct direction to round in.
To illustrate the second kind of error - the result sometimes being wrong due to * 100 and / 100 not truly being inverses of each other - we can do a similar exercise with a very big number:
int main(void) {
double x = 8631192423766613.0;
printf("x: %.1f\n", x);
double res1 = dround(x, 2);
double res2 = round(100 * x) / 100;
printf("Rounded with snprintf: %.1f\n", res1);
printf("Rounded with round, then divided: %.1f\n", res2);
}
Our number now doesn't even have a fractional part; it's an integer value, just stored with type double. So the result after rounding it should be the same number we started with, right?
If you run the program above, you'll see:
x: 8631192423766613.0
Rounded with snprintf: 8631192423766613.0
Rounded with round, then divided: 8631192423766612.0
Oops. Our snprintf method returns the right result again, but the multiply-then-round-then-divide approach fails. That's because the mathematically correct value of 8631192423766613.0 * 100, 863119242376661300.0, is not exactly representable as a double; the closest value is 863119242376661248.0. When you divide that back by 100, you get 8631192423766612.0 - a different number to the one you started with.
Hopefully that's a sufficient demonstration that using roundf for rounding to a number of decimal places is broken, and that you should use snprintf instead. If that feels like a horrible hack to you, perhaps you'll be reassured by the knowledge that it's basically what CPython does.
Also, if you're using C++, you can just create a function like this:
string prd(const double x, const int decDigits) {
stringstream ss;
ss << fixed;
ss.precision(decDigits); // set # places after decimal
ss << x;
return ss.str();
}
You can then output any double myDouble with n places after the decimal point with code such as this:
std::cout << prd(myDouble,n);
There isn't a way to round a float to another float because the rounded float may not be representable (a limitation of floating-point numbers). For instance, say you round 37.777779 to 37.78, but the nearest representable number is 37.781.
However, you can "round" a float by using a format string function.
You can still use:
float ceilf(float x); // don't forget #include <math.h> and link with -lm.
example:
float valueToRound = 37.777779;
float roundedValue = ceilf(valueToRound * 100) / 100;
In C++ (or in C with C-style casts), you could create the function:
/* Function to control # of decimal places to be output for x */
double showDecimals(const double& x, const int& numDecimals) {
int y=x;
double z=x-y;
double m=pow(10,numDecimals);
double q=z*m;
double r=round(q);
return static_cast<double>(y)+(1.0/m)*r;
}
Then std::cout << showDecimals(37.777779,2); would produce: 37.78.
Obviously you don't really need to create all 5 variables in that function, but I leave them there so you can see the logic. There are probably simpler solutions, but this works well for me--especially since it allows me to adjust the number of digits after the decimal place as I need.
Use float roundf(float x).
"The round functions round their argument to the nearest integer value in floating-point format, rounding halfway cases away from zero, regardless of the current rounding direction." C11dr §7.12.9.5
#include <math.h>
float y = roundf(x * 100.0f) / 100.0f;
Depending on your float implementation, numbers that may appear to be half-way are not. as floating-point is typically base-2 oriented. Further, precisely rounding to the nearest 0.01 on all "half-way" cases is most challenging.
void r100(const char *s) {
float x, y;
sscanf(s, "%f", &x);
y = round(x*100.0)/100.0;
printf("%6s %.12e %.12e\n", s, x, y);
}
int main(void) {
r100("1.115");
r100("1.125");
r100("1.135");
return 0;
}
1.115 1.115000009537e+00 1.120000004768e+00
1.125 1.125000000000e+00 1.129999995232e+00
1.135 1.134999990463e+00 1.139999985695e+00
Although "1.115" is "half-way" between 1.11 and 1.12, when converted to float, the value is 1.115000009537... and is no longer "half-way", but closer to 1.12 and rounds to the closest float of 1.120000004768...
"1.125" is "half-way" between 1.12 and 1.13, when converted to float, the value is exactly 1.125 and is "half-way". It rounds toward 1.13 due to ties to even rule and rounds to the closest float of 1.129999995232...
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float, the value is 1.134999990463... and is no longer "half-way", but closer to 1.13 and rounds to the closest float of 1.129999995232...
If code used
y = roundf(x*100.0f)/100.0f;
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float, the value is 1.134999990463... and is no longer "half-way", but closer to 1.13 but incorrectly rounds to float of 1.139999985695... due to the more limited precision of float vs. double. This incorrect value may be viewed as correct, depending on coding goals.
Code definition :
#define roundz(x,d) ((floor(((x)*pow(10,d))+.5))/pow(10,d))
Results :
a = 8.000000
sqrt(a) = r = 2.828427
roundz(r,2) = 2.830000
roundz(r,3) = 2.828000
roundz(r,5) = 2.828430
double f_round(double dval, int n)
{
char l_fmtp[32], l_buf[64];
char *p_str;
sprintf (l_fmtp, "%%.%df", n);
if (dval>=0)
sprintf (l_buf, l_fmtp, dval);
else
sprintf (l_buf, l_fmtp, dval);
return ((double)strtod(l_buf, &p_str));
}
Here n is the number of decimals
example:
double d = 100.23456;
printf("%f", f_round(d, 4));// result: 100.2346
printf("%f", f_round(d, 2));// result: 100.23
I made this macro for rounding float numbers.
Add it in your header / being of file
#define ROUNDF(f, c) (((float)((int)((f) * (c))) / (c)))
Here is an example:
float x = ROUNDF(3.141592, 100)
x equals 3.14 :)
Let me first attempt to justify my reason for adding yet another answer to this question. In an ideal world, rounding is not really a big deal. However, in real systems, you may need to contend with several issues that can result in rounding that may not be what you expect. For example, you may be performing financial calculations where final results are rounded and displayed to users as 2 decimal places; these same values are stored with fixed precision in a database that may include more than 2 decimal places (for various reasons; there is no optimal number of places to keep...depends on specific situations each system must support, e.g. tiny items whose prices are fractions of a penny per unit); and, floating point computations performed on values where the results are plus/minus epsilon. I have been confronting these issues and evolving my own strategy over the years. I won't claim that I have faced every scenario or have the best answer, but below is an example of my approach so far that overcomes these issues:
Suppose 6 decimal places is regarded as sufficient precision for calculations on floats/doubles (an arbitrary decision for the specific application), using the following rounding function/method:
double Round(double x, int p)
{
if (x != 0.0) {
return ((floor((fabs(x)*pow(double(10.0),p))+0.5))/pow(double(10.0),p))*(x/fabs(x));
} else {
return 0.0;
}
}
Rounding to 2 decimal places for presentation of a result can be performed as:
double val;
// ...perform calculations on val
String(Round(Round(Round(val,8),6),2));
For val = 6.825, result is 6.83 as expected.
For val = 6.824999, result is 6.82. Here the assumption is that the calculation resulted in exactly 6.824999 and the 7th decimal place is zero.
For val = 6.8249999, result is 6.83. The 7th decimal place being 9 in this case causes the Round(val,6) function to give the expected result. For this case, there could be any number of trailing 9s.
For val = 6.824999499999, result is 6.83. Rounding to the 8th decimal place as a first step, i.e. Round(val,8), takes care of the one nasty case whereby a calculated floating point result calculates to 6.8249995, but is internally represented as 6.824999499999....
Finally, the example from the question...val = 37.777779 results in 37.78.
This approach could be further generalized as:
double val;
// ...perform calculations on val
String(Round(Round(Round(val,N+2),N),2));
where N is precision to be maintained for all intermediate calculations on floats/doubles. This works on negative values as well. I do not know if this approach is mathematically correct for all possibilities.
...or you can do it the old-fashioned way without any libraries:
float a = 37.777779;
int b = a; // b = 37
float c = a - b; // c = 0.777779
c *= 100; // c = 77.777863
int d = c; // d = 77;
a = b + d / (float)100; // a = 37.770000;
That of course if you want to remove the extra information from the number.
this function takes the number and precision and returns the rounded off number
float roundoff(float num,int precision)
{
int temp=(int )(num*pow(10,precision));
int num1=num*pow(10,precision+1);
temp*=10;
temp+=5;
if(num1>=temp)
num1+=10;
num1/=10;
num1*=10;
num=num1/pow(10,precision+1);
return num;
}
it converts the floating point number into int by left shifting the point and checking for the greater than five condition.
As we all know, floating point numbers can't exactly represent most numbers. I'm not asking a question about the precision of floats or doubles.
In a program, floating point numbers "come from somewhere". Some might originate by promoting an integer, others as numeric literals.
int x = 3;
double xd = x;
float xf = 3.0f;
double xd2 = 3.0;
Of course, some floating point numbers come from calculations involving other numbers.
double yd = std::cos(4.0);
In my problem, I will sometimes read in floating point numbers from a text file, and other times, I will receive them from a complex function that I must treat as a black box. The creator of the text file may choose to enter as many significant figures as they like -- they might only use three, or perhaps eight.
I will then perform some computations using these numbers and I would like to know how many significant figures were implied when they were created.
For argument, consider that I am performing an adaptive piecewise least squares fit to the input points. I will continue splitting my piecewise segments until a certain tolerance is achieved. I want to (in part) base the tolerance on the significant figures of the input data -- don't fit to 10^-8 if the data are rounded to the nearest 10^-3.
Others (below) have asked similar questions (but not quite the same). For example, I'm not particularly concerned with a representation output to the user in a pretty form. I'm not particularly concerned with recovering the same floating point representation from an output text value.
How to calculate the number of significant decimal digits of a c++ double?
How can I test for how many significant figures a float has in C++?
I'd like to calculate the sig figs based purely on the value of the double itself. I don't want to do a bunch of text processing on the original data file.
In most cases, floating point numbers will end up with a large series of 0000000 or 99999999 in the middle of them. An intuitive problem statement is that I'm interested in figuring out where that repeating 0 or 9 sequence begins. However, I'd prefer to not do this with a looping rounded string conversion approach. I'm hoping for a fairly direct and efficient way to figure this out.
Perhaps something as simple at looking at the least significant 'on' bit and then figure out its magnitude?
Ok, I've come up with something like this....
#include <cstdlib>
#include <cstdio>
#include <cfloat>
#include <cmath>
double sigfigs( double x )
{
int m = floor( log10( std::abs( x ) ) );
double pow10i;
for ( int i = m; i > -26; i-- )
{
pow10i = pow( 10, i );
double y = round( x / pow10i ) * pow10i;
if ( std::abs( x - y ) < std::abs( x ) * 10.0 * DBL_EPSILON )
break;
}
return pow10i;
}
int main( )
{
char fmt[10];
sprintf( fmt, "%%.%de", DBL_DIG + 3 );
double x[9] = {1.0, 0.1, 1.2, 1.23, 1.234, 1.2345, 100.2, 103000, 100.3001};
for ( int i = 0; i < 9; i++ )
{
printf( "Double: " );
printf( fmt, x[i] );
printf( " %f is good to %g.\n", x[i], sigfigs( x[i] ) );
}
for ( int i = 0; i < 9; i++ )
{
printf( "Double: " );
printf( fmt, -x[i] );
printf( " %f is good to %g.\n", -x[i], sigfigs( -x[i] ) );
}
exit( 0 );
}
Which gives output:
Double: 1.000000000000000000e+00 1.000000 is good to 1.
Double: 1.000000000000000056e-01 0.100000 is good to 0.1.
Double: 1.199999999999999956e+00 1.200000 is good to 0.1.
Double: 1.229999999999999982e+00 1.230000 is good to 0.01.
Double: 1.233999999999999986e+00 1.234000 is good to 0.001.
Double: 1.234499999999999931e+00 1.234500 is good to 0.0001.
Double: 1.002000000000000028e+02 100.200000 is good to 0.1.
Double: 1.030000000000000000e+05 103000.000000 is good to 1000.
Double: 1.003001000000000005e+02 100.300100 is good to 0.0001.
Double: -1.000000000000000000e+00 -1.000000 is good to 1.
Double: -1.000000000000000056e-01 -0.100000 is good to 0.1.
Double: -1.199999999999999956e+00 -1.200000 is good to 0.1.
Double: -1.229999999999999982e+00 -1.230000 is good to 0.01.
Double: -1.233999999999999986e+00 -1.234000 is good to 0.001.
Double: -1.234499999999999931e+00 -1.234500 is good to 0.0001.
Double: -1.002000000000000028e+02 -100.200000 is good to 0.1.
Double: -1.030000000000000000e+05 -103000.000000 is good to 1000.
Double: -1.003001000000000005e+02 -100.300100 is good to 0.0001.
It mostly seems to work as desired. The pow(10,i) is a bit unfortunate, as is the estimation of the number's base 10 magnitude.
Also, the estimate of the difference between representable doubles is somewhat crude.
Does anyone spot any corner cases where this fails? Does anyone see any obvious ways to improve or optimize this? It would be nice if it was really cheap...
Rob
I suggest dividing the problem into two steps:
Find the minimum number of significant digits for each number.
Analyze the distribution of the numbers from step 1.
For step 1, you can use the method described in the prior answer, or a method based on conversion to decimal string followed by regular expression analysis of the string. The minimum number of digits for 0.1 is indeed 1, as reported by the algorithm in that answer.
Before fixing rules for step 2 you should examine the distributions that result from several different sets of actual numbers whose input significant digits you know. If the problem is solvable at all there should be a peak and sharp drop-off at the required number of digits.
Consider the case (0.1 0.234 0.563 0.607 0.89). The results of step 1 would be:
Digits Count
1 1
2 1
3 3
and count zero for 4 or greater, suggesting 3 significant digits overall.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
strange output in comparison of float with float literal
When I am trying to compare 2 same float values it doesn't print "equal values" in the following code :
void main()
{
float a = 0.7;
clrscr();
if (a < 0.7)
printf("value : %f",a);
else if (a == 0.7)
printf("equal values");
else
printf("hello");
getch();
}
Thanks in advance.
While many people will tell you to always compare floating point numbers with an epsilon (and it's usually a good idea, though it should be a percentage of the values being compared rather than a fixed value), that's not actually necessary here since you're using constants.
Your specific problem here is that:
float a = 0.7;
uses the double constant 0.7 to create a single precision number (losing some precision) while:
if (a == 0.7)
will compare two double precision numbers (a is promoted first).
The precision that was lost when turning the double 0.7 into the float a is not regained when promoting a back to a double.
If you change all those 0.7 values to 0.7f (to force float rather than double), or if you just make a a double, it will work fine - I rarely use float nowadays unless I have a massive array of them and need to save space.
You can see this in action with:
#include <stdio.h>
int main (void){
float f = 0.7; // double converted to float
double d1 = 0.7; // double kept as double
double d2 = f; // float converted back to double
printf ("double: %.30f\n", d1);
printf ("double from float: %.30f\n", d2);
return 0;
}
which will output something like (slightly modified to show difference):
double: 0.6999999|99999999955591079014994
double from float: 0.6999999|88079071044921875000000
\_ different beyond here.
Floating point number are not what you think they are: here are two sources with more information: What Every Computer Scientist Should Know About Floating-Point Arithmetic and The Floating-Point Guide.
The short answer is that due to the way floating point numbers are represented, you cannot do basic comparison or arithmetic and expect it to work.
You are comparing a single-precision approximation of 0.7 with a double-precision approximation. To get the expected output you should use:
if(a == 0.7f) // check a is exactly 0.7f
Note that due to representation and rounding errors it may be very unlikely to ever get exactly 0.7f from any operation. In general you should check if fabs(a-0.7) is sufficiently close to 0.
Don't forget that the exact value of 0.7f is not really 0.7, but slightly lower:
0.7f = 0.699999988079071044921875
The exact value of the double precision representation of 0.7 is a better approximation, but still not exactly 0.7:
0.7d = 0.6999999999999999555910790149937383830547332763671875
a is a float; 0.7 is a value of type double.
The comparison between the two requires a conversion. The compiler will convert the float value to a double value ... and the value resulting from converting a float to a double is not the same as the value resulting from the compiler converting a string of text (the source code) to a double.
But don't ever compare floating point values (float, double, or long double) with ==.
You might like to read "What Every Programmer Should Know About Floating-Point Arithmetic".
Floating point numbers must not be compared with the "==" operator.
Instead of comparing float numbers with the "==" operator, you can use a function like this one :
//compares if the float f1 is equal with f2 and returns 1 if true and 0 if false
int compare_float(float f1, float f2)
{
float precision = 0.00001;
if (((f1 - precision) < f2) &&
((f1 + precision) > f2))
{
return 1;
}
else
{
return 0;
}
}
The lack of absolute precision in floats makes it more difficult to do trivial comparisons than for integers. See this page on comparing floats in C. In particular, one code snippet lifted from there exhibits a 'workaround' to this issue:
bool AlmostEqual2sComplement(float A, float B, int maxUlps)
{
// Make sure maxUlps is non-negative and small enough that the
// default NAN won't compare as equal to anything.
assert(maxUlps > 0 && maxUlps < 4 * 1024 * 1024);
int aInt = *(int*)&A;
// Make aInt lexicographically ordered as a twos-complement int
if (aInt < 0)
aInt = 0x80000000 - aInt;
// Make bInt lexicographically ordered as a twos-complement int
int bInt = *(int*)&B;
if (bInt < 0)
bInt = 0x80000000 - bInt;
int intDiff = abs(aInt - bInt);
if (intDiff <= maxUlps)
return true;
return false;
}
A simple and common workaround is to provide an epsilon with code like so:
if (fabs(result - expectedResult) < 0.00001)
This essentially checks the difference between the values is within a threshold. See the linked article as to why this is not always optimal though :)
Another article is pretty much the de facto standard of what is linked to when people ask about floats on SO.
if you need to compare a with 0.7 than
if( fabs(a-0.7) < 0.00001 )
//your code
here 0.00001 can be changed to less (like 0.00000001) or more (like 0.0001) > It depends on the precision you need.
This question already has answers here:
Floating point inaccuracy examples
(7 answers)
Closed 8 years ago.
int main()
{
float a = 0.7;
float b = 0.5;
if (a < 0.7)
{
if (b < 0.5) printf("2 are right");
else printf("1 is right");
}
else printf("0 are right");
}
I would have expected the output of this code to be 0 are right.
But to my dismay the output is 1 is right why?
int main()
{
float a = 0.7, b = 0.5; // These are FLOATS
if(a < .7) // This is a DOUBLE
{
if(b < .5) // This is a DOUBLE
printf("2 are right");
else
printf("1 is right");
}
else
printf("0 are right");
}
Floats get promoted to doubles during comparison, and since floats are less precise than doubles, 0.7 as float is not the same as 0.7 as double. In this case, 0.7 as float becomes inferior to 0.7 as double when it gets promoted. And as Christian said, 0.5 being a power of 2 is always represented exactly, so the test works as expected: 0.5 < 0.5 is false.
So either:
Change float to double, or:
Change .7 and .5 to .7f and .5f,
and you will get the expected behavior.
The issue is that the constants you are comparing to are double not float. Also, changing your constants to something that is representable easily such as a factor of 5 will make it say 0 is right. For example,
main()
{
float a=0.25,b=0.5;
if(a<.25)
{
if(b<.5)
printf("2 are right");
else
printf("1 is right");
}
else
printf("0 are right");
}
Output:
0 are right
This SO question on Most Effective Way for float and double comparison covers this topic.
Also, this article at cygnus on floating point number comparison gives us some tips:
The IEEE float and double formats were designed so that the numbers
are “lexicographically ordered”, which – in the words of IEEE
architect William Kahan means “if two floating-point numbers in the
same format are ordered ( say x < y ), then they are ordered the same
way when their bits are reinterpreted as Sign-Magnitude integers.”
This means that if we take two floats in memory, interpret their bit
pattern as integers, and compare them, we can tell which is larger,
without doing a floating point comparison. In the C/C++ language this
comparison looks like this:
if (*(int*)&f1 < *(int*)&f2)
This charming syntax means take the address of f1, treat it as an
integer pointer, and dereference it. All those pointer operations look
expensive, but they basically all cancel out and just mean ‘treat f1
as an integer’. Since we apply the same syntax to f2 the whole line
means ‘compare f1 and f2, using their in-memory representations
interpreted as integers instead of floats’.
It's due to rounding issues while converting from float to double
Generally comparing equality with floats is a dangerous business (which is effectively what you're doing as you're comparing right on the boundary of > ), remember that in decimal certain fractions (like 1/3) cannot be expressed exactly, the same can be said of binary,
0.5= 0.1, will be the same in float or double.
0.7=0.10110011001100 etc forever, 0.7 cannot be exactly represented in binary, you get rounding errors and may be (very very slightly) different between float and double
Note that going between floats and doubles you cut off a different number of decimal places, hence your inconsistant results.
Also, btw, you have an error in your logic of 0 are right. You don't check b when you output 0 are right. But the whole thing is a little mysterious in what you are really trying to accomplish. Floating point comparisons between floats and doubles will have variations, minute, so you should compare with a delta 'acceptable' variation for your situation. I've always done this via inline functions that just perform the work (did it once with a macro, but thats too messy). Anyhow, yah, rounding issues abound with this type of example. Read the floating point stuff, and know that .7 is different than .7f and assigning .7 to a float will cast a double into a float, thus changing the exact nature of the value. But, the programming assumption about b being wrong since you checked a blared out to me, and I had to note that :)