FLP02-C. Consider avoiding floating point numbers when precise computation is needed

Computers can only represent a finite number of digits. As a result, it is impossible to precisely represent repeating binary-representation values such as 1/3 or 1/5 with the most common floating point representation: binary floating point.

When precise computation is necessary, consider alternative representations that may be able to completely represent values. For example, if you are performing arithmetic on decimal values and need an exact decimal rounding, represent the values in binary-coded decimal instead of using floating point. Another option is decimal floating-point arithmetic as specified by ANSI/IEEE 754-2007. ISO/IEC WG14 has drafted a proposal to add support for decimal floating-point arithmetic to the C language [[ISO/IEC DTR 24732]].

When precise computation is necessary, carefully and methodically estimate the maximum cumulative error of the computations, regardless of whether decimal or binary is used, to ensure that the resulting error is within tolerances. Consider using numerical analysis to properly understand the numerical properties of the problem. A useful introduction can be found in [[Goldberg 91]].

Noncompliant Code Example

This noncompliant code example takes the mean of ten numbers, and then checks to see if the mean matches the first number. It should, because the ten numbers are all 10.1. Yet, due to the imprecision of floating-point arithmetic, the computed mean does not match the numbers.

#include <stdio.h>

/* Returns the mean value of the array */
float mean(float array[], int size) {
  float total = 0.0;
  int i;
  for (i = 0; i < size; i++) {
    total += array[i];
    printf("array[%d] = %f and total is %f\n", i, array[i], total);
  }
  if (size != 0)
    return total / size;
  else
    return 0.0;
}

enum {array_size = 10};
float array_value = 10.1;

int main(void) {
  float array[array_size];
  float avg;
  int i;
  for (i = 0; i < array_size; i++) {
    array[i] = array_value;
  }

  avg = mean( array, array_size);
  printf("mean is %f\n", avg);
  if (avg == array[0]) {
    printf("array[0] is the mean\n");
  } else {
    printf("array[0] is not the mean\n");
  }
  return 0;
}

On a 64-bit Linux machine using gcc 4.1, this program yields the following output:

array[0] = 10.100000 and total is 10.100000
array[1] = 10.100000 and total is 20.200001
array[2] = 10.100000 and total is 30.300001
array[3] = 10.100000 and total is 40.400002
array[4] = 10.100000 and total is 50.500000
array[5] = 10.100000 and total is 60.599998
array[6] = 10.100000 and total is 70.699997
array[7] = 10.100000 and total is 80.799995
array[8] = 10.100000 and total is 90.899994
array[9] = 10.100000 and total is 100.999992
mean is 10.099999
array[0] is not the mean

Compliant Solution

This code may be fixed by replacing the floating point numbers with integers for the internal additions. Floats are used only when printing results, and when doing the division to compute the mean.

#include <stdio.h>

float nearbyintf( float);

/* Returns the mean value of the array */
float mean(int array[], int size) {
  int total = 0;
  int i;
  for (i = 0; i < size; i++) {
    total += array[i];
    printf("array[%d] = %f and total is %f\n", i, array[i] / 100.0, total / 100.0);
  }
  if (size != 0)
    return ((float) total) / size;
  else
    return 0.0;
}

enum {array_size = 10};
int array_value = 1010;

int main(void) {
  int array[array_size];
  float avg;
  int i;
  for (i = 0; i < array_size; i++) {
    array[i] = array_value;
  }

  avg = mean( array, array_size);
  printf("mean is %f\n", avg / 100.0);
  if (((int) nearbyintf( avg)) == array[0]) {
    printf("array[0] is the mean\n");
  } else {
    printf("array[0] is not the mean\n");
  }
  return 0;
}

On a 64-bit Linux machine using gcc 4.1, this program yields the following expected output:

array[0] = 10.100000 and total is 10.100000
array[1] = 10.100000 and total is 20.200000
array[2] = 10.100000 and total is 30.300000
array[3] = 10.100000 and total is 40.400000
array[4] = 10.100000 and total is 50.500000
array[5] = 10.100000 and total is 60.600000
array[6] = 10.100000 and total is 70.700000
array[7] = 10.100000 and total is 80.800000
array[8] = 10.100000 and total is 90.900000
array[9] = 10.100000 and total is 101.000000
mean is 10.100000
array[0] is the mean

Risk Analysis

Using a representation other than floating point may allow for more precision and accuracy for critical arithmetic.

Recommendation	Severity	Likelihood	Remediation Cost	Priority	Level
FLP02-C	low	probable	high	P2	L3

Automated Detection

Compass/ROSE can detect violations of this recommendation. In particular, it checks to see if the arguments to an equality operator are of a floating point type.

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

References

[[IEEE 754 2006]]
[[ISO/IEC JTC1/SC22/WG11]]
[[ISO/IEC PDTR 24772]] "PLF Floating Point Arithmetic"
[[ISO/IEC DTR 24732]]
[[Goldberg 91]]

FLP01-C. Take care in rearranging floating point expressions 05. Floating Point (FLP) FLP03-C. Detect and handle floating point errors

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