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Narrower primitive arithmetic types can be cast to wider types without any effect on the magnitude of numeric values. However, whereas integer types represent exact values, floating-point types have limited precision. Subclause 6The C Standard, 6.3.1.4 paragraph 2 of the C Standard [ISO/IEC 9899:2011], states:

When a value of integer type is converted to a real floating type, if the value being converted can be represented exactly in the new type, it is unchanged. If the value being converted is in the range of values that can be represented but cannot be represented exactly, the result is either the nearest higher or nearest lower representable value, chosen in an implementation-defined manner. If the value being converted is outside the range of values that can be represented, the behavior is undefined. Results of some implicit conversions may be represented in greater range and precision than that required by the new type (see 6.3.1.8 and 6.8.6.4).

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In this noncompliant example, a large value of type long int is converted to a value of type float without ensuring it is representable in the type:

#include <stdio.h>

int main(void) {
  long int big = 1234567890;
  float approx = big;
  printf("%d\n", (big - (long int)approx));
  return 0;
}

When compiled with GCC 4.8.1 on LinuxFor most floating-point hardware, the value closest to 1234567890 that is representable in type float is 1234567844; consequently, this program prints the value -46.

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This compliant solution replaces the type float with a double. Furthermore, it uses an assertion to guarantee that the double type can represent any long int without loss of precision for implementations. (See see INT35-C. Use correct integer precisions for the definition and rationale of the PRECISION() macro and MSC11-C. Incorporate diagnostic tests using assertions):

#include <assert.h>
#include <stdio.h>
#include <float.h>
#include <limits.h>

extern size_t popcount(uintmax_t); /* See INT35-C */
#define PRECISION(umax_value) popcount(umax_value) 

int main(void) {
  assert(PRECISION(INTLONG_MAX) <= DBL_MANT_DIG * log2(DBL_MANT_DIG));
  long int big = 1234567890;
  double approx = big;
  printf("%d\n", (big - (long int)approx));
  return 0;
}

On the same platformimplementation, this program prints 0, implying that the integer value 1234567890 is representable in type double without change.

Risk Assessment

Conversion from integral types to floating-point types without sufficient precision can lead to loss of precision (loss of least significant bits).

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Related Vulnerabilities

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

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Bibliography

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