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#include <limits.h>
 
void f(signed int si_a, signed int si_b) {
  signed int sum;
  if (((si_b > 0) && (si_a > (INT_MAX - si_b))) ||
      ((si_b < 0) && (si_a < (INT_MIN - si_b)))) {
    /* Handle error */
  } else {
    sum = si_a + si_b;
  }
  /* ... */
}

Subtraction

Compliant Solution (GNU)

This compliant solution uses the GNU extension __builtin_sadd_overflow, available with GCC, Clang, and ICC:

void f(signed int si_a, signed int si_b) {
  signed int sum;
  if (__builtin_sadd_overflow(si_a, si_b, &sum)) {
    /* Handle error */
  }
  /* ... */
}

Subtraction

Subtraction is between two operands of arithmetic type, two pointers to qualified or unqualified versions of compatible object types, or a pointer to an object type and an integer type. This rule applies only to subtraction between Subtraction is between two operands of arithmetic type, two pointers to qualified or unqualified versions of compatible object types, or a pointer to an object type and an integer type. This rule applies only to subtraction between two operands of arithmetic type. (See ARR36-C. Do not subtract or compare two pointers that do not refer to the same array, ARR37-C. Do not add or subtract an integer to a pointer to a non-array object, and ARR30-C. Do not form or use out-of-bounds pointers or array subscripts for information about pointer subtraction.)

Decrementing is equivalent to subtracting 1.

Noncompliant Code Example

-C. Do not add or subtract an integer to a pointer to a non-array object, and ARR30-C. Do not form or use out-of-bounds pointers or array subscripts for information about pointer subtraction.)

Decrementing is equivalent to subtracting 1.

Noncompliant Code Example

This noncompliant code example can result in a signed integer overflow during the subtraction of the signed operands si_a and si_b:

void func(signed int si_a, signed int si_b) {
  signed int diff = si_a - si_b;
  /* ... */
}

Compliant Solution

This compliant solution tests the operands of the subtraction to guarantee there is no possibility of signed overflow, regardless of representationThis noncompliant code example can result in a signed integer overflow during the subtraction of the signed operands si_a and si_b:

#include <limits.h>
 
void func(signed int si_a, signed int si_b) {
  signed int diff =;
  if ((si_b > 0 && si_a < INT_MIN + si_b) ||
      (si_b < 0 && si_a - > INT_MAX + si_b;)) {
    /* ...Handle error */
}

Compliant Solution

This compliant solution tests the operands of the subtraction to guarantee there is no possibility of signed overflow, regardless of representation:

#include <limits.h>
 
void func(signed int  } else {
    diff = si_a, signed- int si_b) {;
  }

  /* ... */
}

Compliant Solution (GNU)

This compliant solution uses the GNU extension __builtin_ssub_overflow, available with GCC, Clang, and ICC:

void func(signed intsigned int diff;
  if ((si_b > 0 && si_a, <signed INT_MINint + si_b) ||{
  signed int diff;
  if (si_b < 0 && si_a > INT_MAX +__builtin_ssub_overflow(si_a, si_b, &diff)) {
    /* Handle error */
  } else {
    diff = si_a - si_b; Handle error */
  }

  /* ... */
}

Multiplication

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The assertion fails if long long has less than twice the precision of int. The  PRECISION() macro and popcount() function provide the correct precision for any integer type. (See INT35-C. Use correct integer precisions.)

Compliant Solution

The following portable compliant solution can be used with any conforming implementation, including those that do not have an integer type that is at least twice the precision of int:

#include <limits.h>
 
void func(signed int si_a, signed int si_b) {
  signed int result;  
  if (si_a > 0) {  /* si_a is positive */
    if (si_b > 0) {  /* si_a and si_b are positive */
      if (si_a > (INT_MAX / si_b)) {
        /* Handle error */
      }
    } else { /* si_a positive, si_b nonpositive */
      if (si_b < (INT_MIN / si_a)) {
        /* Handle error */
      }
    } /* si_a positive, si_b nonpositive */
  } else { /* si_a is nonpositive */
    if (si_b > 0) { /* si_a is nonpositive, si_b is positive */
      if (si_a < (INT_MIN / si_b)) {
        /* Handle error */
      }
    } else { /* si_a and si_b are nonpositive */
      if ( (si_a != 0) && (si_b < (INT_MAX / si_a))) {
        /* Handle error */
      }
    } /* End if si_a and si_b are nonpositive */
  } /* End if si_a is nonpositive */

  result = si_a * si_b; {
        /* Handle error */
      }
    } /* End if si_a and si_b are nonpositive */
  } /* End if si_a is nonpositive */

  result = si_a * si_b;
}

Compliant Solution (GNU)

This compliant solution uses the GNU extension __builtin_smul_overflow, available with GCC, Clang, and ICC:

void func(signed int si_a, signed int si_b) {
  signed int result;
  if (__builtin_smul_overflow(si_a, si_b, &result)) {
    /* Handle error */
  }
}

Division

Division is between two operands of arithmetic type. Overflow can occur during two's complement signed integer division when the dividend is equal to the minimum (negative) value for the signed integer type and the divisor is equal to −1. Division operations are also susceptible to divide-by-zero errors. (See INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors.)

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The left-shift operator takes two integer operands. The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. 

The C Standard,  66.5.78, paragraph 4 [ISO/IEC 9899:20112024], states

If E1 has a signed type and nonnegative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

In almost every case, an attempt to shift by a negative number of bits or by more bits than exist in the operand indicates a logic error. These issues are covered by INT34-C. Do not shift an expression by a negative number of bits or by greater than or equal to the number of bits that exist in the operand.

Noncompliant Code Example

This noncompliant code example performs a left shift, after verifying that the number being shifted is not negative, and the number of bits to shift is valid.  The PRECISION() macro and popcount() function provide the correct precision for any integer type. (See INT35-C. Use correct integer precisions.) However, because this code does no overflow check, it can result in an unrepresentable value. 

#include <limits.h>
#include <stddef.h>
#include <inttypes.h>
 
extern size_t popcount(uintmax_t);
#define PRECISION(umax_value) popcount(umax_value) 

void func(signed long si_a, signed long si_b) {
  signed long result;
  if ((si_a < 0) || (si_b < 0) ||
      (si_b >= PRECISION(ULONG_MAX))) {
    /* Handle error */
  } else {
    result = si_a << si_b;
  } 
  /* ... */
}

Compliant Solution

This compliant solution eliminates the possibility of overflow resulting from a left-shift operation:

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Related Vulnerabilities

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

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Union( CWE-190, CWE-191) = Union( INT30-C, INT32-C)

Intersection( INT30-C, INT32-C) == Ø

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Union( CWE-190, CWE-191) = Union( INT30-C, INT32-C)

Intersection( INT30-C, INT32-C) == Ø

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