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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

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

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

Compliant Solution (C23, Checked Integers)

This compliant solution uses the new-to-C23 checked integer arithmetic to safely perform integer addition:

#include <stdckdint.h>

void f(signed int si_a, signed int si_b) {
  int sum;
  if (ckd_add(&sum, si_a, si_b)) {
    /* 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 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

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

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

  /* ... */
}

Multiplication

Multiplication is between two operands of arithmetic type.

Noncompliant Code Example

Compliant Solution (C23, Checked Integers)

This compliant solution uses the new-to-C23 checked integer arithmetic to safely perform integer subtractionThis noncompliant code example can result in a signed integer overflow during the multiplication of the signed operands si_a and si_b:

#include <stdckdint.h>

void funcf(signed int si_a, signed int si_b) {
  signedint diff;
 int result =if (ckd_sub(&diff, si_a, * si_b;)) {
    /* ...Handle error */
}

Compliant Solution

The product of two operands can always be represented using twice the number of bits than exist in the precision of the larger of the two operands. This compliant solution eliminates signed overflow on systems where long long is at least twice the precision of int:

  }
  /* ... */
}

Multiplication

Multiplication is between two operands of arithmetic type.

Noncompliant Code Example

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

#include <stddef.h>
#include <assert.h>
#include <limits.h>
#include <inttypes.h>
 
extern size_t popcount(uintmax_t);
#define PRECISION(umax_value) popcount(umax_value) 
  
void func(signed int si_a, signed int si_b) {
  signed int result;
 = signed long long tmp;
  assert(PRECISION(ULLONG_MAX) >= 2 * PRECISION(UINT_MAX));
  tmp = (signed long long)si_a * (signed long long)si_a * si_b;
 
  /*
 ... */
}

Compliant Solution

The product of two operands can always be represented using twice the number of bits than exist in the precision of the larger of the two operands. This compliant solution eliminates signed overflow on systems where long long is at least twice the precision of int:

#include <stddef.h>
#include <assert.h>
#include <limits.h>
#include <inttypes.h>
 
extern size_t popcount(uintmax_t);
#define PRECISION(umax_value) popcount(umax_value) 
  
void func(signed int si_a, signed int si_b  * If the product cannot be represented as a 32-bit integer,
   * handle as an error condition.
   */
  if ((tmp > INT_MAX) || (tmp < INT_MIN)) {
  signed  /* Handle error */
  } else {
    result = (int)tmp;
  }
  /* ... */
}

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:

int result;
  signed long long tmp;
  assert(PRECISION(ULLONG_MAX) >= 2 * PRECISION(UINT_MAX));
  tmp = (signed long long)si_a * (signed long long)si_b;
 
  /*
   * If the product cannot be represented as a 32-bit integer,
   * handle as an error condition.
   */
  if ((tmp > INT_MAX) || (tmp < INT_MIN)) {
    /* Handle error */
  } else {
    result = (int)tmp;
  }
  /* ... */
}

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

#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_ba <> (INT_MINMAX / si_ab)) {
        /* Handle error */
      }
    } else { /* si_a positive, si_b nonpositive */
  }  else { /*if (si_a is nonpositive */
    if (si_b > 0) { /* si_a is nonpositive, si_b is positive */
      if (si_a < (INT_MIN / si_ba)) {
        /* Handle error */
      }
    } else { /* si_a andpositive, si_b are nonpositive */
  } else {  if (/* si_a is nonpositive */
    if (si_ab !=> 0) && { /* si_a is nonpositive, si_b is positive */
      if (si_ba < (INT_MAXMIN / si_ab))) {
        /* Handle error */
      }
    } else { /* End if si_a and si_b are nonpositive */
  }  /* End if ( (si_a is nonpositive!= 0) && (si_b < (INT_MAX / si_a))) {
        /* Handle error */
      }
    } result =/* End if si_a *and si_b;
}

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.)

Noncompliant Code Example

This noncompliant code example prevents divide-by-zero errors in compliance with  INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors but does not prevent a signed integer overflow error in two's-complement. 

 are nonpositive */
  } /* End if si_a is nonpositive */

  result = si_a * si_b;
}

Compliant Solution (C23, Checked Integers)

This compliant solution uses the new-to-C23 checked integer arithmetic to safely perform integer subtraction:

#include <stdckdint.h>

void func(signed int si

void func(signed long s_a, signed longint ssi_b) {
  signedint long resultproduct;
  if (s_b == 0ckd_mul(&product, si_a, si_b)) {
    /* Handle error */
  } else {
    result = s_a / s_b;
  }
  /* ... */
}

Implementation Details

On the x86-32 architecture, overflow results in a fault, which can be exploited as a  denial-of-service attack.

Compliant Solution

This compliant solution eliminates the possibility of divide-by-zero errors or signed overflow:

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.)

Noncompliant Code Example

This noncompliant code example prevents divide-by-zero errors in compliance with  INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors but does not prevent a signed integer overflow error in two's-complement. 

#include <limits.h>
 
void func(signed long s_a, signed long s_b) {
  signed long result;
  if ((s_b == 0) || ((s_a == LONG_MIN) && (s_b == -1))) {
   {
    /* Handle error */
  } else {
     result = s_a / s_b;
  }
   /* ... */
}

Remainder

Implementation Details

On the x86-32 architecture, overflow results in a fault, which can be exploited as a  denial-of-service attack.

Compliant Solution

This compliant solution eliminates the possibility of The remainder operator provides the remainder when two operands of integer type are divided. Because many platforms implement remainder and division in the same instruction, the remainder operator is also susceptible to arithmetic overflow and division by zero. (See INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors .)

Noncompliant Code Example

Many hardware architectures implement remainder as part of the division operator, which can overflow. Overflow can occur during a remainder operation when the dividend is equal to the minimum (negative) value for the signed integer type and the divisor is equal to −1. It occurs even though the result of such a remainder operation is mathematically 0. This noncompliant code example prevents divide-by-zero errors in compliance with INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors but does not prevent integer overflow:

or signed overflow:

#include <limits.h>
 
void func(signed long s_a, signed long s_b) {
  signed long result;
  if ((s_b == 0) || ((s_a == LONG_MIN) && 

void func(signed long s_a, signed long s_b) {
  signed long result;
  if (s_b == 0-1))) {
    /* Handle error */
  } else {
     resultresult = s_a %/ s_b;
  }
   /* ... */
}

Implementation Details

Remainder

The remainder operator provides the remainder when two operands of integer type are divided. Because many platforms implement remainder and division in the same instructionOn x86-32 platforms, the remainder operator for signed integers is implemented by the idiv instruction code, along with the divide operator. Because LONG_MIN / −1 overflows, it results in a software exception with LONG_MIN % −1 as well.

Compliant Solution

This compliant solution also tests the remainder operands to guarantee there is no possibility of an overflow:

is also susceptible to arithmetic overflow and division by zero. (See INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors.)

Noncompliant Code Example

Many hardware architectures implement remainder as part of the division operator, which can overflow. Overflow can occur during a remainder operation when the dividend is equal to the minimum (negative) value for the signed integer type and the divisor is equal to −1. It occurs even though the result of such a remainder operation is mathematically 0. This noncompliant code example prevents divide-by-zero errors in compliance with INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors but does not prevent integer overflow:

void func(signed long s_a, signed long s_b) {
  signed long result;
  if

#include <limits.h>
 
void func(signed long s_a, signed long s_b) {
  signed long result;
  if ((s_b == 0 ) || ((s_a == LONG_MIN) && (s_b == -10))) {
    /* Handle error */
  } else {
    result result = s_a % s_b;
  }  
   /* ... */
}

Left-Shift Operator

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, 6.5.7, paragraph 4 [ISO/IEC 9899:2011], 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.

Implementation Details

On x86-32 platforms, the remainder operator for signed integers is implemented by the idiv instruction code, along with the divide operator. Because LONG_MIN / −1 overflows, it results in a software exception with LONG_MIN % −1 as well.

Compliant Solution

This compliant solution also tests the remainder operands to guarantee there is no possibility of an overflow:

#include <limits.h>
 
void func(signed long s_a, signed long s_b) {
  signed long result;
  if ((s_b == 0 ) || ((s_a == LONG_MIN) && (s_b == -1))) {
    /* Handle error */
  } else {
    result = s_a % s_b;
  }  
  /* ... */
}

Left-Shift Operator

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, 6.5.8, paragraph 4 [ISO/IEC 9899:2024], 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 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:

#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)) ||
      (si_a > (LONG_MAX >> si_b))) {
    /* Handle error */
  } else {
    result = si_a << si_b;
  } 
  /* ... */
}

Unary Negation

The unary negation operator takes an operand of arithmetic type. Overflow can occur during two's complement unary negation when the operand is equal to the minimum (negative) value for the signed integer type.

Noncompliant Code Example

This noncompliant code example can result in a signed integer overflow during the unary negation of the signed operand s_a:

void func(signed long s_a) {
  signed long result = -s_a;#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)) ||
      (si_a > (LONG_MAX >> si_b))) {
    /* Handle error */
  } else {
    result = si_a << si_b;
  } 
  /* ... */
}

Unary Negation

The unary negation operator takes an operand of arithmetic type. Overflow can occur during two's complement unary negation when the operand is equal to the minimum (negative) value for the signed integer type.

Noncompliant Code Example

This noncompliant code example can result in a signed integer overflow during the unary negation of the signed operand s_a:

void func(signed long s_a) {
  signed long result = -s_a;
  /* ... */
}

Compliant Solution

This compliant solution tests the negation operation to guarantee there is no possibility of signed overflow:

#include <limits.h>
 
void func(signed long s_a) {
  signed long result;
  if (s_a == LONG_MIN) {
    /* Handle error */
  } else {
    result = -s_a;
  }
  /* ... */
}

Risk Assessment

Integer overflow can lead to buffer overflows and the execution of arbitrary code by an attacker.

Automated Detection

Compliant Solution

This compliant solution tests the negation operation to guarantee there is no possibility of signed overflow:

#include <limits.h>
 
void func(signed long s_a) {
  signed long result;
  if (s_a == LONG_MIN) {
    /* Handle error */
  } else {
    result = -s_a;
  }
  /* ... */
}

Risk Assessment

Integer overflow can lead to buffer overflows and the execution of arbitrary code by an attacker.

Automated Detection

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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