Signed integer overflow is undefined behavior (see undefined behavior 36 in Annex J of the C Standard). Consequently, implementations have considerable latitude in how they deal with signed integer overflow. (see See MSC15-C. Do not depend on undefined behavior.) . An implementation that defines signed integer types as being modulo, for example, need not detect integer overflow. Implementations may also trap on signed arithmetic overflows, or simply assume that overflows will never happen and generate object code accordingly.  It is also possible for the same conforming implementation to emit code that exhibits different behavior in different contexts. For example, an implementation may determine that a signed integer loop control variable declared in a local scope cannot overflow and may emit efficient code on the basis of that determination, while the same implementation may determine that a global variable used in as a similar context will wrap.

For these reasons, it is important to ensure that operations on signed integers do no not result in overflow. (See MSC15-C. Do not depend on undefined behavior. ) Of particular importance are operations on signed integer values that originate from a tainted source and are used as:

• integer Integer operands of any pointer arithmetic, including array indexing;
• the The assignment expression for the declaration of a variable length array;
• the The postfix expression preceding square brackets [] or the expression in square brackets [] of a subscripted designation of an element of an array object; and
• function Function arguments of type size_t or rsize_t (for example, an argument to a memory allocation function).

Most integer operations can result in Integer operations will overflow if the resulting value cannot be represented by the underlying representation of the integer. The following table indicates which operators operations can result in overflow:.

The following sections examine specific operations that are susceptible to integer overflow. When operating on small integer types (smaller with less precision than int), integer promotions are applied. The usual arithmetic conversions may also be applied to (implicitly) convert operands to equivalent types before arithmetic operations are performed. Programmers should understand integer conversion rules before trying to implement secure arithmetic operations. (See INT02-C. Understand integer conversion rules.)

Implementation Details

GNU GCC invoked with the -fwrapv command-line option defines the same modulo arithmetic for both unsigned and signed integers.

...

GNU GCC invoked without either the -fwrapv or the -ftrapv option may simply assume that signed integers never overflow and may generate object code accordingly. AnchorAdditionAddition

Addition

Atomic Integers

The C Standard defines the behavior of arithmetic on atomic signed integer types to use two's complement representation with silent wraparound on overflow; there are no undefined results. Although defined, these results may be unexpected and therefore carry similar risks to unsigned integer wrapping. (See INT30-C. Ensure that unsigned integer operations do not wrap.) Consequently, signed integer overflow of atomic integer types should also be prevented or detected. 

Addition

Addition is between two operands of arithmetic type or between a pointer to an object type and an integer type. This rule applies only to addition between two operands of arithmetic type. (See ARR37-C. Do not add or subtract an integer to Addition is between two operands of arithmetic type or between a pointer to an object type and an integer type. (See 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.)

Incrementing is equivalent to adding 1.

Noncompliant Code Example

This noncompliant code example may can result in a signed integer overflow during the addition of the signed operands si_a and si_b. If this behavior is unanticipated, it can lead to an exploitable vulnerability.:

void func(signed int si_a, signed int si_b) {
  signed int sum = si_a + si_b;

  /* ... */
}

Compliant Solution

This compliant solution tests ensures that the suspect addition operation to ensure no cannot overflow occurs , regardless of representation:

#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;
  }
  /* ... */
}

...

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. 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. If this behavior is unanticipated, the resulting value may be used to allocate insufficient memory for a subsequent operation or in some other manner that can lead to an exploitable vulnerability.

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 representation:

#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;
  }

  /* ... */
}

...

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, 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 representation:

#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

This noncompliant code example can result in a signed integer overflow during the multiplication of the signed operands si_a and si_b. If this behavior is unanticipated, the resulting value may be used to allocate insufficient memory for a subsequent operation or in some other manner that can lead to an exploitable vulnerability.

void func(signed int si_a, signed int si_b) {
  signed int result = si_a * si_b;

  /* ... */
}

Compliant Solution (GNU)

This compliant solution guarantees there is no possibility of signed overflow on systems where long long is at least twice the width of intuses the GNU extension __builtin_ssub_overflow, available with GCC, Clang, and ICC:

#include <stddef.h>
#include <assert.h>
#include <limits.h>
#include <inttypes.h>
 
size_t popcount(uintmax_t numvoid func(signed int si_a, signed int si_b) {
  size_tsigned widthint = 0diff;
  while (num != 0if (__builtin_ssub_overflow(si_a, si_b, &diff)) {
    if/* (numHandle % 2 == 1) {
      width++;error */
    }
    num >>= 1;
  }
  return width;
}
#define UWIDTH(umax_value) popcount(umax_value)
  
/* ... */
}

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:

void func(signed int si_a, signed int si_b) {
  signed int result;
 = signedsi_a long long tmp* si_b;
  assert(UWIDTH(ULLONG_MAX) >=
         2 * UWIDTH(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 width of int. The purpose of the UWIDTH() macro and the function popcount() is explained in INT19-C. Correctly compute integer widths.

On systems where this assertion would fail, this compliant solution may be used to ensure signed overflow does not occur:

/* ... */
}

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) {
  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_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

#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 isand nonpositive, si_b isare positive */
      if (si_a <> (INT_MINMAX / si_b)) {
        /* Handle error */
      }
    } else { /* si_a andpositive, si_b are nonpositive */
      if ( (si_a != 0) && (si_b < (INT_MAXMIN / si_a))) {
        /* Handle error */
      }
    } /* End if si_a andpositive, si_b are nonpositive */
  } else { /* End if si_a is nonpositive */

  result =    if (si_b > 0) { /* si_a is *nonpositive, 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 can result in a divide-by-zero error during the division of the signed operands s_a and s_b. It can also result in a signed integer overflow error on twos-complement platforms. The x86-32 architecture, for example, requires that both conditions result in a fault, which can easily result in a denial-of-service attack.

void func(signed long s_a, signed long s_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))) {
  signed long result;
  if (s_b == 0) {
    /* Handle error */
      }
 else  {
  } /* resultEnd =if ssi_a /and ssi_b; are nonpositive */
  }
  /* ... */
}

Compliant Solution

 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 ICCThis compliant solution tests the suspect division operation to guarantee there is no possibility of divide-by-zero errors or signed overflow:

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

...

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 instruction, the remainder operator has the same possibilities of arithmetic overflow and division by 0. (See INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors.)

Noncompliant Code Example

Many hardware platforms 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. This 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 a signed integer overflow error in two's-complement. 

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

Implementation Details

On the x86 platforms, the remainder operator for signed integers is implemented by the idiv instruction code, along with the divide operator. Because LONG_MIN / -1 overflows, this code will throw a floating-point exception on LONG_MIN % -1.

On Microsoft Visual Studio 2013, taking the remainder of LONG_MIN by −1 results in abnormal termination on x86 and x64. On GCC/Linux, taking the remainder of LONG_MIN by −1 produces a floating-point exception. However, on GCC 4.2.4 and newer, with optimization enabled, taking the remainder of LONG_MIN by −1 yields the value 0.

Compliant Solution

This compliant solution tests the remainder operand to guarantee there is no possibility of an overflow or divide-by-zero error:

-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:

#include <limits.h>
 
void func(signed long s_a, signed long s_b) {
  signed long 

#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

The remainder operator provides the remainder when

...

Left-Shift Operator

The left-shift operator takes two operands of integer type . 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.

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 bug (logic error). These issues are covered by INT34-C. Do not shift a negative number of bits or more bits than exist in the operand.

Noncompliant Code Example

...

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:

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

void func(signed long si_void func(signed long s_a, signed long sis_b) {
  signed long result;
  if ((si_a < 0) || (si_b <s_b == 0) ||{
    /*  (si_b >= UWIDTH(ULONG_MAX)) {
    /* Handle error Handle error */
  } else {
    result result = sis_a <<% sis_b;
  }
  
  /* ... */
}

Compliant Solution

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

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

void func(signed long si_a 
void func(signed long s_a, signed long sis_b) {
  signed long result;
  if ((sis_ab <== 0 ) || (si(s_ba < 0) ||
      (si== LONG_MIN) && (s_b >= UWIDTH(ULONG_MAX)) ||
      (si_a > (LONG_MAX >> si_b)))== -1))) {
    /* Handle error */
  } else {
    result = sis_a <<% sis_b;
  }
  
  /* ... */
}

The UWIDTH() macro provides the correct width for an unsigned integer type (see INT19-C. Correctly compute integer widths). 

...

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 suspect 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;
  }
  /* ... */
}

Atomic Integers

The C Standard defines the behavior of arithmetic on atomic signed integer types to use two's complement representation with silent wraparound on overflow; there are no undefined results. However, although defined, these results may be unexpected and therefore carry similar risks to unsigned integer wrapping (see INT30-C. Ensure that unsigned integer operations do not wrap). Consequently, signed integer overflow of atomic integer types should also be prevented or detected. 

This section includes an example for the addition of atomic integer types only. For other operations, tests similar to the precondition tests for two’s complement integers used for nonatomic integer types can be used.

Noncompliant Code Example

This noncompliant code example using atomic integers can result in unexpected signed integer overflow:

#include <stdatomic.h>
 
atomic_int i;

void func(int si_a) {
  atomic_init(&i, 42); 
  atomic_fetch_add(&i, si_a);

  /* ... */
}

Compliant Solution

This compliant solution tests the operands to guarantee there is no possibility of signed overflow. It loads the value stored in the atomic integer and tests for possible overflow before performing the addition. However, this code contains a race condition where i can be modified after the load, but prior to the atomic store. This solution is only compliant if i is guaranteed to only be access by a single thread. See CON43-C. Do not assume that a group of calls to independently atomic methods is atomic for more information.

#include <limits.h>
#include <stdatomic.h>
 
atomic_int i;

void func(int si_a) {
  int si_b = atomic_load(&i);

  if (((si_b > 0) && (si_a > (INT_MAX - si_b))) ||
      ((si_b < 0) && (si_a < (INT_MIN - si_b)))) {
    /* Handle error */
  } else {
    atomic_store(&i, si_a + si_b);
  }

  /* ... */
}

Risk Assessment

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

Automated Detection

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Tool

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Version

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Checker

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Description

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Fortify SCA

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5.0

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Can detect violations of this rule with CERT C Rule Pack. Specifically, it checks to ensure that the operand of a unary negation is compared to the type's minimum value immediately before the operation

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LDRA tool suite

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43 D
493 S
494 S

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0278
0296
0297
2800

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Related Vulnerabilities

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

Related Guidelines

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INT02-C. Understand integer conversion rules
INT19-C. Correctly compute integer widths
INT33-C. Ensure that division and remainder operations do not result in divide-by-zero errors
INT34-C. Do not shift a negative number of bits or more bits than exist in the operand
ARR30-C. Do not form or use out of bounds pointers or array subscripts
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
MSC15-C. Do not depend on undefined behavior
CON08-C. Do not assume that a group of calls to independently atomic methods is atomic

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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 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;
  /* ... */
}

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.

Related Guidelines

Key here (explains table format and definitions)

CERT-CWE Mapping Notes

Key here for mapping notes

CWE-20 and INT32-C

See CWE-20 and ERR34-C

CWE-680 and INT32-C

Intersection( INT32-C, MEM35-C) = Ø

Intersection( CWE-680, INT32-C) =

• Signed integer overflows that lead to buffer overflows

CWE-680 - INT32-C =

• Unsigned integer overflows that lead to buffer overflows

INT32-C – CWE-680 =

• Signed integer overflows that do not lead to buffer overflows

CWE-191 and INT32-C

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

Intersection( INT30-C, INT32-C) == Ø

Intersection(CWE-191, INT32-C) =

• Underflow of signed integer operation

CWE-191 – INT32-C =

• Underflow of unsigned integer operation

INT32-C – CWE-191 =

• Overflow of signed integer operation

CWE-190 and INT32-C

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

Intersection( INT30-C, INT32-C) == Ø

Intersection(CWE-190, INT32-C) =

• Overflow (wraparound) of signed integer operation

CWE-190 – INT32-C =

• Overflow of unsigned integer operation

INT32-C – CWE-190 =

• Underflow of signed integer operation

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Bibliography

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