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The result of the / operator is the quotient from the division of the first arithmetic operand by the second arithmetic operand. Division operations are susceptible to divide-by-zero errors. Overflow can also 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. (See INT32-C. Ensure that operations on signed integers do not result in overflow.)

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) {
  if ((s_a == LONG_MIN) && (s_b == -1)) {
    /* Handle error */
  } else {
    signed long result = s_a / s_b;
  }
  /* ... */
}

Compliant Solution

This 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 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;
  }

  /* ... */
}

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Remainder

The modulo remainder operator provides the remainder when two operands of integer type are divided. 

Noncompliant Code Example

This noncompliant code example can result in a divide-by-zero error during the modulo operation on the signed operands s_a and s_b. Furthermore, many hardware platforms implement modulo as part of the division operator, which can overflow. Overflow can occur during a modulo 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 modulo operation is mathematically 0.

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

Implementation Details

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

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Compliant Solution

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

#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;
  }
  
  /* ... */
}

Compliant Solution (Absolute Value)

This compliant solution is based on the fact that both the The division and modulo remainder operators truncate toward 0, as specified in subclause 6.5.5, footnote 105, of the C Standard [ISO/IEC 9899:2011], which guarantees that

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However, the minimum signed value modulo remainder −1 results in undefined behavior because the minimum signed value divided by -1 is not representable.

#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 {
    if ((s_b < 0) && (s_b != LONG_MIN)) {
      s_b = -s_b;
    }
    result = s_a % s_b;
  }

  /* ... */
}

Risk Assessment

A divide by zero can result in abnormal program termination and denial of service.

Automated Detection

Related Vulnerabilities

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

Related Guidelines

Bibliography

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