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The following sections examine specific operations that are susceptible to integer overflow. The specific tests that are required to guarantee that the operation does not result in an integer overflow depend on the signedness of the integer types. When operating on small types (smaller than int), integer conversion rules apply. The usual arithmetic conversions may also be applied to (implicitly) convert operands to equivalent types before arithmetic operations are performed. Make sure you understand implicit conversion rules before trying to implement secure arithmetic operations (see INT02-A. Understand integer conversion rules).anchor

Addition

Addition Additionis between two operands of arithmetic type or between a pointer to an object type and an integer type. Incrementing is equivalent to adding one.

Non-Compliant Code Example (Unsigned)

This code may result in an unsigned integer overflow during the addition of the unsigned operands ui1 and ui2. If this behavior is unexpected, the resulting value may be used to allocate insufficient memory for a subsequent operation or in some other manner that could lead to an exploitable vulnerability.

unsigned int ui1, ui2, sum;

sum = ui1 + ui2;

Compliant Solution (Unsigned)

This compliant solution tests the suspect addition operation to guarantee there is no possibility of unsigned overflow.

unsigned int ui1, ui2, sum;

if (UINT_MAX - ui1 < ui2) {
  /* handle error condition */
}
sum = ui1 + ui2;

Non-Compliant Code Example (Signed)

This code may result in a signed integer overflow during the addition of the signed operands si1 and si2. If this behavior is unanticipated, it could lead to an exploitable vulnerability.

int si1, si2, sum;

sum = si1 + si2;

Compliant Solution (Two's Complement Signed)

This compliant solution tests the addition operation to ensure no overflow occurs, assuming two's complement representation.

signed int si1, si2, sum;

if ( ((si1^si2) | (((si1^(~(si1^si2) & (1 << (sizeof(int)*CHAR_BIT-1))))+si2)^si2)) >= 0) {
   /* handle error condition */
}

sum = si1 + si2;

Compliant Solution (General Signed)

This compliant solution tests the suspect addition operation to ensure no overflow occurs regardless of representation.

signed int si1, si2, sum;

if (((si1>0) && (si2>0) && (si1 > (INT_MAX-si2))) ||
    ((si1<0) && (si2<0) && (si1 < (INT_MIN-si2)))) {
   /* handle error condition */
}

sum = si1 + si2;

This solution is more readable but contains branches and consequently may be less efficient than the solution that is specific to two's complement representation.

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