CodeQL queries 1.23

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Name: Multiplication result converted to larger type

Description: A multiplication result that is converted to a larger type can be a sign that the result can overflow the type converted from.

ID: cpp/integer-multiplication-cast-to-long

Kind: problem

Severity: warning

Precision: high

Query: IntMultToLong.ql
 * @name Multiplication result converted to larger type
 * @description A multiplication result that is converted to a larger type can
 *              be a sign that the result can overflow the type converted from.
 * @kind problem
 * @problem.severity warning
 * @precision high
 * @id cpp/integer-multiplication-cast-to-long
 * @tags reliability
 *       security
 *       correctness
 *       types
 *       external/cwe/cwe-190
 *       external/cwe/cwe-192
 *       external/cwe/cwe-197
 *       external/cwe/cwe-681

import cpp
import semmle.code.cpp.controlflow.SSA

 * Holds if `e` is either:
 *  - a constant
 *  - a char-typed expression, meaning it's a small number
 *  - an array access to an array of constants
 *  - flows from one of the above
 * In these cases the value of `e` is likely to be small and
 * controlled, so we consider it less likely to cause an overflow.
predicate likelySmall(Expr e) {
  e.getType().getSize() <= 1
  exists(SsaDefinition def, Variable v |
    def.getAUse(v) = e and

 * Gets an operand of a multiply expression (we need the restriction
 * to multiply expressions to get the correct transitive closure).
Expr getMulOperand(MulExpr me) { result = me.getAnOperand() }

 * Gets the number of non-constant operands of a multiply expression,
 * exploring into child multiply expressions rather than counting them
 * as an operand directly.  For example the top level multiply here
 * effectively has two non-constant operands:
 * ```
 *   (x * y) * 2
 * ```
int getEffectiveMulOperands(MulExpr me) {
  result = count(Expr op |
      op = getMulOperand*(me) and
      not op instanceof MulExpr and
      not likelySmall(op)

from MulExpr me, Type t1, Type t2
  t1 = me.getType().getUnderlyingType() and
  t2 = me.getConversion().getType().getUnderlyingType() and
  t1.getSize() < t2.getSize() and
    t1.getUnspecifiedType() instanceof IntegralType and
    t2.getUnspecifiedType() instanceof IntegralType
    t1.getUnspecifiedType() instanceof FloatingPointType and
    t2.getUnspecifiedType() instanceof FloatingPointType
  ) and
  // exclude explicit conversions
  me.getConversion().isCompilerGenerated() and
  // require the multiply to have two non-constant operands
  // (the intuition here is that multiplying two unknowns is
  // much more likely to produce a result that needs significantly
  // more bits than the operands did, and thus requires a larger
  // type).
  getEffectiveMulOperands(me) >= 2 and
  // exclude varargs promotions
  not exists(FunctionCall fc, int vararg |
    fc.getArgument(vararg) = me and
    vararg >= fc.getTarget().getNumberOfParameters()
  ) and
  // exclude cases where the type was made bigger by a literal
  // (compared to other cases such as assignment, this is more
  // likely to be a trivial accident rather than suggesting a
  // larger type is needed for the result).
  not exists(Expr other, Expr e |
    other = me.getParent().(BinaryOperation).getAnOperand() and
    not other = me and
      e = other or
      e = other.(BinaryOperation).getAnOperand*()
    ) and
    e.(Literal).getType().getSize() = t2.getSize()
select me,
  "Multiplication result may overflow '" + me.getType().toString() + "' before it is converted to '"
    + me.getFullyConverted().getType().toString() + "'."

This rule finds code that converts the result of an integer multiplication to a larger type. Since the conversion applies after the multiplication, arithmetic overflow may still occur.

The rule flags every multiplication of two non-constant integer expressions that is (explicitly or implicitly) converted to a larger integer type. The conversion is an indication that the expression would produce a result that would be too large to fit in the smaller integer type.


Use a cast to ensure that the multiplication is done using the larger integer type to avoid overflow.


int i = 2000000000;
long j = i * i; //Wrong: due to overflow on the multiplication between ints, 
                //will result to j being -1651507200, not 4000000000000000000

long k = (long) i * i; //Correct: the multiplication is done on longs instead of ints, 
                       //and will not overflow