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V5010. OWASP. The variable is increment…
Analyzer Diagnostics
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Contents

V5010. OWASP. The variable is incremented in the loop. Undefined behavior will occur in case of signed integer overflow.

March 3, 2021

The analyzer has detected a potential signed integer overflow in a loop. Overflowing signed variables leads to undefined behavior.

Consider the following example:

int checksum = 0;
for (....) {
  checksum += ....;
}

This is an abstract algorithm to calculate a checksum. It implies the possibility of overflowing the 'checksum' variable, but since this variable is signed, an overflow will result in undefined behavior. The code above is incorrect and must be rewritten.

You should use unsigned types whose overflow semantics are well-defined.

Fixed code:

unsigned checksum = 0;
for (....) {
  checksum += ...
}

Some programmers believe that there is nothing bad about signed overflow and that they can predict their program's behavior. This is a wrong assumption because there are many possible outcomes.

Let's examine how errors of this type occur in real-life programs. One developer left a post on the forum complaining about GCC's acting up and incorrectly compiling his code in optimization mode. He included the code of a string checksum function that he used in his program:

int foo(const unsigned char *s)
{
  int r = 0;
  while(*s) {
    r += ((r * 20891 + *s *200) | *s ^ 4 | *s ^ 3) ^ (r >> 1);
    s++;
  }
  return r & 0x7fffffff;
}

His complaint is that the compiler does not generate code for the bitwise AND (&), which makes the function return negative values although it should not.

The developer believes this has to do with some bug in the compiler, but in fact it is his own fault since he wrote incorrect code. The function does not work properly because of undefined behavior occurring in it.

The compiler sees that the 'r' variable is used to calculate and store a sum. It assumes that this variable cannot overflow because that would be considered undefined behavior, which the compiler should not investigate and take into account whatsoever. So, the compiler assumes that since the 'r' variable cannot store a negative value after the loop terminates, then the 'r & 0x7fffffff' operation, which sets off the sign bit, is unnecessary, so it simply returns the value of the 'r' variable from the function.

It is defects like this that diagnostic V5010 is designed for. To fix the code, you should simply use an unsigned variable to calculate the checksum.

Fixed code:

int foo(const unsigned char *s)
{
  unsigned r = 0;
  while(*s) {
    r += ((r * 20891 + *s *200) | *s ^ 4 | *s ^ 3 ) ^ (r >> 1);
    s++;
  }
  return (int)(r & 0x7fffffff);
}

References:

This diagnostic is classified as:

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