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V1026. The variable is incremented in the loop. Undefined behavior will occur in case of signed integer overflow.
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 notices that a certain sum is calculated in the 'r' variable. According to the C and C++ standards, an overflow of the signed variable 'r' cannot occur. Otherwise, the program contains undefined behavior that the compiler should not consider.
So, the compiler considers that since the 'r' variable does not overflow after the end of the loop, it cannot become negative. Therefore, the 'r & 0x7fffffff' operation to reset the signed bit is senseless, and the compiler removes it. It returns the value of the 'r' variable from the function.
The V1026 diagnostic is designed to detect such errors. 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:
- Andrey Karpov. Undefined behavior is closer than you think.
- Will Dietz, Peng Li, John Regehr, and Vikram Adve. Understanding Integer Overflow in C/C++.
This diagnostic is classified as:
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