[klee-dev] Question about how Klee performs symbolic execution

[Cristian Cadar]

Hi, this question was asked on the mailing list before, but might not be 
easy to find.  In a nutshell, when KLEE encounters a symbolic memory 
access, it forks one path for each object which can be referenced there.

Cristian

On 10/11/2016 01:10, Rutledge, Richard L wrote:
> I hope someone can clarify up a little klee-fundlement on my part regarding how Klee performs symbolic execution. Consider the following example program:
>
> --------------------------------------------------------
> #include <klee/klee.h>
>
> #define MAXSTR 10
>
> int test(char *str, int offset) {
>
>   if (str[offset] == '\n') {
>     return 1;
>   }
>   return 0;
> }
>
> int main(int argc, char *argv[]) {
>
>   char str[MAXSTR];
>   int offset;
>
>   klee_make_symbolic(str, MAXSTR, "str");
>   offset = klee_int("offset");
>   klee_assume(offset >= 0);
>   klee_assume(offset < MAXSTR);
>
>   int result = test(str, offset);
>
>   return result;
> }
> --------------------------------------------------------
>
> There are two paths through this program. As expected, Klee finds exactly two paths and generates two test cases.
>
> However, if I delete to two klee_assume() statements, then Klee finds 186 paths/test cases.
>
> What leads to the path explosion? Without the assumes, the program admits undefined behavior, but does not introduce any new paths. It still only has two and execution should fork into two states at the test() if statement. What condition forks the other 184 states?
>
> Cheers and Thanks!
> Rick Rutledge
>
>
>
>
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