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Use when code uses complex arithmetic expressions with modulo operations in conditional statements, especially involving multiple variables
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Use when printf statements output strings without trailing newlines
Use when array is accessed with direct user input as index (0-based indexing)
Use when the program accesses a precomputed table using an offset (e.g., arr[n-1]) that introduces an arithmetic operation between the symbolic input and the index
Use when loop variables iterate over character values (e.g., ASCII codes) to represent different operations or choices
Use when code uses a loop to search for an input value in an array and then uses the found index for further computation
Use when code reads a fixed number of characters into an array and only accesses individual elements
| name | avoid-complex-modulo-conditions |
| description | Use when code uses complex arithmetic expressions with modulo operations in conditional statements, especially involving multiple variables |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When code uses complex arithmetic expressions with modulo operations in conditional statements, especially involving multiple variables
Refactor to use floating-point arithmetic with rounding checks instead of integer arithmetic with modulo operations
Complex modulo expressions create non-linear constraints that are difficult for SMT solvers to reason about, while floating-point comparisons with rounding create simpler constraints that KLEE can explore more efficiently
Before:
#include <stdio.h>
typedef long long ll;
int main(void) {
ll n, i, j;
scanf("%lld", &n);
for(i = 1; i <= 3500; ++i) for(j = 1; j <= 3500; ++j) if(4 * i * j - i * n - j * n > 0 && !((n * i * j) % (4 * i * j - i * n - j * n))) {
printf("%lld %lld %lld", i, j, (n * i * j) / (4 * i * j - i * n - j * n));
return 0;
}
}
After:
#include <stdio.h>
#include <math.h>
int main(void){
double n,i,j,k,ans,ap;
scanf("%lf",&n);
ans = 4/n;
for (i=1;i<5000;i++)
{
for (j=1;j<5000;j++)
{
ap = ans - 1/i - 1/j;
if (ap >0)
{
ap = 1.0/ap;
ap = roundl(ap * 10000000) / 10000000;
if ((int)(ap) == ap)
{
printf("%d %d %d",(int)(i),(int)(j),(int)(ap));
return 0;
}
}
}
}
return 0;
}
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