avoid-loop-based-search
Use when code uses a loop to search for a value that satisfies an equation or constraint
التثبيت باستخدام Codex أو Claude انسخ هذا Prompt والصقه في Codex أو Claude أو مساعد آخر ليراجع صفحة Skill ويثبّتها لك.
القائمة
Use when code uses a loop to search for a value that satisfies an equation or constraint
التثبيت باستخدام Codex أو Claude انسخ هذا Prompt والصقه في Codex أو Claude أو مساعد آخر ليراجع صفحة Skill ويثبّتها لك.
استنادا إلى تصنيف SOC المهني
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-loop-based-search |
| description | Use when code uses a loop to search for a value that satisfies an equation or constraint |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When code uses a loop to search for a value that satisfies an equation or constraint
Replace the loop with a direct mathematical calculation to find the solution, using inverse operations or algebraic manipulation
Loops create many path constraints that KLEE must explore sequentially, while direct calculations produce simpler constraints that can be solved more efficiently by the constraint solver
Before:
#include <stdio.h>
int n;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
if (i * 108 / 100 == n) {
printf("%d\n", i);
return 0;
}
}
printf(":(\n");
}
After:
#include <stdio.h>
#include <math.h>
int main(int argc, char const *argv[])
{
int n,m;
scanf("%d",&n);
m=ceil(n/1.08);
if(n==(int)(m*1.08)) printf("%d",m);
else printf(":(");
return 0;
}
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