avoid-early-return-in-loops
Use when code uses early return statements inside nested loops to exit upon finding a condition
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 early return statements inside nested loops to exit upon finding a condition
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-early-return-in-loops |
| description | Use when code uses early return statements inside nested loops to exit upon finding a condition |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When code uses early return statements inside nested loops to exit upon finding a condition
Replace early returns with state modification (e.g., setting a flag variable) and check the state after loops complete
Early returns create path termination that prevents KLEE from exploring remaining loop iterations, while state modification allows KLEE to symbolically explore all loop paths and generate more diverse test cases
Before:
#include<stdio.h>
int main(void) { int qq, n, i, u;scanf("%d", &n); for(i = 1;i <= 9;i++) {for(u = 1;u <= 9;u++) { qq = i * u; if(qq == n){puts("Yes"); return 0;}}}puts("No");return 0;}
After:
#include <stdio.h>
int n;
int main() {
scanf("%d", &n);
for (int i = 1; i < 10; i++) {
for (int j = 1; j < 10; j++) {
if (n == i*j)n = 0;
}
}
if (n)printf("No\n");
else printf("Yes\n");
}
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