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Use when code declares large static/global arrays that are accessed with symbolic indices
Codex 또는 Claude로 설치 이 Prompt를 복사해 Codex, Claude 또는 다른 어시스턴트에 붙여 넣으면 Skill 페이지를 검토하고 설치를 진행할 수 있습니다.
SOC 직업 분류 기준
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-large-static-arrays |
| description | Use when code declares large static/global arrays that are accessed with symbolic indices |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When code declares large static/global arrays that are accessed with symbolic indices
Replace static array declarations with dynamic computation of values on-demand, computing array elements inline during output rather than pre-storing them
Large static arrays force KLEE to track symbolic memory states for all array elements, creating complex constraints when accessed with symbolic indices, while computing values on-demand only creates constraints for the specific values actually used
Before:
#include<stdio.h>
#define SZ 1001
int k, n, A[SZ][SZ];
int main() {
int i, j;
scanf("%d", &k);
if (k <= 500) {
printf("%d\n", k);
for (i = 1; i <= k; i++) {
for (j = 1; j <= k; j++) {
printf("%d ", i);
}
printf("\n");
}
}
else {
n = 500;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
A[i][j] = (i % 2 == 0) ? (i + j) % (n) : (i + j) % n + n;
if (A[i][j] >= k) {
A[i][j] -= n;
}
}
}
printf("%d\n", n);
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
printf("%d ", 1 + A[i][j]);
}
printf("\n");
}
}
}
After:
#include<stdio.h>
int main()
{
int k;
scanf("%d",&k);
if(k==1)
{
printf("1\n1\n");
return 0;
}
int n=(((k+3)>>2)<<1);
printf("%d\n",n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
int val=(i+j)%n+1;
if(i&1)
{
val+=n;
if(val>k)
{
val-=n;
}
}
printf("%d",val);
if(j==n-1)
{
putchar('\n');
}
else
{
putchar(' ');
}
}
}
return 0;
}
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