القائمة
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
التثبيت باستخدام Codex أو Claude انسخ هذا Prompt والصقه في Codex أو Claude أو مساعد آخر ليراجع صفحة Skill ويثبّتها لك.
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 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
Use when code uses an array to store a single character/value that is only accessed at one index
استنادا إلى تصنيف SOC المهني
| name | align-array-indexing-with-input |
| description | 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 |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
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
Reindex the precomputed array so the symbolic input directly maps to the array slot (e.g., store dp[n] instead of x[n-1]), removing the subtraction at the access site
KLEE models array accesses as constraints over the index expression; eliminating arithmetic on the symbolic index produces simpler, more directly solvable constraints and avoids extra path conditions for boundary cases like n-1 underflow, letting the solver enumerate valid inputs faster and reach more branches.
Before:
#include<stdio.h>
int main(){
int x[32]={1,2,4};
for(int i=3;i<31;i++)
x[i]=x[i-1]+x[i-2]+x[i-3];
int n;
while(1){
scanf("%d",&n);if(n==0)break;
int d=x[n-1];
int y=d/365/10+1;
printf("%d\n",y);
}
return 0;
}
After:
#include<stdio.h>
int main(){
int i=2,n,dp[31];
dp[0]=dp[1]=1;
dp[2]=2;
while(++i<31)
dp[i]=dp[i-1]+dp[i-2]+dp[i-3];
while(scanf("%d",&n),n)
printf("%d\n",dp[n]/(10*365)+1);
}
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