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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
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Basé sur la classification professionnelle SOC
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
| 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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