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Use when integer values are updated using floating-point multiplication (e.g., y = y * 1.01) inside a loop condition that KLEE must symbolically reason about
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-float-arithmetic-in-integer-loops |
| description | Use when integer values are updated using floating-point multiplication (e.g., y = y * 1.01) inside a loop condition that KLEE must symbolically reason about |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When integer values are updated using floating-point multiplication (e.g., y = y * 1.01) inside a loop condition that KLEE must symbolically reason about
Replace the floating-point scaling with equivalent integer arithmetic (e.g., n = n + n/100), keeping all variables and operations in integer domain
KLEE's constraint solver (STP/Z3) handles bitvector/integer constraints far more efficiently than floating-point ones; floating-point operations are often concretized or poorly modeled, blocking path exploration and reducing achievable branch coverage
Before:
#include<stdio.h>
int main(){
long long x,y=100;
scanf("%lld",&x);
for(int i=1;;++i){
y=y*1.01;
if(y>=x){
printf("%d\n",i);
break;
}
}
}
After:
#include <stdio.h>
long long n, m, i, j, k;
int main(){
scanf("%lld", &m);
n = 100;
while(n < m){
n = n + n/100;
k++;
}
printf("%lld\n", k);
return 0;
}
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