القائمة
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
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-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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