Find the K-Beauty of a Number Unexpected Runtime Error - c++

Question

I was recently doing the first question in the Leetcode Biweekly Competition 78, and I received an unexpected runtime error which I couldn't understand, especially since I had written similar code before which worked fine. I'm quite new to programming and these competitions, please tell me what this Runtime Error means and how I could change my code to fix it.
class Solution {
public:
int divisorSubstrings(int num, int k) {
string b=to_string(num);
string a="";
int x;
int ans=0;
for(int i=0;i<=b.size()-k;++i){
for(int j=i;i<i+k;++j){
a+=b[j];
}
x=stoi(a);
if(num%x==0){
++ans;
}
}
return ans;
}
};
And the error:
=================================================================
==33==ERROR: AddressSanitizer: stack-buffer-overflow on address 0x7ffca3ed0900 at pc 0x000000343d81 bp 0x7ffca3ed0890 sp 0x7ffca3ed0888
READ of size 1 at 0x7ffca3ed0900 thread T0
#2 0x7fe89b85d0b2 (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
Address 0x7ffca3ed0900 is located in stack of thread T0 at offset 96 in frame
This frame has 4 object(s):
[32, 40) '__endptr.i'
[64, 96) 'b' <== Memory access at offset 96 overflows this variable
[128, 160) 'a'
[192, 193) 'ref.tmp'
HINT: this may be a false positive if your program uses some custom stack unwind mechanism, swapcontext or vfork
(longjmp and C++ exceptions *are* supported)
Shadow bytes around the buggy address:
0x1000147d20d0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d20e0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d20f0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d2100: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d2110: 00 00 00 00 f1 f1 f1 f1 f8 f2 f2 f2 00 00 00 00
=>0x1000147d2120:[f2]f2 f2 f2 00 00 00 00 f2 f2 f2 f2 f8 f3 f3 f3
0x1000147d2130: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d2140: 00 00 00 00 00 00 00 00 00 00 00 00 f1 f1 f1 f1
0x1000147d2150: 01 f2 04 f2 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d2160: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x1000147d2170: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
Shadow gap: cc
==33==ABORTING
If you need the question, it is at https://leetcode.com/contest/biweekly-contest-78/problems/find-the-k-beauty-of-a-number/
Thanks

Answer 1

Your problem is this line of code:
for(int j=i;i<i+k;++j){
You have two habits you should break. First, you don't use white space. That makes the error in this line much harder to read. Second, you use very short variable names. That ALSO makes the error in this line harder to read.
That for-loop loops forever. The problem is the center clause:
i < i + k
Notice how obvious it is when I add spaces? This problem will get worse as you get older and your eyes get older. The code begins to resemble a wall of unreadable text. Old farts like me won't be able to read your code.
So please, add a little white space. I would have written that line like this:
for (int j = i; j < i + k; ++j) {
Yes, it takes more horizontal space. Space is cheap. Bugs are expensive.
Note that I still think this code is going to go out of range of b's size, so you might still have issues.

Answer 2

Here's the accepted solution modified from your snippet. Review the changes made.
class Solution {
public:
int divisorSubstrings(int num, int k) {
string b = to_string(num);
int ans = 0;
for(int i = 0; i <= b.size() - k; i++) { // the error causing crash
string a = ""; // keep declation close to it's usage, compiler will optimize declaration
for(int j = i; j < i + k; j++) a += b[j];
int x = stoi(a);
if (!x) continue; // you might not want to devide by 0
if( num % x == 0 ) ans++;
}
return ans;
}
};