Garbage value of iterator value in C++

Question

I am solving a problem where I need to return the last index of '1' in a given string. If not present then return -1. I wrote the following simple code, but for input string input "0" it fails. I tried to debug bu using GDB and I noticed that once loop statement of index() function runs once then a garbage value is assigned to iteration variable i.

#include <iostream>
#include <string>
using namespace std;

int index(string &str) {
    int result = -1;
    for(auto i = str.length() - 1; i >= 0; --i) {
        if(str[i] == '1')
            return i;
    }
    return result;
}
int main() {
    int T;
    cin >> T;
    cin.ignore();
    while(T--) {
        string str;
        cin >> str;
        cout << index(str) << endl;
    }
    return 0;
}

What exactly is the problem?

Notice the value of i in the second iteration.

Does this answer your question? [Traversing a vector in reverse direction with size\_t values](https://stackoverflow.com/questions/27402237/traversing-a-vector-in-reverse-direction-with-size-t-values) — Asteroids With Wings, Mar 21 '20 at 15:55
@HolyBlackCat How could that be? The OP is looping until it's < 0. — , Mar 21 '20 at 15:55
@Chipste No, they're looping until it's < 0. Which it never is. So it's looping until some random unspecified value looks like '1', and then some massive `i` is returned. — Asteroids With Wings, Mar 21 '20 at 15:56
Take a look at the image I added and notice the value of `i` in the second iteration. It must be -1 now but It's some different value. — Arun Suryan, Mar 21 '20 at 15:59
Your screenshot is really hard to read. You should just copy and paste the text of your shell session when posting here. — David Grayson, Mar 21 '20 at 16:03
@ArunSuryan The point was to not use a screenshot at all. [We don't like screenshots](https://meta.stackoverflow.com/questions/285551/why-not-upload-images-of-code-on-so-when-asking-a-question). — HolyBlackCat, Mar 21 '20 at 16:25

Asteroids With Wings · Accepted Answer · 2020-03-21T16:01:32.127

Your program has undefined behaviour here:

for(auto i = str.length() - 1; i >= 0; --i) {
    if(str[i] == '1')
        return i;
}

The length() of an std::string has an unsigned type, and since you used auto that means your i is unsigned too (to be precise, a std::size_t).

Such values never go below zero. They wrap around to the maximum value of the type (a very large number!).

So, your loop condition does nothing; it's always true. Instead, out-of-bounds accesses str[i] then occur until one of the unspecified values that results happens to look like '1'. Then, the massive i is returned.

It is possible to loop backwards through a standard container or string, but you have to be careful about it. My recommendation to you is to use iterators; in this case, reverse iterators.

for (auto it = str.rcbegin(); it != str.rcend(); ++it)
{
    if (*it == '1')
        return std::distance(it, str.rcend());
}

I got your point. But for taking `i` as following warning occurs: Narrowing conversion from 'unsigned long' to signed type 'int' is implementation-defined. What is a good way to tackle this? — Arun Suryan, Mar 21 '20 at 16:01
@ArunSuryan I added a link to some material for you to read. Take your time and study it. — Asteroids With Wings, Mar 21 '20 at 16:04

Garbage value of iterator value in C++

What exactly is the problem?

1 Answers1