Why does the following loop run infinite times

Question

This seems like a very normal piece of code ...but I don't know why its showing abnormal behaviour .

#include <bits/stdc++.h>
using namespace std ;

int main()
{
string s ;
cin>>s;

for(int i =2 ; i < s.length()-2 ; i++)
{
    cout<<"AAAA"<<endl;
}

}

For an input of a string of lenght 1 like 'B' ....The expected output should be nothing ..but instead the output is 'AAAA' infinite times ..can't figure out why the loop runs infiniye times

You might notice you receive a warning for signed/unsigned mismatch there. Warnings are important. — Retired Ninja, Jun 08 '19 at 17:24
Well my compiler didn't give any warnings ..but yeah that solved the issue ..thanks — , Jun 08 '19 at 18:11

score 2 · Answer 1 · answered Jun 08 '19 at 17:25

2

Not infinite. It will run from 2 (start), to s.length() - 2 (end). Since s is empty, it's size is 0 and it's type is unsigned, substracting 2 from 0 will produce 2^32 - 2 (or 2^64 - 2), which is quite large number.

answered Jun 08 '19 at 17:25

Radosław Cybulski

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score 1 · Answer 2 · answered Jun 08 '19 at 17:26

Because s.length() is unsigned, when you subtract 2 it becomes very large (about 2^32 or 2^64) for values of s.length() less than 2. So you need to cast s.length() to int to make this work:

for(int i =2 ; i < int(s.length())-2 ; i++)

And cycle is infinite, because int is signed, so it can't be more than 2^31 or 2^63 (1 bit is reserved for sign).

Why does the following loop run infinite times

2 Answers2