CODEWITHSUNDEEP
c++data-structuresstackqueue
AP7
AP7

Reputation: 45

AddressSanitizer:DEADLYSIGNAL

I am getting this error when I use `st.pop() in following code in leetcode:

AddressSanitizer:DEADLYSIGNAL
=================================================================
==31==ERROR: AddressSanitizer: SEGV on unknown address (pc 0x0000003a1925 bp 0x7ffd7b62dce0 sp 0x7ffd7b62dcd0 T0)
==31==The signal is caused by a READ memory access.
==31==Hint: this fault was caused by a dereference of a high value address (see register values below).  Dissassemble the provided pc to learn which register was used.
    #8 0x7f3d81a1e0b2  (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
AddressSanitizer can not provide additional info.
==31==ABORTING

The code is as follows:

class Solution {
public:
    string removeOuterParentheses(string S) {
        stack<char> st;
        int count=0;
        string ns;
        for(int i=0;i<S.size();i++)
        {
            if(S[i] == 40 && count++ > 0)
            {
                st.push(S[i]);
                ns+=S[i];
            }
            if(S[i] == 41 && count-- > 0)
            {
                st.pop();
                ns+=S[i];
            }
        }
        return ns;
    }
};

Upvotes: 1

Views: 13285

Answers (2)

AP7
AP7

Reputation: 45

Thanks for helping out. It was trying to pop an empty stack.

Error - if(S[i] == 41 && count-- > 0)

Correction - if(S[i] == 41 && count-- > 1)

Upvotes: 1

StefanKssmr
StefanKssmr

Reputation: 1226

If you have an input string "(some text)" your condition S[i] == 40 && count++ > 0 will always evaluate to false. When S[i]=='(' is true then count++ > 0 is false as count gets incremented after the comparison. However, count gets incremented such that the second condition S[i] == ')' && count-- > 0 will eventually be true. Then you hit st.pop(). As st is empty, it will cause a problem.

Also, if you fix your code, e.g. by writing S[i] == 40 && ++count > 0, it won't remove parentheses as the name implies.

Upvotes: 2

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