Runtime error in c++. Relocation protocol version %d

Question

Please help me with following problem of transforming expression of spoj here is the link. It is giving runtime error.

     #include <iostream>
     #include <stack>
#include <string>
using namespace std;
int main()
{
    int testcases;
    cin >> testcases;
    while(testcases-->0)
    {
        string s;
        cin >> s;
        cout << s;
        stack<string> st;
        for(int i=0;i<s.length();i++)
        {
            if(s.at(i)=='(')
                continue;
            else
                if(s.at(i)==')')
                {
                    string s2=st.top();
                    st.pop();
                    string expression=st.top();
                    st.pop();
                    string s1=st.top();
                    st.pop();
                    string tba=s1+s2+expression+"";
                    st.push(tba);
                    cout << tba << endl ;
                    }
            else
                st.push(s.at(i)+"");
                }

        string ss=st.top();
        cout << ss;
        }

    }

And the error coming is not understandable. following is the error with input in first and second line.

1
(a+(b*c))
(a+(b*c))do relocation protocol version %d.
o relocation protocol version %d.
uery failed for %d bytes at address %p
udo relocation protocol version %d.
do relocation protocol version %d.
o relocation protocol version %d.
uery failed for %d bytes at address %pery failed for %d bytes at address %p
udo relocation protocol version %d.
do relocation protocol version %d.
o relocation protocol version %d.
uery failed for %d bytes at address %pery failed for %d bytes at address %p

Answer 1

Looks like you've fallen victim to a Java-ism.

char ch = 'a';
string str = value + "";

does not make string str "a".

This is because "" is not a string. It's actually a pointer to a constant character array, a const char *, and because it is a pointer, pointer arithmetic takes place.

"" is some place in memory. Let's assume address 10000. 'a' has a numerical value. Let's stick with ASCII and use 97. value + ""; is telling the compiler to go to address 10000 + 97, and nobody knows what it will find at address 10097. But it is a const char *, and string has a constructor that will take a const char * and try to turn it into a string. Whatever happens to be at 10097 will be used to make that string. This could crash the program. It could also grab garbage from the land of the string literals, and this looks to be what's happening to OP.

Solution:

Construct a string.

string str(1, ch);

or in OP's case,

st.push(string(1, s.at(i)));

Documentation on string constructor See constructor 2.

But watch out. You have a bunch of other logic errors best resolved by using the debugging software that almost certainly came with your development environment.