while reversing string 'my.name.is' i'm getting output as 'is@.name.my'. I don't understand from where '@' came from

Question

Given a String of length S, reverse the whole string without reversing the individual words in it. Words are separated by dots.

Input: The first line contains T denoting the number of testcases. T testcases follow. Each case contains a string S containing characters.

Output: For each test case, in a new line, output a single line containing the reversed String.

Constraints: 1 <= T <= 100 1 <= |S| <= 2000

Example: Input:

i.like.this.program.very.much

Output: much.very.program.this.like.i

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


int main() {
    //code
    int t;cin>>t;
    while(t--) {
        string s;
        cin>>s;
        stack<string> st;
        int siz = s.size();
        char c[siz];
        for(int i =0;i<siz;i++) {
            c[i] = s[i];
        }
        char *token = strtok(c,".");
        while (token != NULL) 
        { 
            st.push(token);
            st.push("."); 
            token = strtok(NULL, "."); 
        }
        st.pop();
        while(!st.empty()) {
            cout<<st.top();
            st.pop();
        }
        cout<<"\n"; 

    }
    return 0;
}

Doesn't compile on my machine. `#include ` is not part of C++. `char c[siz];` is not C++. — Eljay, May 14 '20 at 12:30
@Eljay check this link https://ide.geeksforgeeks.org/ka5lZW9hip and try for above test case. — hrithik mahesh, May 14 '20 at 12:34
@Eljay It's a GCC header that's [common to see](https://stackoverflow.com/q/31816095/327083) in academic exercises and competitions. It's a mass include of most of the standard library headers. — J..., May 14 '20 at 12:38
When you copy the string into your buffer you don't null terminate it so you get whatever garbage happens to be there. https://ideone.com/LB7P10 — Retired Ninja, May 14 '20 at 12:39
[Why should I not #include ?](https://stackoverflow.com/a/31816096/4641116) — Eljay, May 14 '20 at 12:40

score 1 · Accepted Answer · answered May 14 '20 at 12:56

1

When you do

    char c[siz];
    for(int i =0;i<siz;i++) {
        c[i] = s[i];
    }

you don't append a 0 to c[] to mark the end of the string.

If you used siz + 1 as an array size, and put a zero (null character) at the end, it would work.

But still, you should not use VLAs.

answered May 14 '20 at 12:56

Jeffrey

11,063
1
21
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can you share your implementation without VLAs – hrithik mahesh May 14 '20 at 13:00
`char c[2001];` would do (see problem statement). Or `std::string c(siz,' ')` with `c.c_str()` – Jeffrey May 14 '20 at 13:35

score 0 · Answer 2 · answered May 14 '20 at 13:56

0

    char c[siz];
    c.push(0);
    for(int i =0;i<siz+1;i++) {
        c[i] = s[i];
    }

You could add a null character to the end of the string and increase the string size by 1 to avoid size violation errors.

answered May 14 '20 at 13:56

Rafael Corzo

1

while reversing string 'my.name.is' i'm getting output as 'is@.name.my'. I don't understand from where '@' came from

2 Answers2