Why cout is not giving output for string?

Question

My code is not giving st in output. There is no compilation error but there's no output either.

#include <bits/stdc++.h>
using namespace std;
int main() {
    string s;
    cin>>s;
    string st;
    int j=0;
    transform(s.begin(),s.end(),s.begin(),::tolower);
    
    for(int i=0;i<s.size();i++) {
        if(s[i]!='a' && s[i]!='e' && s[i]!='o' && s[i]!='u' &&s[i]!='i') {
            st[j++]='.';
            st[j++]=s[i];
        }
    }
    cout<<st<<endl;
}

https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice — πάντα ῥεῖ, May 23 '21 at 21:08
... and even worse: https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h — πάντα ῥεῖ, May 23 '21 at 21:09
`st` is an empty string so won't print anything, assigning to its characters is undefined behaviour — Alan Birtles, May 23 '21 at 21:09

Paul Sanders · Accepted Answer · 2021-05-23T22:32:56.673

2

st[j++]=... writes beyond the end of the allocated space for the string and is undefined behaviour. It also overwrites the string's terminating nul character.

Instead, you want:

st.push_back (...);

or:

st += ...;

edited May 23 '21 at 22:32

answered May 23 '21 at 21:09

Paul Sanders

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score 2 · Answer 2 · answered May 23 '21 at 21:17

transform(s.begin(),s.end(),s.begin(),::tolower);

This is actually wrong. ::tolower is deceptively difficult to use. The problem is:

The parameter of ::tolower is int.
The behaviour is undefined if the argument is not representable as unsigned char.
char may be a signed type, and thus its value isn't necessarily representable as unsigned char. Passing such char value would result in undefined behaviour.

The solution is to convert the char typed character into unsigned char before passing into ::tolower. Example:

std::transform(s.begin(), s.end(), s.begin(), [](unsigned char c) {
    return ::tolower(c);
);

Note that ::tolower processes one code unit at a time, and thus cannot work correctly with variable width character encodings such as Unicode.

The other, more obvious bug is:

string st;

That's an empty string.

st[j++]='.';

Here, you access the empty string outside of its bounds. The behaviour of the program is undefined.

A solution is to instead use:

st += '.';

score 0 · Answer 3 · answered May 23 '21 at 21:43

Try using st.push_back() while dealing with things in which out of bound error can occur like string and vectors. That way you won't need to declare the size of array and can add elements easily. Do something like

string s ="wpfk"; // Taken anything, you can take by cin too.
string st;

and

for(int i=0;i<s.size();i++)
    {
        if(s[i]!='a' && s[i]!='e' && s[i]!='o' && s[i]!='u' &&s[i]!='i')
        {
            st.push_back('.');
            st.push_back(s[i]);
        }
    }

Why cout is not giving output for string?

3 Answers3