String concatenation not working while concatenating with character

Question

I attempted the following: The count-and-say sequence is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, ... 1 is read off as one 1 or 11. 11 is read off as two 1s or 21.

21 is read off as one 2, then one 1 or 1211.

Given an integer n, generate the nth sequence.

I actually wrote some code to solve this problem and it was not generating the right answer.The problem in the code is that i tried to add a character to a string using the "+" operator.

string Solution::countAndSay(int A)
{
    string s = "11";
    string str;

    string ans[A + 2];
    ans[1] = "1";
    ans[2] = "11";

    int count = 1;

    for (int j = 3; j <= A; j++) {
        str = "";
        count = 1;

        for (int i = 0; i < s.size() - 1; i++) {
            if (s[i] == s[i + 1])
                count++;
            else {
                str += to_string(count) + s[i];
                count = 1;
            }
        }

        if (s[s.size() - 1] != s[s.size() - 2]) {
            str += "1" + s[s.size() - 1];
        }
        else
            str += to_string(count) + s[s.size() - 1];

        ans[j] = str;
        s = str;
    }
    return ans[A];
}

When i changed the code inside the if condition below in the nested loop it worked. The change i did was:

if(s[s.size()-1]!=s[s.size()-2]){
    str+="1";
    str.push_back(s[s.size() - 1)];
}

I just want to know what was wrong in earlier code. Can we not use a "+" operator to add characters to a string?

`"1" + s[s.size() - 1];` doesn't add characters to a string, because `"1"` is not a string ;) Instead it performs pointer arithmethic, because `"1"` is a pointer. See also here: https://stackoverflow.com/questions/39198251/string-and-int-concatenation-in-c?noredirect=1&lq=1 — Lukas-T, Oct 12 '20 at 10:54
Does this answer your question? [string and int concatenation in C++](https://stackoverflow.com/questions/39198251/string-and-int-concatenation-in-c) — Lukas-T, Oct 12 '20 at 10:54

score 1 · Accepted Answer · answered Oct 12 '20 at 11:05

str += "1" + s[s.size() - 1];

Does not mean

str += "1";
str += s[s.size() - 1];

But

str += ("1" + s[s.size() - 1]);

where the right-hand-side expression is not at all what you wanted.

The expression "1" is not a char, but a string literal containing the two characters {'1', 0} (it's a C string, so there is a null terminator).
Then, "1" + ... isn't an append like your top-level operation, because C strings don't have overloaded operators, but just decay to a pointer.
Now you're doing pointer arithmetic, so s[s.size() - 1] is promoted from a char to an integer, and added to the address of the first character in your string literal.

Unless the character was 0 or 1 (and I don't mean '0' or '1'), the result is off the end of your string literal, and behaviour will be undefined.
str += (...) is now appending a C-string, which it is illegal to even look at (due to the UB above).

The reason that

str += to_string(count) + s[i];

works, is that the left-hand side of to_string(count) + s[i] is a std::string with an overloaded operator+. You can get the same effect by writing

str += string{"1"} + s[s.size() - 1];

so you also have a std::string instead of a C string literal as the left-hand side of the operator+.

String concatenation not working while concatenating with character

1 Answers1