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: stackoverflow.com/questions/39198251/… — Lukas-T
– Lukas-T, Commented Oct 12, 2020 at 10:54
Does this answer your question? string and int concatenation in C++ — Lukas-T
– Lukas-T, Commented Oct 12, 2020 at 10:54

Useless · Accepted Answer · 2020-10-12 11:05:04Z

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+.

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