how adding string and numbers in cpp works using + operator?

Question

I've used cpp for quite a while, I was known that we cannot add string and numbers(as + operator is not overloaded for that). But , I saw a code like this.

#include <iostream>
using namespace std;
int main() {
    string a = "";
    a += 97;
    cout << a;
}

this outputs 'a' and I also tried this.

string a ="";
a=a+97;

The second code gives a compilation error(as invalid args to + operator, std::string and int). I don't want to concatenate the string and number. What is the difference? Why does one work but not the other?

I was expecting that a+=97 is the same as a=a+97 but it appears to be different.

https://stackoverflow.com/questions/191757/how-to-concatenate-a-stdstring-and-an-int — simon, Aug 02 '21 at 14:02
The first `string` is obviously not `std::string`. You haven't provided enough information, so it is only possible to guess. I suspect it is a `typedef` for some pointer type (e.g. `const char *`) so the `a += 97` does pointer arithmetic. It may also be a macro that has similar effect (but other gotchas) to a `typedef`. — Peter, Aug 02 '21 at 14:03
@Peter: "The first string is obviously not std::string" -- Huh? How do you arrive at that conclusion? — Fred Larson, Aug 02 '21 at 14:04
@Peter it can be `std::string` : https://godbolt.org/z/TnbsjcfE6 — 463035818_is_not_an_ai, Aug 02 '21 at 14:04
"I don't want to concatenate the string and number" what do you want instead? — 463035818_is_not_an_ai, Aug 02 '21 at 14:05
the answer to your question taken literally is no. `+=` and `+` followed by `=` typically do the same, but not necessarily. Operators can be overloaded to do different things — 463035818_is_not_an_ai, Aug 02 '21 at 14:07
@463035818_is_not_a_number I want to understand how its working — Nakul Gopal, Aug 02 '21 at 14:08
I know, and I suggest to rephrase it a little, because the quesiton you did ask is based on a false premise — 463035818_is_not_an_ai, Aug 02 '21 at 14:08
@Peter ` #include using namespace std; int main() { string a =""; // a+=97; a=a+97; cout< — Nakul Gopal, Aug 02 '21 at 14:09

Fred Larson · Accepted Answer · 2021-08-02T14:56:02.480

7

The first snippet works because std::string overrides operator+= to append a character to a string. 97 is the ASCII code for 'a', so the result is "a".

The second snippet does not work because there is no + operator defined that accepts a std::string and an int, and no conversion constructor to make a std::string out of an int or char. There two overloads of the + operator that take a char, but the compiler cannot tell which one to use. The match is ambiguous, so an error is reported.

edited Aug 02 '21 at 14:56

answered Aug 02 '21 at 14:09

Fred Larson

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Shouldn't overload 3 of [`operator+`](https://en.cppreference.com/w/cpp/string/basic_string/operator+) take a `char` as second argument? But I guess it doesn't work, because `97` is an `int` and no `char`? – Lukas-T Aug 02 '21 at 14:11
1

@churill: Ah yes, you're correct. But the compiler cannot select that because it's not an unambiguous match. I'll update. – Fred Larson Aug 02 '21 at 14:13
1

`a = a + char(97);` would be a workaround. – Ted Lyngmo Aug 02 '21 at 14:16
Hmm, but `std::string::operator+=()` is also overloaded for `char*`. So why no ambiguity there? I'm not sure I have this completely correct yet. – Fred Larson Aug 02 '21 at 14:18
@FredLarson There's no `char*`? – Ted Lyngmo Aug 02 '21 at 14:19
@TedLyngmo: Overload 3 of `operator+=()` is `constexpr basic_string& operator+=( const CharT* s );` – Fred Larson Aug 02 '21 at 14:21
@FredLarson Yes, I mean, there is no `char*` in OP:s code. Perhaps I misunderstood. – Ted Lyngmo Aug 02 '21 at 14:22
@TedLyngmo: My premise was that there was ambiguity whether to convert the `int` to a `char` or a `char*`, but both operators have overloads for both. So now I'm doubting myself. – Fred Larson Aug 02 '21 at 14:23
1

@FredLarson `int` to `char*` won't happen implicitly. – Ted Lyngmo Aug 02 '21 at 14:24
@TedLyngmo: So I bet there's a different overload that's ambiguous in the `operator+()` case. – Fred Larson Aug 02 '21 at 14:25
@FredLarson No, it's not ambiguity. It's that the template parameter doesn't match so it can't find _any_ overload (if I'm not confusing myself here :-)). – Ted Lyngmo Aug 02 '21 at 14:29
@TedLyngmo: I'm looking over the template vomit. I get mostly "mismatched types", but for two overloads I get "deduced conflicting types". Trying to understand the overloads in question. – Fred Larson Aug 02 '21 at 14:31
@FredLarson :-) Yeah, I was looking at [`clang++`'s output](https://godbolt.org/z/d5od5TPrM) – Ted Lyngmo Aug 02 '21 at 14:33
@TedLyngmo: Yes, clang shows the same thing as my g++. The conflict seems to be between the const reference overload and the rvalue reference overload. Fascinating. – Fred Larson Aug 02 '21 at 14:40

how adding string and numbers in cpp works using + operator?

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