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Home/ Questions/Q 8968961
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T17:31:39+00:00 2026-06-15T17:31:39+00:00

I want to test what if string append char’s size, and below is the

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I want to test what if string append char’s size, and below is the outcome.

I know that the string end with the null character, but why the outcome is like that?

#include <iostream>
#include <string>
using namespace std;
int main(){
    string a = "" + 'a';    //3
    string b = "" + '1';    //2
    string c = "a" + 'a';   //2
    string d = "1" + '1';   //3
    string e = "\0" + 'a';  //20
    string f = "\0" + '1';  //1
    string g = "a" + '\0';  //1
    string h = "1" + '\0';  //1
    string i = "" + '\0';   //0
    string j = "" + '\0';   //0
    cout << a.size() << endl;
    cout << b.size() << endl;
    cout << c.size() << endl;
    cout << d.size() << endl;
    cout << e.size() << endl;
    cout << f.size() << endl;
    cout << g.size() << endl;
    cout << h.size() << endl;
    cout << i.size() << endl;
    cout << j.size() << endl;
    return 0;
}
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  1. Editorial Team
    Editorial Team
    2026-06-15T17:31:40+00:00Added an answer on June 15, 2026 at 5:31 pm

    Your code is not doing what you think.

    String literals decay to const char *, and char is an integer type. If you try to sum them, the compiler finds that the simplest way to make sense of that stuff is to convert chars to ints, so the result is performing pointer arithmetic over the string literals – e.g. ""+'a' goes to the 97th character in memory after the beginning of the string literal "" (if 'a' is represented by 97 on your platform).

    This results in garbage being passed to the string constructor, which will store inside the string being constructed whatever it finds at these locations of memory until it founds a \0 terminator. Hence the “strange” results you get (which aren’t reproducible, since the exact memory layout of the string table depends from the compiler).

    Of course all this is undefined behavior as far as the standard is concerned (you are accessing char arrays outside their bounds, apart from the cases where you add \0).

    To make your code do what you mean, at least one of the operands must be of type string:

    string c = string("a") + 'a';
    

    or

    string c = "a" + string("a");
    

    so the compiler will see the relevant overloads of operator+ that involve std::string.

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