[Solved] Difference between std::string [] operator and at()

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Your second program is malformed. It’s not using std::string::operator[] at all.

string * ps = new string(str);
ps[0] = 'n';

Not only does std::string support the [] operator, every pointer to a type also supports the [] operator. It’s how C style arrays work, and it’s what you’re doing in the code above.

ps isn’t a string. It’s a pointer. And ps[0] is the one string, not unlike *ps.

You probably wanted this instead:

#include <iostream>
#include <string>

using namespace std;

void remodel(string & str){
    string * ps = new string(str);
    (*ps)[0] = 'n';
    // or: ps->operator[](0) = 'n';
    cout<<*ps<<endl;
    delete ps;
}

int main(){
    string str = "Hello world";

    remodel(str);

    cin.get();
    return 0;
}

Or, more idiomatically, use this instead:

#include <iostream>
#include <string>

using namespace std;

void remodel(string & str){
    string ps = str;
    ps[0] = 'n';
    cout<<ps<<endl;
}

int main(){
    string str = "Hello world";

    remodel(str);

    cin.get();
    return 0;
}

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solved Difference between std::string [] operator and at()