CODEWITHSUNDEEP
c++
tin tan
tin tan

Reputation: 49

Concatenating char variables

I tried making a small program using the libraries "iostream" and "String" to display a given string backwardly as the output on the command prompt. I used a recursive returning-value (string) function to perform the whole process of getting the given string in backward and returning it to the main function to be displayed on screen, as you can see below: 

#include <iostream>
#include <string>
using namespace std;
string rev(string, int); 

int main() {
    string let; 
    cout << "Enter your string: "; 
    cin >> let; 
    cout << "The string in reverse is: " << rev(let, let.length());
    cout << endl;
    return 0;
}

string rev(string x, int y) {
    if (y != 0 )
        return x[y - 1] + rev(x, y - 1);
    else
        return "\0";
}

What I don't get about the process, is that while the concatenation performed on the rev function, recursively, and with the char variables works correctly and returns the string in backward to the main function, trying to concatenate the char variables normally like this gives rubbish as the output:

#include <iostream>
#include <string>
using namespace std;

int main() {
    string hd; 
    string ah = "foo";
    hd = ah[2] + ah[1] + ah[0];
    cout << hd << endl; 
    return 0;
}

And even if I add to the "hd" chain "\0", it still gives rubbish.

Upvotes: 0

Views: 117

Answers (3)

JVApen
JVApen

Reputation: 11317

How about making use of everything that's already available?

string rev(const string &x) { return string{x.rbegin(), x.rend()}; }

Reverse iterators allow you to reverse the string, and the constructor of string with 2 iterators, constructs an element by iterati from begin to end.

Upvotes: 0

Bathsheba
Bathsheba

Reputation: 234635

Writing instead

hd = ""s + ah[2] + ah[1] + ah[0];

will, informally speaking, put + into a "concatenation mode", achieving what you want. ""s is a C++14 user-defined literal of type std::string, and that tells the compiler to use the overloaded + operator on the std::string class on subsequent terms in the expression. (An overloaded + operator is also called in the first example you present.)

Otherwise, ah[2] + ah[1] + ah[0] is an arithmetic sum over char values (each one converted to an int due to implicit conversion rules), with the potential hazard of signed overflow on assignment to hd.

Upvotes: 1

user1316208
user1316208

Reputation: 687

Your first example implicitly converts characters to strings and uses appropriate operator +

While your second example is adding up characters https://en.cppreference.com/w/cpp/string/basic_string/operator_at
returns reference to character at position

Upvotes: 1

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