Returning a string variable in C++

Question

I'm attempting to make a basic user interface that is not case-sensitive for the sake of convenience. To do this, I've made a converter class that makes the string uppercase, but I've stumbled on an issue. After using the class, an if statement in main() is supposed to interpret the message from the converter, but it only reads what the original input was, not it's uppercase counterpart, and I've attempted to return the converted string directly from the converter, but it will not let me.

#include "stdafx.h"
#include <iostream>
#include <string>

using namespace std;

string response;

//converts responses to upper-case
void convert(string response) {
    for (int i = 0; i < response.length(); i++) {
        response[i] = toupper(response[i]);
    }
}

//main dialogue
int main() {

    cout << "How are you?: ";
    getline(cin, response);
    convert(response);
    if (response == "GOOD") {
        cout << "Response 1./l";
    }
        else {
        cout << "Response 2./l";
    }
}

I'm very new to C++, so I apologize if the mistake was an easy one to fix or if I have difficulty understanding the solution.

dmg · Accepted Answer · 2016-12-16 05:29:13Z

2

Aside from the need to pass a reference instead of a value, you should try to use C++-11 features:

void convert(string &response) {
    for (auto &c: response) {
         c = toupper(c);
    }
}

it is much cleaner, and simpler.

answered Dec 16, 2016 at 5:29

dmg

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Comments

John3136 · Accepted Answer · 2016-12-16 05:42:17Z

Look up "pass by value" and "pass by reference" - you have "pass by value" but you are expecting "pass by reference"

In C++: void convert(string& response) {

In your case things are slightly "odd" because as pointed out in the comments by @NeilLocketz, you have a global response, the local response in the method - which is actually the global one since you use that as the calling parameter. If you want to do things properly, you probably don't want response to be global.

Note that the accepted answer still has more memory copies than this. The real key is to understand pass by value and pass by reference and use whichever one is appropriate to your circumstances.

NoseKnowsAll · Accepted Answer · 2016-12-16 05:39:24Z

1

Another option is to change your function header so that it returns a string. That is:

string convert(const string &inResponse) {
    string outResponse(inResponse);
    for (int i = 0; i < inResponse.length(); i++) {
        outResponse[i] = toupper(inResponse[i]);
    }
    return outResponse;
}

And then you use the returned string in your main function like:

....
// response is input, outputResponse is output:
string outputResponse = convert(response);
....

edited Dec 16, 2016 at 5:39

answered Dec 16, 2016 at 5:24

NoseKnowsAll

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3 Comments

NoseKnowsAll Over a year ago

@John3136 Good point about avoiding copying whenever possible.

Zyxlm Over a year ago

I was unaware classes could be identified as strings, which would have probably fixed my issue. In the end, I put the getline(cin, response); in the converter itself and just called the converter whenever I needed a response.

John3136 Over a year ago

I was unaware classes could be identified as strings What does that mean? And now you have input combined with modification - that is poor software engineering. Several flavors of good workable solutions have been presented here, but you're ignoring them all?

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