How to add an multidigit int and a char, where the output will be int + char

Question

this is what I have so far

Header file reads:

    string MyAdd(int A, char B)
    {
        char C[10];
        itoa(A,C,10);
        C[1] = '+';
        C[2] = B;
        C[3] = '\0';

        return C;
    }

Calling program:

cout << "\n\t\tThe sum of 1 and X is = " << MeLoad.MyAdd(1   ,'X'   );

Where the output is:

The sum of 1 and X is = 1 + X

But it will only work with single digit integers

so if i try to add 1000 + X, it will still return 1 + X and I have no idea why.

Can someone point me in the right direction? any help is appreciated,thanks.

Using VS 2010

Why do you need an add function for printing only? Cout can already do what you want. Please show more use cases. — László Papp, Apr 27 '14 at 09:49

Sergey Kalinichenko · Accepted Answer · 2014-04-27T09:56:54.050

1

First, let's fix your existing code:

Your code assumes that itoa's output ends at character 1. Use strlen to find the actual ending position, and then write at that position, like this:

itoa(A,C,10);
int p = strlen(C);
C[p++] = '+';
C[p++] = B;
C[p] = '\0';

Next, let's consider an alternative to this: you could use sprintf to fit your code on a single line, like this:

sprintf(C, "%d+%c", A, B);

Finally, let's make your code C++:

Your code above might as well be in C, because it does not take advantage of C++ facilities. Both approaches above are limited, because you must pre-allocate the buffer C at some maximum capacity. C++ gives you a stringstream class, which lets you not worry about preallocation:

string MyAdd(int A, char B) {
    stringstream ss;
    ss << A << "+" << B;
    return ss.str();
}

edited Apr 27 '14 at 09:56

answered Apr 27 '14 at 09:51

Sergey Kalinichenko

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you are missing first argument in `sprintf("%d+%c", A, B);` – Grijesh Chauhan Apr 27 '14 at 09:54
Also, why sprintf in C++? :O – László Papp Apr 27 '14 at 09:55
1

@LaszloPapp I prefer "portable" to "ancient". – Sergey Kalinichenko Apr 27 '14 at 09:59
OK, it is your problem. Now that there is also a more modern answer (including compatibility), the readers will not suffer, but in general, they could. – László Papp Apr 27 '14 at 10:01

László Papp · Answer 2 · 2014-04-27T09:57:33.947

You are essentially misusing C++. You have high-level functionality for all what you are trying to do.

If you want to be portable to work with old compilers, I would use sstringstream for the conversion, otherwise see below.

string MyAdd(int A, char B)
{
    stringstream ss;
    ss << A << B;
    return ss.str();
}

That being said, I would personally really utilize C++11 here with to_string, so I would be writing something like this:

string MyAdd(int A, char B)
{
    return to_string(A) + B;
}

See my complete test code below.

main.cpp

#include <string>
#include <iostream>

using namespace std;

string MyAdd(int A, char B)
{
    return to_string(A) + B;
}

int main()
{
    cout << MyAdd(0, 'A') << endl;
    return 0;
}

Build and run

g++ -std=c++11 main.cpp && ./a.out

Output

0A

========================================================================

Here, you can find my pre-c++11 version if you need it for compatiblity:

main.cpp

#include <string>
#include <iostream>
#include <sstream>

using namespace std;

string MyAdd(int A, char B)
{
    stringstream ss;
    ss << A << B;
    return ss.str();
}

int main()
{
    cout << MyAdd(0, 'A') << endl;
    return 0;
}

Build and run

g++ main.cpp && ./a.out

Output

0A

How to add an multidigit int and a char, where the output will be int + char

2 Answers2

main.cpp

Build and run

Output

main.cpp

Build and run

Output