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I have two chars and I want to create a string concatenating them:

char a = '1';
char b = '2';

string s = "(" + a + "," + b + ")";

What's the easiest way to achieve that? Since the first element "(" is a string, by concatenating the elements from left to right it should work, since each char will be casted into a string and appended.

However the compiler doesn't seem to like it.

error: invalid operands of types 'const char*' and 'const char [2]' to binary 'operator+'

How can I achieve that?

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  • See this similar question. You get your specific error because in "(" + a, the a is converted to an int and the add gives a pointer. Then you try to add "," to it. Commented Apr 22, 2020 at 21:11
  • I like a stringstream for stuff like this: std::ostringstream strm; strm << "(" << a << "," << b << ")"; Commented Apr 22, 2020 at 21:12

2 Answers 2

Reset to default
5

"(" is not a std::string. It is a char[2] C string array. Make it std::string by using the s literal:

using namespace std::string_literals;
std::string s = "("s + a + ","s + b + ")"s;

This can still fail if you try to do something like this:

std::string s = a + b + "."s; // error

In this case you can simply start with an empty string:

std::string s = ""s + a + b + "."s;

Another option is to use std::ostringstream to build the string:

std::ostringstream oss;
oss << "(" << a << "," << b << ")";
std::string s3 = oss.str();
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3 Comments

Do built-in literals technically count as user-defined literals? As far as I know UDL suffixes must start with a _, for starter.
i did not know about the s literal, thanks! I like this solution because is compact, clear and doesn't need extra lines of complexity.
std::format("({},{})", a, b); godbolt
2

You could just write

char a = '1';
char b = '2';

std::string s = std::string( "(" ) + a + "," + b + ")";

Or

char a = '1';
char b = '2';

string s;

for ( char c : { '(', a, ',', b, ')' } )
{
    s += c;
}

Here is a demonstrative program.

#include <iostream>
#include <string>

int main() 
{
    char a = '1';
    char b = '2';

    std::string s = std::string( "(" ) + a + "," + b + ")";

    std::cout << "s = " << s << '\n';

    std::string t;

    for ( char c : { '(', a, ',', b, ')' } )
    {
        t += c;
    }

    std::cout << "t = " << t << '\n';

    return 0;
}

The program output is

s = (1,2)
t = (1,2)

Or you could use just a constructor like

std::string s( { '(', a, ',', b, ')' } );

or the method assign

std::string s;
s.assign( { '(', a, ',', b, ')' } );

or append

std::string s;
s.append( { '(', a, ',', b, ')' } );

Here is another demonstrative program.

#include <iostream>
#include <string>

int main() 
{
    char a = '1';
    char b = '2';

    std::string s1( { '(', a, ',', b, ')' } );

    std::cout << "s1 = " << s1 << '\n';

    std::string s2;
    s2.assign( { '(', a, ',', b, ')' } );

    std::cout << "s2 = " << s2 << '\n';

    std::string s3( "The pair is " );
    s3.append( { '(', a, ',', b, ')' } );

    std::cout << "s3 = " << s3 << '\n';

    return 0;
}

Its output is

s1 = (1,2)
s2 = (1,2)
s3 = The pair is (1,2)
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1 Comment

Your for loop could be replaced with a single call to std::string::append() instead: s.append({ '(', a, ',', b, ')' });

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