如何返回包含string / int变量的字符串

Question

For example, if I have this little function:

string lw(int a, int b) {    
    return "lw $" + a + "0($" + b + ")\n";
}

....and make the call lw(1,2) in my main function I want it to return "lw $1, 0($2)" .

But I keep getting an error: invalid operands of types 'const char*' and 'const char [11]' to binary 'operator+'

What am I doing wrong? I pretty much copied an example from class, and changed it to suit my function.

Answer 1

You are trying to concatenate integers to strings, and C++ can't convert values of such different types. Your best bet is to use std::ostringstream to construct the result string:

#include <sstream>

// ...

string lw(int a, int b)
{
    ostringstream os;
    os << "lw $" << a << "0($" << b << ")\n";
    return os.str();
}

If you have Boost , you can use Boost.Lexical_cast :

#include <boost/lexical_cast.hpp>

// ...

string lw(int a, int b)
{
    return
        string("lw $") +
        boost::lexical_cast<std::string>(a) +
        string("0($") +
        boost::lexical_cast<std::string>(b) +
        string(")\n");
}

---

And now with C++11 and later there is std::to_string :

string lw(int a, int b)
{
    return
        string("lw $") +
        std::to_string(a) +
        string("0($") +
        std::to_string(b) +
        string(")\n");
}

Answer 2

#include <sstream>

string lw(int a, int b) {    
    std::string s;
    std::stringstream out;
    out << "lw $" << a << "0($" << b << ")" << endl;
    s = out.str();
    return s;
}

Answer 3

Use an ostringstream:

#include <sstream>
...
string lw(int a, int b) {
    std::ostringstream o;
    o << "lw $" << a << "0($" << b << ")\n";
    return o.str();
}

Answer 4

You cannot add string literals (like "hello") to integers. That is what compiler says to you. This is partial answer to your question. Please see how to accomplish what you want in another posts.

Answer 5

To understand this question, you have to know that in C++, string literals like "lw $" are treated as const char[] as inherited from the C language. However this means that you only get operators that are defined for arrays, or in this case the array-degrades-to-pointer case.

So what happens is you have a string literal, and then add an integer to it, creating a new pointer. Then you attempt to add another string literal which again degrades to char* . You can't add two pointers together, which then generates the error you're seeing.

You're trying to format integers into a string format with some delimiting text. In C++ the canonical way of doing that is with stringstreams:

#include <sstream>

string lw(int a, int b)
{
    std::ostringstream os;
    os << "lw $" << a << "0($" << b << ")\n";
    return os.str();
}