How can I return string as char* from a function in cpp?

Question

for instance,

if str = "ab" is passed in a function returnString()

and we have a function defination like

string returnString(string str) 
{
    str+='c';
    return str;
}

Output :- abc

but if we have function as

char* returnString(string str) 
{
    str+='c';
    return str;
}

will give the error like :

[Error] cannot convert 'std::string' {aka 'std::__cxx11::basic_string<char>'} to 'char*' in return***

How can I resolve this?

Answer 1

You can use str.data() or str.c_str() to get a pointer to the first character, but you should never return this outside of your function . You are inviting a very swift access violation once your string gets destroyed (look carefully, str is a local copy on the function's stack), and that's if you're lucky.

If you're trying to interface with a C API, you can strdup that pointer before exiting the function, but make sure you understand that the new pointer needs to be free d at some later point by you.

Answer 2

There is no implicit conversion from the type std::string to the type char * or const char * .

So the compiler issues the error

[Error] cannot convert 'std::string' {aka 'std::__cxx11::basic_string<char>'} to 'char*' in return

You could declare and define the function the following way.

const char * returnString( std::string &str ) 
{
    str += 'c';
    return str.c_str();
}

The function changes the passed object of the type std::string . The returned pointer will be valid while the referenced object of the type std::string is alive and is not changed.

Otherwise you need to allocate dynamically a character array. In this case the function could be defined for example like

#include <string>
#include <cstring>

char * returnString( const std::string &str ) 
{
    auto n = str.length();

    char *p = new char[ n + 2 ];

    std::memcpy( p, str.c_str(), n );

    p[n] = 'c';
    p[n+1] = '\0';

    return p;
}

You will need to remember to delete the allocated character array when it will not be required any more using the operator delete [] .