C++ returning string

Question

Is it possible to do this

std::string str(const char* s)
{  
     return std::string(s);
} 

int main() {  
    char* strz = (char*)str("asd").c_str();  
}

Instead of:

int main(){  
    std::string temp = str("asd");  
    char* strz = (char*)temp.c_str();  
}

I know it should be const char* strz but I need it only within block of code(and without new/delete). After returning string from method it look for reference(if cant find it, deletes string) and then calls c_str(). I have a lot of char's(independent from me) and I could use second solution but it takes too much code.

Please indent your code 4 spaces instead of marking every single line as code using the inline code syntax (``)! — Niklas B.
– Niklas B., Commented Apr 22, 2012 at 17:00
^ i.e. highlight all of it and press ctrl-K or hit the code format button — Qaz
– Qaz, Commented Apr 22, 2012 at 17:03
Your str function is unneccesary. Do you have it just to shorten the code? — Zyx 2000
– Zyx 2000, Commented Apr 22, 2012 at 17:16
@Zyx2000 no, it's just illustrative, practically I'm using it to convert text between MBCS and Unicode — user1112008
– user1112008, Commented Apr 22, 2012 at 17:25

Luchian Grigore · Accepted Answer · 2012-04-22 17:02:38Z

6

If you use your second option -

std::string temp = str("asd");
char* strz = (char*)temp.c_str();

You're possibly running into undefined behavior. It's illegal to modify the contents of strz. You'll have to use strcpy to get a mutable array of char's. Your first isn't any better either, moreover, it's redundant, since you can directly use string's constructor.

Anyway, to get a char* from a string you can do:

char* get(const std::string& str)
{
   char* ret = new char[str.length() + 1]();
   strcpy(ret,str.c_str());
   return ret;
}

You'll have to delete[] the memory yourself when you're done with it.

answered Apr 22, 2012 at 17:02

Luchian Grigore

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3 Comments

user1112008 Over a year ago

I need to use this cast only temporarily(not even writeable), later it is copied(but copied from char*)

Luchian Grigore Over a year ago

@user1112008 if it's not writable, why not have a const char*?

user1112008 Over a year ago

As I told it's independent from me :( I could edit structure which holds it on "my side" but on "other side" it will be read as char*. I'm writing extension for existing application(sending is synchronous).

bames53 · Accepted Answer · 2012-04-22 17:13:45Z

1

There's a problem even beyond the cast from const char* to char*

char* strz = (char*)str("asd").c_str();

This creates a temporary std::string which is destroyed at the end of this line. And that invalidates the char* strz.

It sounds to me like you should just be holding on to the returned std::string rather than treating it as a temporary:

std::string strz = str("asd");
...
foo(strz.data()); // use strz

answered Apr 22, 2012 at 17:13

bames53

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Comments

anatolyg · Accepted Answer · 2012-04-22 17:24:36Z

Let me guess what you wanted to ask and answer it :) Feel free to ignore if i guessed wrong :)

So you have some function that returns std::string:

std::string str(...);

And you have some old code that uses char* (so old that it doesn't even declare its argument as const char*, even though it doesn't modify the string), which you want to use with the result:

int ugly_func1(char*) {...}
int ugly_func2(char*) {...}
int ugly_func3(char*) {...}

int main()
{
    std::string temp = str(...);
    char* temp1 = (char*)temp.c_str();
    ugly_func1(temp1);
    ugly_func2(temp1);
    ugly_func3(temp1);
}

So, no, you must use the syntax with the temp variable. You can't write this:

int main()
{
    char* temp = (char*)str(...).c_str();
    ugly_func1(temp);
    ugly_func2(temp);
    ugly_func3(temp);
}

This is illegal code (e.g. destroying the return-value of str() before calling ugly_func1 on it). The nasty thing is that it may sometimes appear to work.

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