printf question with a const char* variable

michael picture michael · Mar 5, 2010 · Viewed 7.5k times · Source

I am stuck in a printf problem. I would appreciate if I can get some help here: In the below code, I can see the font family get displaced correctly in first printf(), but if I set it to variable, i only get an empty string. How can I put it in a variable and have the right values? I just don't want to type 'font.family().family().string().utf8().data()' everywhere?

I did this in the same method:

void myMethod() {
      const char* fontFamily = font.family().family().string().utf8().data();
      // get displayed correctly
      printf ("drawText1 %s \n", font.family().family().string().utf8().data());
      // get an empty string
      printf ("drawText2 %s \n", fontFamily);
}

And the signature of 'data()' is

class CString {
public:
    CString() { }
    CString(const char*);
    CString(const char*, unsigned length);
    CString(CStringBuffer* buffer) : m_buffer(buffer) { }
    static CString newUninitialized(size_t length, char*& characterBuffer);

    const char* data() const;
 //...

}

The signature of utf8() is

class String {
 CString utf8() const;
}

Thank you.

Answer

Mark Ransom picture Mark Ransom · Mar 5, 2010

Something in the chain of font.family().family().string().utf8().data() is returning a temporary object. In your first printf, the temporary object doesn't go out of scope until the printf returns. In the second printf, the temporary has been destroyed after the pointer assignment was made, and the pointer is now invalid. You're seeing a classic example of "undefined behavior".

There are two ways to fix this. Either make a copy of the data before the temporary gets destroyed, or make a reference to the temporary. The copy is probably easiest and clearest, as long as the class has a copy operator. Assuming that utf8() generates a temporary CString, this would be

CString fontFamily = font.family().family().string().utf8();
printf ("drawText2 %s \n", fontFamily.data());