Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

I have a method defined as below:

const std::string returnStringMethod()
{
    std::string myString;

    // populate myString

    return myString;
}

Now, in the caller, I was doing something like this:

const char * ptr = returnStringMethod().c_str(); 

As I can see this is returning some truncated string which I did not expect. However, the folllowing works fine:

std::string str = returnStringMethod();
const char * ptr = str.c_str();

Can someone please help me understand whats happening here? .

PS: We build code once a week. I tested this when I was submitting my code last week and things were fine. So, I really wanted to know what I might be missing here.

thanks, Pavan.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
531 views
Welcome To Ask or Share your Answers For Others

1 Answer

The first is undefined behavior, the temporary returnStringMethod() is valid only until the trailing ;, so the internal string (returned by c_str()) is destroyed, so you're left with a dangling pointer.

The second version is valid because str will be destroyed when its scope ends, and its scope is the same as ptr (at least in your example).

For example, the following would also be wrong:

const char * ptr = NULL;
{
    std::string str = returnStringMethod();
    ptr = str.c_str();
}

After }, ptr is no longer valid.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...