C++ temporary string lifetime

问题

Apologies as I know similar-looking questions exist, but I'm still not totally clear. Is the following safe?

void copyStr(const char* s)
{
    strcpy(otherVar, s);
}

std::string getStr()
{
    return "foo";
}

main()
{
    copyStr(getStr().c_str());
}

A temporary std::string will store the return from getStr(), but will it live long enough for me to copy its C-string elsewhere? Or must I explicitly keep a variable for it, eg

std::string temp = getStr();
copyStr(temp.c_str());

回答1:

Yes, it's safe. The temporary from getStr lives to the end of the full expression it appears in. That full expression is the copyStr call, so it must return before the temporary from getStr is destroyed. That's more than enough for you.

回答2:

A temporary variable lives until the end of the full expression. So, in your example,

copyStr(getStr().c_str());

getStr()’s return value will live until the end of copyStr. It is therefore safe to access its value inside copyStr. Your code is potentially still unsafe, of course, because it doesn’t test that the buffer is big enough to hold the string.

回答3:

You have to declare a temporary variable and assign to it the return result:

...
{
  std::string tmp = getStr();
  //tmp live
  ...
}
//tmp dead
...

A temporary var must have a name, and will live in the scope where it has been declared in.

Edit: it is safe, you pass it by copied value (your edit was heavy, and nullified my above answer. The above answer concern the first revision)

copyStr(getStr().c_str()); will pass the rvalue to copyStr()

来源：https://stackoverflow.com/questions/24488122/c-temporary-string-lifetime

标签

c++

string

temporary