Life Scope of Temporary Variable
c++
Solution
See this question. The standard guarantees that a temporary lives until the end of the expression of which it is a part. Since the entire function invocation is the expression, the temporary is guaranteed to persist until after the end of the function invocation expression in which it is a part.
Problem
``` #include <cstdio> #include <string> void fun(const char* c) { printf("--> %s\n", c); } std::string get() { std::string str = "Hello World"; return str; } int main() { const char *cc = get().c_str(); // cc is not valid at this point. As it is pointing to // temporary string internal buffer, and the temporary string // has already been destroyed at this point. fun(cc); // But I am surprise this call will yield valid result. // It seems that the returned temporary string is valid within // scope (...) // What my understanding is, scope means {...} // Is this valid behavior guarantee by C++ standard? Or it depends // on your compiler vendor implementations? fun(get().c_str()); getchar(); } ``` The output is : ``` --> --> Hello World ``` Hello, may I know the correct behavior is guarantee by C++ standard, or it depends on your compiler vendor implementations? I have tested this under VC2008 and VC6. Works fine for both.