What happens if you write following code?
string& foo()
{
return "Hello World";
}
cout << foo() << endl;
A: 1. Will give an error as Hello World is created as unnamed character pointer to const. this is being assigned to non-const reference that is not allowed. Could not alter '"Hello World"' to 'std::string&'
2. const string& foo1()
{
return "Hello World";
}
It will give a warning .As you are returning a reference to temporary, which will die instantly when the expression is completed?
classsize.C:7: warning: returning reference into temporary output : Aborted. Segment fault.
3. Char *foo1()
{
return "Hello World";
}
Returning the address of character literal that is created on the static memory.
In C++, the compiler let the use of string literals to initialize character arrays. A string literal contain zero or more characters surrounded through double quotation marks ("). A string represents a sequence of characters that, taken mutually, form a null-terminated string. The compiler developed static storage space for the string, null-terminates it, & puts the address of this space into the char* variable. The sort of a literal string is an array of const chars.
char* szMyString = "Hello world.";
szMyString[3] = 'q'; // undefined, altering static buffer!!!
In the following instance, the compiler puts a null-character automatically at the end of the literal string of characters "Hello world". Then it creates a storage space for the resulting string - it is an array of const chars. Then it puts the initial address of this array into the szMyString variable. We will attempt to change this string (wherever it is stored) by accessing it through an index into szMyString. It is a Bad Thing; the standard does not say where the compiler puts literal strings. They can go anywhere, perhaps in some place in memory that you shouldn't be altering.