I am a C++ newbie. Although many similar questions have been asked and answered, I still find these concepts confusing.
I know
char c='a' // declare a single char c and assign value 'a' to it
char * str = "Test"; // declare a char pointer and pointing content str,
// thus the content can't be modified via point str
char str1[] = "Test"; // declare a char array str1 and assign "Test" to it
// thus str1 owns the data and can modify it
my first question is char * str creates a pointer, how does char * str = "Test"; work? assign a string literal to a pointer? It doesn’t make sense to me although it is perfectly legal, I think we can only assign an address to a pointer, however "Test" is a string literal not an address.
Second question is how come the following code prints out "Test" twice in a row?
char str2[] = {'T','e','s','t'}; // is this line legal?
// intializing a char array with initilizer list, seems to be okay to me
cout<<str2<<endl; // prints out "TestTest"
why cout<<str2<<endl; prints out "TestTest"?
>Solution :
char * str = "Test"; is not allowed in C++. A string literal can only be pointed to by a pointer to const. You would need const char * str = "Test";.
If your compiler accepts char * str = "Test"; it is likely outdated. This conversion has not been allowed since C++11 (which came out over 10 years ago).
how does char * str = "Test"; work?
String literals are implicitly convertible to a pointer to the start of the literal. In C++ arrays are implicitly convertible to pointer to their first element. For example int x[10] is implicitly convertible to int*, the conversion results in &(x[0]). This applies to string literals, their type is a const array of characters (const char[]).
how come the following code prints out "Test" twice in a row?
In C++ most features related to character strings assume the string is null terminated, which is implied in string literals. You would need {'T','e','s','t','\0'} to be equivalent to "Test".