char*
and char[]
are different types, but it's not immediately apparent in all cases. This is because arrays decay into pointers, meaning that if an expression of type char[]
is provided where one of type char*
is expected, the compiler automatically converts the array into a pointer to its first element.
Your example function printSomething
expects a pointer, so if you try to pass an array to it like this:
char s[10] = "hello";
printSomething(s);
The compiler pretends that you wrote this:
char s[10] = "hello";
printSomething(&s[0]);