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]);
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