3 ms·

The declaration and use syntax of pointers is not 'virtually identical.' int* a; could be used to set int* b = a; or int c = * a; or int* d = &a; Also for arra

• Peaker 14y ago
  You are misunderstanding. I'll start from what you said about arrays: > because int a[10];, is of course a pointer This is a common misconception about C. int a[10] defines an array, not a pointer. typeof(a) is int[10], and not (int ptr). The thing is, whenever you take the value of an array in C, it is degraded to a pointer to the array's first element, and this confuses people to no end that arrays are pointers. For example, if we declare: int a[10]; int *b; Then sizeof(a) and sizeof(b) are very different. typeof(&a) is pointer-to-int[10], whereas typeof(&b) is pointer-to-pointer-to-int. void f(int (*x)[10]) { ... } If you call: f(&a); // will type-check f(&b); // will not So the syntax for declaring arrays is fine, and when you declare an array, no pointer is being declared at all. This is C syntax for type declarations: <base type> <type declaration here> The "<type declaration here>" part uses virtually the same syntax as use syntax. For example, if x is a pointer to a function that returns a pointer to an array of floats, you could reach a float via this syntax: (*(*x)())[1] You could declare x using this syntax: float (*(*x)())[10]; The main differences are: * Array size instead of index * There's no requirement to dereference a function pointer to call it, but there is a requirement to use the dereference syntax to declare it. * Array values can be used as pointers to their first elements (due to the automatic degradation), but must be declared as arrays.