What does `((void (*)())0x1000)();` mean?
c, pointers, void
Solution
`C` declarations are decoded from inside out using a simple rule: start from the identifier and check on the right side for `[]` (array) or `()` (function) then check on the left side for the type of the values (stored in the array or returned by the function), without crossing the parentheses; escape from the parentheses and repeat.
For example:
void (*p)()
`p` is (nothing on the right) a pointer (on the left, don't cross the parentheses) to (escape the parentheses, read the next level) a function (right) that returns nothing (left).
When the identifier (`p` in this case) is missing, all that remains is a type declaration.
A type enclosed in parentheses, put in front of a value is a type cast.
(void (*)())0x1000
converts the number `0x1000` to a pointer to a function that doesn't return anything (see what's outside the parentheses in the paragraph about the declaration of `p` above).
On the next level, the expression above (a pointer to a function can be used in the same way as a function name) is used to execute the code pointed at.
See below the entire expression de-composed:
(
(
void (*)() /* type: pointer to function that doesn't return anything */
)0x1000 /* value 0x1000 treated as a value of the type declared above */
) /* enclose in parentheses to specify the order of evaluation */
(); /* the pointer above used as a function name to run the code */
Problem
Here is a code that purpose is to set the program counter to jump to address `0x1000`. I know what it does but I don't understand how. It is related to my lack of C language knowledge. May be you can enlighten me. Here is the statement/function (I even don't know what it is :)) ``` ((void (*)())0x1000)(); ``` I thing it is pointer to a functions that returns `void` and accepts no argument. Please correct me if I am wrong.