What is *(uint32_t*)?

Question

I have a hard time understanding *(uint32_t*).

Let's say that I have:

uint32_t* ptr;
uint32_t num
*(uint32_t*)(ptr + num); // What does this do? Does it 

Answer 1

uint32_t is a numeric type that guarantees 32 bits. The value is unsigned, meaning that the range of values goes from 0 to 2³² - 1.

This

uint32_t* ptr;

declares a pointer of type uint32_t*, but the pointer is uninitialized, that is, the pointer does not point to anywhere in particular. Trying to access memory through that pointer will cause undefined behaviour and your program might crash.

This

uint32_t num;

is just a variable of type uint32_t.

This

*(uint32_t*)(ptr + num);

ptr + num returns you a new pointer. It is called pointer arithmetic. It's like regular arithmetic, only that compiler takes the size of types into consideration. Think of ptr + num as the memory address based on the original ptr pointer plus the number of bytes for num uint32_t objects.

The (uint32_t*) x is a cast. This tells the compiler that it should treat the expression x as if it were a uint32_t*. In this case, it's not even needed, because ptr + num is already a uint32_t*.

The * at the beginning is the dereferencing operator which is used to access the memory through a pointer. The whole expression is equivalent to

ptr[num];

Now, because none of these variables is initialized, the result will be garbage.

However, if you initialize them like this:

uint32_t arr[] = { 1, 3, 5, 7, 9 };
uint32_t *ptr = arr;
uint32_t num = 2;

printf("%u\n", *(ptr + num));

this would print 5, because ptr[2] is 5.