C type casting doubts when type are converted many times here and there-CodePudding

What is this type casting doing?

ph = (struct Proghdr *)((uint8_t *) ELFHDR   ELFHDR->e_phoff);

Here, ELFHDR is already defined as #define ELFHDR ((struct Elf *) 0x10000).

I want to ask: At the end, the value in ph will be (struct Proghdr *) then what is the need for the (uint8_t *)?

I will be grateful if someone solves my problem related to type-casting.

CodePudding user response：

Pointer arithmetic in C is performed in units whose size is that of the pointed-to type. So, for example, if p is a pointer to a double, then p = p 1 will add the size (in bytes) of a double to the address stored in p.

In your case, the e_phoff member of the pointed-to structure contains a byte offset for some particular data; however, simply adding that value to the ELFHDR pointer itself (which is of Elf* type) would add that value multiplied by the size of the Elf structure. So, in order to properly use that byte offset, the pointer is first cast to a uint8_t*, so that any pointer arithmetic is performed in terms of raw (single) bytes.

CodePudding user response：

Remember the equivalence p[i] == *( p i ).

Given pointer p and integer i, p i advances pointer p by i times the size of *p.

For example, you see p advanced by 8 times the time size of a uint32_t in the following:

uint32_t a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
uint32_t *p = a;
p = p   8;
printf( "%" PRIu32 "\n", *p );  // 8

So if you wanted to advance the pointer by i bytes, you'd need a pointer to a byte (char).

uint32_t a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
uint32_t *p = a;
p = (uint32_t *)( (char *)p   8 );
printf( "%" PRIu32 "\n", *p );  // 2

uint8_t is used as an equivalent of char in the snippet you posted.

Demo on Compile Explorer