CODEWITHSUNDEEP
c
Seonix
Seonix

Reputation: 13

Why does this program print 23 after casting?

int main() {
uint32_t x;
uint32_t* p = (uint32_t*)malloc(sizeof(uint32_t));
uint32_t array[9] = {42, 5, 23, 82, 127, 21, 324, 3, 8};
*p = *(uint32_t*) ((char*) array + 8);
printf("l: %d\n", *p);

return 0;

}

Why does *p print the value of the second index of array after *p = *(uint32_t*) ((char*) array + 8); ?

Upvotes: 1

Views: 63

Answers (2)

John Bollinger
John Bollinger

Reputation: 181849

Perhaps the question revolves around operator precedence and associativity. This expression ...

*(uint32_t*) ((char*) array + 8)

... is equivalent to ...

*((uint32_t*) (((char*) array) + 8))

... NOT ...

*((uint32_t*) ((char*) (array + 8)))  // not equivalent

Typecast operators have very high precedence, higher than binary +, in particular. Pointer arithmetic works in units of the pointed-to type, and the type of ((char *) array) is, or course, char *. Its pointed to type is char. Therefore, (((char*) array) + 8) evaluates to a pointer to the 8th char past the beginning of the array.

Your implementation has 8-bit chars, which is exceedingly common (but not universal). Type uint32_t has exactly 32 bits, and so comprises four 8-bit chars. It follows that (((char*) array) + 8) points to the first char of the third uint32_t in array, as the 8 bytes skipped are exactly the bytes of the first two uint32_ts.

When that pointer is converted back to type uint32_t *, it points to the whole third uint32_t of the array, and the value of that element, 23, is read back by dereferencing the pointer with the unary * operator.

Upvotes: 3

0___________
0___________

Reputation: 68059

sizeof(uint32_t) is 4

sizeof(char) is 1 => 8 * sizeof(char) == 8

8 / sizeof(uint32_t) = 2

then ((char *)array) + 8 == array + (8 / sizeof(uint32_t))

p = array + (8 / sizeof(uint32_t)) => p = array + 2

then *p == array[2] == 23

Upvotes: 2

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