对双变量的引用进行类型转换

Question

I have a question about typecasting the reference of a variable of type double. What is exactly happening when &d becomes typecasted into an unsigned char* ? How is typecasting the address of a variable valid and what is it actually doing?

#include <stdio.h>
 
int main(void) {
    // examining object representation is a legitimate use of cast
    double d = 3.14;
    printf("The double %.2f(%a) is: ", d, d);
    for(size_t n = 0; n < sizeof d; ++n)
        printf("0x%02x ", ((unsigned char*)&d)[n]); // <--
}

And how do the array brackets work in this case?

Answer 1

&d represents a pointer to the location at which the value of the double is stored. A pointer is simply a memory address (likely a 64-bit value) which identifies a location in memory, regardless of the type of data that is being pointed to.

This makes the cast from a double * to an unsigned char * possible as the representation of both is simply a word of memory. The array notation then dereferences this pointer and adds n times the size of a char (1 byte), and then reads the next char-many bits.

The array notation a[n] is equivalent to *(a+n) . It is simply adding an offset to the pointer referred to by a .