通过附加两个int的二进制表示来创建double

Question

I'm trying to create a double by appending the binary representations of two int s. This is what I have now:

int i = arc4random();
*(&i+1) = arc4random();
double d = *(double*)&i;

I hoped the double pointer would also use the value in the &i+1 address, and this is indeed the case. When I print the binary representations of i and *(&i+1) and compare them to the binary representation of d , d is composed by appending *(&i+1) and i . So *(&i+1) comes first?! Why is this the case?

EDIT: also take a look at the answers of Juan Catalan and Mark Segal to know what's the right way of doing what I did using unions.

Answer 1

Use a union instead. You still could generate a NaN But won't have memory management issues.

This is the whole purpose of a union , share a memory block among several different types.

Make sure in your architecture sizeof(double) is equal to twice sizeof(int)

Answer 2

You could either use a union, like this:

typedef union {
    double d;
    struct {
        int a;
        int b;
    } integers;
} double_generator;

int main(void) {
    double_generator g;
    g.integers.a = arc4random();
    g.integers.b = arc4random();

    // g.d is now a "randomly generated double"

    return 0;
}

Or "manually", possibly unsafely, do the job yourself:

int main(void) {
    double d;
    *((int *)&d) = arc4random();
    *(((int *)&d) + 1) = arc4random();
    printf("%f", d);
    return 0;
}

In both cases, you simply address the same memory as if it was int or double . This could generate every possible double value, including NaN or Infinity .

However, this code assumes that sizeof(double) is (at least) twice as sizeof(int) . If it isn't, you must introduce different code to handle that.

Answer 3

When I print the binary representations of i and *(&i+1) and compare them to the binary representation of d, d is composed by appending *(&i+1) and i. So *(&i+1) comes first?! Why is this the case?

the actual ordering of the bytes depends upon the 'Endian-ness' of the underlying hardware architecture.

with little Endian architecture:

1) the lowest address byte contains the least significant 8 bits of the total variable  
2) the highest address byte contains the most significant 8 bits of the total variable.