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Create double by appending binary representations of two ints

I'm trying to create a double by appending the binary representations of two ints. 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.

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Rugen Heidbuchel Avatar asked Mar 15 '23 21:03

Rugen Heidbuchel

1 Answers

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)

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Juan Catalan Avatar answered Apr 26 '23 17:04

Juan Catalan