这是什么意思？ （int＆）a

Question

define a float variable a, convert a to float & and int &, what does this mean? After the converting , a is a reference of itself? And why the two result is different?

#include <iostream>
using namespace std;
int
main(void)
{
    float a = 1.0;
    cout << (float &)a <<endl;
    cout << (int &)a << endl;

    return 0;
}


thinkpad ~ # ./a.out 
1
1065353216

Answer 1

cout << (float &)a <<endl;
cout << (int &)a << endl;

The first one treats the bits in a like it's a float. The second one treats the bits in a like it's an int. The bits for float 1.0 just happen to be the bits for integer 1065353216.

It's basically the equivalent of:

float a = 1.0;
int* b = (int*) &a;
cout << a << endl;
cout << *b << endl;

(int &) a casts a to a reference to an integer. In other words, an integer reference to a. (Which, as I said, treats the contents of a as an integer.)

Edit: I'm looking around now to see if this is valid. I suspect that it's not. It's depends on the type being less than or equal to the actual size.

Answer 2

It means undefined behavior:-).

Seriously, it is a form of type punning. a is a float , but a is also a block of memory (typically four bytes) with bits in it. (float&)a means to treat that block of memory as if it were a float (in other words, what it actually is); (int&)a means to treat it as an int . Formally, accessing an object (such as a ) through an lvalue expression with a type other than the actual type of the object is undefined behavior, unless the type is a character type. Practically, if the two types have the same size, I would expect the results to be a reinterpretation of the bit pattern.

In the case of a float , the bit pattern contains bits for the sign, an exponent and a mantissa. Typically , the exponent will use some excess-n notation, and only 0.0 will have 0 as an exponent. (Some representations, including the one used on PCs, will not store the high order bit of the mantissa, since in a normalized form in base 2, it must always be 1. In such cases, the stored mantissa for 1.0 will have all bits 0.) Also typically (and I don't know of any exceptions here), the exponent will be stored in the high order bits. The result is when you "type pun" a floating point value to a an integer of the same size, the value will be fairly large, regardless of the floating point value.

Answer 3

The values are different because interpreting a float as an int & (reference to int ) throws the doors wide open. a is not an int , so pretty much anything could actually happen when you do that. As it happens, looking at that float like it's an int gives you 1065353216 , but depending on the underlying machine architecture it could be 42 or an elephant in a pink tutu or even crash.

Note that this is not the same as casting to an int , which understands how to convert from float to int . Casting to int & just looks at bits in memory without understanding what the original meaning is.