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What does it mean? (int &) a

define the variable float a, convert a to float and and int &, what does this mean? After conversion, a is a reference to yourself? And why are the two results 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
+5
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3 answers
cout << (float &)a <<endl;
cout << (int &)a << endl;

The first handles the bits as a float. The second considers bits in like an int. The bits for float 1.0 are simply bits for the integer 1065353216.

This is basically the equivalent of:

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

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

Edit: Now I look back to find out if this is true. I suspect this is not the case. It depends on which type is smaller or equal to the actual size.

+14
source

undefined: -).

, . a float, a ( ) . (float&)a , float ( , ); (int&)a int. (, a) lvalue , , - undefined, . , , , .

float , . , , 0.0 0 . ( , , , 2 1. 1.0 0.) ( ), . , "" " " , , .

+9

, a float int & ( int) . a int, , . , float, int, 1065353216, 42 .

, , int, , float int. int & , , .

+2

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