Why is sizeof(unsigned double) equal to 4?

Question

A colleague of mine asked if there is unsigned double, and I said there isn't, but I still checked it, and this compiles in Microsoft Visual C++ 2010:

unsigned double a; double b; printf("size_a=%d size_b=%d", (int) sizeof(a), (int) sizeof(b)); 

It outputs size_a=4 size_b=8. That is, four bytes for unsigned double, and eight bytes for double.

Answer 1

unsigned double is invalid. This is also true in MSVC. When compiling the above code in MSCV 2010 with warnings enabled you get:

warning C4076: 'unsigned' : can not be used with type 'double'

The compiler actually ignores double after unsigned, making your a actually an unsigned int.

If you try the following:

unsigned double a = 1.0; 

You actually get two warnings:

warning C4076: 'unsigned' : can not be used with type 'double' warning C4244: 'initializing' : conversion from 'double' to 'unsigned int', possible loss of data 

Interestingly, there is no C4076 warning in MSDN for VS2010. It is present only for VS2005 and VS2008.

Answer 2

If you set the warning level higher (/W3 in my test), you will get an appropriate warning:

warning C4076: 'unsigned' : can not be used with type 'double'

If you then use the debugger to inspect the variable, all becomes clear:

You can see that the variable is in fact an unsigned int