Question

I have this sample code which generates the following warning (VS2008 compiler with SP1):

warning C4146: unary minus operator
applied to unsigned type, result still
unsigned

Code:

void f(int n)
{
}

int main()
{
    unsigned int n1 = 9;
    f(-n1);
}

But since function f is taking it’s parameter as an int shouldn’t this code compile without any warnings?

Answer 1

Standard 5.3.1/7

The operand of the unary – operator
shall have arithmetic or enumeration
type and the result is the negation of
its operand. Integral promotion is
performed on integral or enumeration
operands. The negative of an unsigned
quantity is computed by subtracting
its value from 2n, where n is the
number of bits in the promoted
operand. The type of the result is the
type of the promoted operand.

And the paragraph on Integral Promotion 4.5/1

An rvalue of type char, signed char,
unsigned char, short int, or unsigned
short int can be converted to an
rvalue of type int if int can
represent all the values of the source
type; otherwise, the source rvalue can
be converted to an rvalue of type
unsigned int.

i.e. an unsigned int will not be promoted to an int.