Why does 'uint64_t i = -UINT64_MAX' have value '1'?

From:
Dom Jackson <nospam@mci2000.com>
Newsgroups:
comp.lang.c++,comp.lang.c
Date:
Sat, 23 Jun 2007 19:22:25 +0100
Message-ID:
<fsnq731dpl53npvjjqbb9njt4gdnih3eie@4ax.com>
What is the value of -(max-val-of-an-unsigned-type)? According to my
compiler, -UINT32_MAX = 1, and -UINT64_MAX is also 1 (test code
below).

From the C99 LRM, p79:

"The result of the unary - operator is the negative of its (promoted)
operand. The integer promotions are performed on the operand, and the
result has the promoted type."

As far as I can make out from p43, uint32_t and uint64_t are unchanged
by the integer promotions. So how do you take the negative of an
unsigned operand, and how can the compiler get the result '1' without
using a temporary of a higher precision? Any thoughts?

Thanks -

Dom

--------------------------------------------------------------
[the code below is C++ to simplify data output, and produces this
result]:

i is ffffffff; j is 1; k is 1; l is ffffffffffffffff; -l is 1; m is 1

--------------------------------------------------------------

#define __STDC_LIMIT_MACROS
#include <stdint.h>
#include <iostream>
#include <iomanip>

int main() {
uint32_t i = UINT32_MAX;
int64_t j;
uint64_t k;
uint64_t l = UINT64_MAX;
uint64_t m = -UINT64_MAX;

j = -i;
k = -i;

std::cout << std::hex << std::setfill('0')
<< "i is " << i
<< "; j is " << j
<< "; k is " << k
<< "; l is " << l
<< "; -l is " << -l
<< "; m is " << m
<< std::endl;

return 0;
}

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