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Fix mpfp::operator^= to resolve operator overloading problem

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This commit is contained in:
Soonho Kong 2013-08-14 23:47:36 -07:00
parent 2d74ff5fe0
commit eb29a67395
1 changed files with 2 additions and 3 deletions
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5
src/util/numerics/mpfp.h
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@ -438,8 +438,7 @@ public:
// friend mpfp operator/(mpz const & a, mpfp b) { return a.rdiv(b); }
// friend mpfp operator/(mpq const & a, mpfp b) { return a.rdiv(b); }
mpfp & pow(unsigned long int k, mpfr_rnd_t rnd = get_mpfp_rnd()) { mpfr_pow_ui(m_val, m_val, k, rnd); return *this; }
mpfp & operator^=(unsigned long int k) { return pow(k); }
mpfp & operator^=(unsigned long int k) { power(k); return *this; }
friend mpfp operator^(mpfp a, unsigned long int k) { return a ^= k; }
mpfp & operator++() { return operator+=(1lu); }
@ -507,7 +506,7 @@ public:
static void inv(mpfp & v) { v.inv(); }
static void reset(mpfp & v) { v = 0.0; }
// v <- v^k
static void power(mpfp & v, unsigned k) { v.pow(k); }
static void power(mpfp & v, unsigned k) { v.power(static_cast<unsigned long>(k)); }
static void abs(mpfp & v) { v.abs(); }
static void ceil(mpfp & v) { v.ceil(); }
static void floor(mpfp & v) { v.floor(); }