std::div, std::ldiv, std::lldiv, std::imaxdiv
<cstdlib>
(constexpr since C++23)
(constexpr since C++23)
<cinttypes>
(constexpr since C++23)
Computes both the quotient and the remainder of the division of the numerator x by the denominator y.
std::div for std::intmax_t is provided in <cinttypes> if and only if std::intmax_t is an extended integer type.The quotient is the algebraic quotient with any fractional part discarded (truncated towards zero). The remainder is such that quot * y + rem == x.
(until C++11)The quotient is the result of the expression x / y. The remainder is the result of the expression x % y.
(since C++11)Contents
[edit] Parameters
[edit] Return value
If both the remainder and the quotient can be represented as objects of the corresponding type (int, long, long long, std::intmax_t , respectively), returns both as an object of type std::div_t, std::ldiv_t, std::lldiv_t, std::imaxdiv_t defined as follows:
std::div_t
struct div_t { int quot; int rem; };
or
struct div_t { int rem; int quot; };
std::ldiv_t
struct ldiv_t { long quot; long rem; };
or
struct ldiv_t { long rem; long quot; };
std::lldiv_t
struct lldiv_t { long long quot; long long rem; };
or
struct lldiv_t { long long rem; long long quot; };
std::imaxdiv_t
struct imaxdiv_t { std::intmax_t quot; std::intmax_t rem; };
or
struct imaxdiv_t { std::intmax_t rem; std::intmax_t quot; };
If either the remainder or the quotient cannot be represented, the behavior is undefined.
[edit] Notes
Until CWG issue 614 was resolved (N2757), the rounding direction of the quotient and the sign of the remainder in the built-in division and remainder operators was implementation-defined if either of the operands was negative, but it was well-defined in std::div.
On many platforms, a single CPU instruction obtains both the quotient and the remainder, and this function may leverage that, although compilers are generally able to merge nearby / and % where suitable.
[edit] Example
#include <cassert> #include <cmath> #include <cstdlib> #include <iostream> #include <sstream> #include <string> std::string division_with_remainder_string(int dividend, int divisor) { auto dv = std::div(dividend, divisor); assert (dividend == divisor * dv.quot + dv.rem); assert (dv.quot == dividend / divisor); assert (dv.rem == dividend % divisor); auto sign = [](int n){ return n > 0 ? 1 : n < 0 ? -1 : 0; }; assert ((dv.rem == 0) or (sign(dv.rem) == sign(dividend))); return (std::ostringstream () << std::showpos << dividend << " = " << divisor << " * (" << dv.quot << ") " << std::showpos << dv.rem).str(); } std::string itoa(int n, int radix /*[2..16]*/) { std::string buf; std::div_t dv{}; dv.quot = n; do { dv = std::div(dv.quot, radix); buf += "0123456789abcdef"[std::abs(dv.rem)]; // string literals are arrays } while (dv.quot); if (n < 0) buf += '-'; return {buf.rbegin(), buf.rend()}; } int main() { std::cout << division_with_remainder_string(369, 10) << '\n' << division_with_remainder_string(369, -10) << '\n' << division_with_remainder_string(-369, 10) << '\n' << division_with_remainder_string(-369, -10) << "\n\n"; std::cout << itoa(12345, 10) << '\n' << itoa(-12345, 10) << '\n' << itoa(42, 2) << '\n' << itoa(65535, 16) << '\n'; }
Output:
+369 = +10 * (+36) +9 +369 = -10 * (-36) +9 -369 = +10 * (-36) -9 -369 = -10 * (+36) -9 12345 -12345 101010 ffff
[edit] See also
(function) [edit]