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Showing content from http://en.cppreference.com/w/cpp/language/../numeric/math/div.html below:

std::div, std::ldiv, std::lldiv, std::imaxdiv - cppreference.com

std::div_t     div( int x, int y );

(1) (constexpr since C++23)

std::ldiv_t    div( long x, long y );

(2) (constexpr since C++23)

std::lldiv_t   div( long long x, long long y );

(3) (since C++11)
(constexpr since C++23)

std::ldiv_t   ldiv( long x, long y );

(4) (constexpr since C++23)

std::lldiv_t lldiv( long long x, long long y );

(5) (since C++11)
(constexpr since C++23) (6) (since C++11)
(constexpr since C++23) (7) (since C++11)
(constexpr since C++23)

Computes both the quotient and the remainder of the division of the numerator x by the denominator y.


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) [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; };

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 [edit] External links

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