Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

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

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

std::div_t     div( int x, int y );

std::ldiv_t    div( long x, long y );

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

std::ldiv_t   ldiv( long x, long y );

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

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.

The quotient is the result of the expression x / y. The remainder is the result of the expression x % y.

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:

struct div_t { int quot; int rem; };

or

struct div_t { int rem; int quot; };

struct ldiv_t { long quot; long rem; };

or

struct ldiv_t { long rem; long quot; };

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.

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.

#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

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4