Asked  7 Months ago    Answers:  5   Viewed   47 times

Why do such operations:

std::cout << (-7 % 3) << std::endl;
std::cout << (7 % -3) << std::endl;

give different results?




From ISO14882:2011(e) 5.6-4:

The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral operands the / operator yields the algebraic quotient with any fractional part discarded; if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

The rest is basic math:

(-7/3) => -2
-2 * 3 => -6
so a%b => -1

(7/-3) => -2
-2 * -3 => 6
so a%b => 1

Note that

If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined.

from ISO14882:2003(e) is no longer present in ISO14882:2011(e)

Tuesday, June 1, 2021
answered 7 Months ago

Magic Enum header-only library provides static reflection for enums (to string, from string, iteration) for C++17.

#include <magic_enum.hpp>

enum Color { RED = 2, BLUE = 4, GREEN = 8 };

Color color = Color::RED;
auto color_name = magic_enum::enum_name(color);
// color_name -> "RED"

std::string color_name{"GREEN"};
auto color = magic_enum::enum_cast<Color>(color_name)
if (color.has_value()) {
  // color.value() -> Color::GREEN

For more examples check home repository

Where is the drawback?

This library uses a compiler-specific hack (based on __PRETTY_FUNCTION__ / __FUNCSIG__), which works on Clang >= 5, MSVC >= 15.3 and GCC >= 9.

Enum value must be in range [MAGIC_ENUM_RANGE_MIN, MAGIC_ENUM_RANGE_MAX].

  • By default MAGIC_ENUM_RANGE_MIN = -128, MAGIC_ENUM_RANGE_MAX = 128.

  • If need another range for all enum types by default, redefine the macro MAGIC_ENUM_RANGE_MIN and MAGIC_ENUM_RANGE_MAX.

  • MAGIC_ENUM_RANGE_MIN must be less or equals than 0 and must be greater than INT16_MIN.

  • MAGIC_ENUM_RANGE_MAX must be greater than 0 and must be less than INT16_MAX.

  • If need another range for specific enum type, add specialization enum_range for necessary enum type.

    #include <magic_enum.hpp>
    enum number { one = 100, two = 200, three = 300 };
    namespace magic_enum {
    template <>
      struct enum_range<number> {
        static constexpr int min = 100;
        static constexpr int max = 300;
Tuesday, June 1, 2021
answered 7 Months ago

I found something that at least begins to answer my own question. The following two links have wmv files from Microsoft that demonstrate using a C# class in unmanaged C++.

This first one uses a COM object and regasm:

This second one uses the features of C++/CLI to wrap the C# class: I have been able to instantiate a c# class from managed code and retrieve a string as in the video. It has been very helpful but it only answers 2/3rds of my question as I want to instantiate a class with a string perimeter into a c# class. As a proof of concept I altered the code presented in the example for the following method, and achieved this goal. Of course I also added a altered the {public string PickDate(string Name)} method to do something with the name string to prove to myself that it worked.

wchar_t * DatePickerClient::pick(std::wstring nme)
    IntPtr temp(ref);// system int pointer from a native int
    String ^date;// tracking handle to a string (managed)
    String ^name;// tracking handle to a string (managed)
    name = gcnew String(nme.c_str());
    wchar_t *ret;// pointer to a c++ string
    GCHandle gch;// garbage collector handle
    DatePicker::DatePicker ^obj;// reference the c# object with tracking handle(^)
    gch = static_cast<GCHandle>(temp);// converted from the int pointer 
    obj = static_cast<DatePicker::DatePicker ^>(gch.Target);
    date = obj->PickDate(name);
    ret = new wchar_t[date->Length +1];
    interior_ptr<const wchar_t> p1 = PtrToStringChars(date);// clr pointer that acts like pointer
    pin_ptr<const wchar_t> p2 = p1;// pin the pointer to a location as clr pointers move around in memory but c++ does not know about that.
    wcscpy_s(ret, date->Length +1, p2);
    return ret;

Part of my question was: What is better? From what I have read in many many efforts to research the answer is that COM objects are considered easier to use, and using a wrapper instead allows for greater control. In some cases using a wrapper can (but not always) reduce the size of the thunk, as COM objects automatically have a standard size footprint and wrappers are only as big as they need to be.

The thunk (as I have used above) refers to the space time and resources used in between C# and C++ in the case of the COM object, and in between C++/CLI and native C++ in the case of coding-using a C++/CLI Wrapper. So another part of my answer should include a warning that crossing the thunk boundary more than absolutely necessary is bad practice, accessing the thunk boundary inside a loop is not recommended, and that it is possible to set up a wrapper incorrectly so that it double thunks (crosses the boundary twice where only one thunk is called for) without the code seeming to be incorrect to a novice like me.

Two notes about the wmv's. First: some footage is reused in both, don't be fooled. At first they seem the same but they do cover different topics. Second, there are some bonus features such as marshalling that are now a part of the CLI that are not covered in the wmv's.


Note there is a consequence for your installs, your c++ wrapper will not be found by the CLR. You will have to either confirm that the c++ application installs in any/every directory that uses it, or add the library (which will then need to be strongly named) to the GAC at install time. This also means that with either case in development environments you will likely have to copy the library to each directory where applications call it.

Wednesday, June 2, 2021
answered 7 Months ago

The integer division there is just taking the floor of the number obtained at the end.

10/3  -> floor(3.33)  ->  3
-10/3 -> floor(-3.33) -> -4

(Why it floors)

The modulo operation on the other hand is following the mathematical definition.

Thursday, June 3, 2021
answered 6 Months ago

I don't want to bother you with some complex mathematical concepts, so i'll try to keep it simple. When we say that a = b (mod c), we simply say that a-b is a multiple of c. This means that when we want to know what is the value of a mod c, saying that it is a or a-c or a+c or a+1000*c is true. Thus, your 2 formulas are valid.

But what you want is to know the answer that a computer would give to you, right ? Well, it depends of the language you are using. With Java for example, a mod b has the sign of a and has is absolute value strictly inferior to b. This means that with a = 7, b = 3 and N = 5, (a-b)%N = 4, but your two expressions will return -1.

What I would suggest you to do if you want to do arithmetics with modulos is to create your own mod function, so it always give you a positive integer for example. This way, your 2 expressions will always be equal to the original one.

An example here in pseudocode :

function mod (int a, int N)
  return (a%N+N)%N
Tuesday, October 19, 2021
Eduardo Sousa
answered 2 Months ago
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