Как преобразовать int в double c

Cast int variable to double

I am a beginner C# programmer, and I am trying to create a calculator. I can’t seem to figure out how to cast an int variable to a double . This is what I have so far:

How can I make this work?

EDIT: Both result and value are int variables.

5 Answers 5

is not good. The result of a casting expression is always an rvalue, i. e. something that cannot be assigned to. Related: you can’t change the type of an expression (you can cast it, but that won’t really change its type, just act as another type temporarily). Once you declared your variable as, say, an int , you won’t be able to store a double in it — however you cast the division, etc. it will always be truncated in the end.

You most likely have to introduce a double temporary variable to store the result of the division.

If both of your variables are not double s, assign them into a double variable and then divide.

I presume you are trying to make rezultat a double and I presume it’s not declared as one and you just can’t do that. Your resulting variable that will hold the result must also be a double or you will just get a whole number not rounded.

This depends on the type of the rezultat variable. If it’s double , then you don’t have to do anything, integer division won’t be used in any case. But if it’s int , then your cast doesn’t make any sense, you can’t store a double value in an int variable.

So, the correct solution depends on what exactly do you want to do. But if your goal is to have the result of the actual division as a double , you will need some double variable for that. And if you have that, your if won’t make any sense anymore, just use double division in all cases.

resultat must be a double to store the result of rezultat / value . Otherwise, if both resultat and value are int , you won’t get floating point numbers. For example, 5 / 3 = 1 , but (double)5 / 3 = 1.666667 . Notice that the value 1.6666667 is just a double .

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Convert Int to Double in C++

Typecasting is a procedure for transforming data between different data types. This data conversion method is also termed type conversion or type coercion. Casting allows the data type to be modified rather than the data itself. C++ supports typecasting for all the data types. This article will discuss one typecasting method: the conversion from int data type to double data type. We will explore different approaches that simply convert int to double data type.

Two Categories of Conversion in C++

Implicit Type Conversion

The implicit type conversion is done spontaneously. In this conversion, the user has no input, and the compiler does the conversion entirely on its own. When there are many types of data in an expression, conversion is frequently performed. However, there is a risk of data loss, sign loss, or data overflow with this form of conversion.

Explicit Type Conversion

The Explicit type is defined by the user and is sometimes referred to as “type-casting.” Depending on the needs, the user translates or converts data of one data type to another. These types of conversions are more secure.

Example 1

The following example is the working of implicit type conversion, which converts int to double data type. The program starts from the header files. We have two header files in the header section. Then there’s the code implementation’s main function. In the main function, we have defined a variable as “IntNumber” of integer data type and initialized it with the integral value.

Likewise, we have created another variable as “DoubleNumber” of data type double, but it is not initialized. Now, we have to convert the int data type value to the double data type value. So, we have assigned the int variable “IntNumer” to a double data type variable “DoubleNumber”. This approach is the implicit type conversion. The values of int data type and double data type will be printed through the C++ cout command.

Before assigning the int value to the double variable, the compiler transforms it to double automatically. As you can see in the image, the int cannot include a decimal value, and the digits after the decimal point are truncated in this example.

Example 2

We have seen the conversion of int to double in the preceding example. Now, we will convert double to int, which is also done through implicit conversion.

In the first step, we have included iostream and std namespace files. Then, we invoked the program’s main function, where we declared an integer data type variable “IntVal”. Also, we have declared another variable of double data type as “DoubleVal” and initialized it with the double value as it contains a decimal part. For the conversion, we have assigned the double variable to the int variable. The implicit type conversion value will be printed on the console screen when executing the program.

You can see the double value and the conversion of double data in the int data as the output.

Example 3

As we can see from the previous example, converting from one data type to another is a chance of data loss. This occurs when larger-type data is transformed into smaller-type data. To overcome the data loss problem, we performed explicit type conversion. The explicit type casting is done through the C-style typecasting. As the name implies, it favors a C language form of casting. Cast notation is another term for it.

The program begins with the header section, which includes the C++ files. In the next step, the main function is created where the variable is defined as “IntNumber” of data type “int” and stores the integral value. Another variable is declared as “DoubleNumber” of data type “double”.

Then, we have used the C-style type casting method, which uses the type double notation with the int variable and assigns it to the double variable, “DoubleNumber”. These values will be printed by using the C++ cout command.

The decimal part is truncated as the int data type has no data type.

Example 4

The notation function can also move data between different data types. The following program uses function style casting to convert int to double data type.

We have the main function where we have created an int data type variable as “Integer_x” and initialized it with the numeric value “20”. The double type variable is also defined as “Double_y”. Then, we used function typecasting to convert int data to double type. We have passed the int variable “Integer_x” as the function and defined it with the data type “double”. The variable “Double y” is subsequently assigned to the function.

The outcome of the function style casting performed on int type to convert into double data type displayed on the following image:

Example 5

For type conversion of double to int, we implemented both C-style type conversion and function-style casting. This will show the results of the two ways of explicit typecasting.

In the main function block of this program, we have a double data type variable as “DoubleX” and initialized with the decimal value. Then, we created another two variables, “Integer1” and “Integer2”, and assigned a data type int. The int variable “Integer1” is used for C-style conversion, and the int variable “Integre2” is used for function style conversion. The converted values into integers from double data will be printed after performing both the conversion.

As both ways of explicit conversion do the same thing, they produce the same following results:

Conclusion

We will conclude our discussion of typecasting of int to double in C++. We have learned about the implicit and explicit conversions of int data to double and double data to int data type used in C++. However, conversions or typecasting can only be used carefully and only when necessary to avoid data loss and other problems.

Как преобразовать int в double c

Если в арифметических операциях участвуют значения разных типов, то компилятор неявно пытается привести их к одному типу. Кроме того, когда мы присваиваем переменной какое-либо значение, это значение всегда приводится к типу переменной. Например:

Переменной d, которая представляет тип int, присваивается значение типа char, поэтому компилятор выполняет приведение значения от типа char к типу int.

В то же время не всегда преобразования могут быть безопасными, поскольку разные типы имеют разное внутреннее представление. И просто так перейти от одного представления к другому без потери точности данных не всегда возможно.

Рассмотрим, какие преобразования применяет компилятор при арифметических операциях:

Если один из операндов имеет тип long double , то второй операнд тоже будет преобразован в тип long double

Если предыдущий пункт не выполняется и если один из операндов имеет тип double , то второй операнд тоже будет преобразован к типу double

Если предыдущий пункт не выполняется и если один из операндов имеет тип float , то второй операнд тоже будет преобразован к типу float

Если предыдущий пункт не выполняется и если один из операндов имеет тип unsigned long int , то второй операнд тоже будет преобразован к типу unsigned long int

Если предыдущий пункт не выполняется и если один из операндов имеет тип long , то второй операнд тоже будет преобразован к типу long

Если предыдущий пункт не выполняется и если один из операндов имеет тип unsigned , то второй операнд тоже будет преобразован к типу unsigned

Если предыдущий пункт не выполняется то оба операнда приводятся к типу int

В выражении number1 + number2 число number2 представляет тип double , поэтому число number1 будет автоматически приводиться к числу double. И результат операции сложения также будет представлять тип double .

Операция преобразования

С помощью специальной операции преобразования мы можем явным образом привести данные к нужному типу. Например:

В выражении int c = a / b; результат деления будет целочисленный — 2, при котором дробная часть будет отброшена, так как оба операнда операции представляют целые числа.

В выражении double d = a / b; результат деления будет представлять вещественное число — 2.00000, но так как оба операнда являются целыми числами, то опять же результат операции будет представлять целое число 2, и только поле выполнения деления произойдет присвоение результата переменной d с приведением значения 2 от типа int к типу double.

В выражении double e = (double)a / (double)b применяется явное преобразование данных к типу double, поэтому и результат деления будет представлять вещественное число — 2.50000.

Для выполнения операции приведении в скобках указывается тот тип, к которому надо привести значение:

В ряде случаев преобразования сопровождаются потерей информации, например, когда числа большей разрядности (скажем размером 4 байт) получаем число меньшей разрядности (например, в 2 байта). Без потери информации проходят следующие цепочки преобразований:

char -> short -> int -> long

unsigned char -> unsigned short -> unsigned int -> unsigned long

float -> double -> long double

При всех остальных преобразованиях, которые не входят в эти цепочки, мы можем столкнуться с потерей точности данных. Так, в примере выше преобразование от int к char не является безопасным, поэтому к таким преобразованиям следует относиться с осторожностью. Например:

Здесь две ситуации небезопасных преобразований. В первом случае число типа int , которое равно 300, присваивается переменной типа char . В итоге переменная code будет равна 44. Почему? Число 300 в двоичной системе:

Оставляем только первый младший байт:

И у нас получается число 44 в десятичной системе.

Во втором случае число 100000 (которое по умолчанию представляет тип int ), усекается до разрядности типа short — до двух байт.

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Built-in numeric conversions (C# reference)

C# provides a set of integral and floating-point numeric types. There exists a conversion between any two numeric types, either implicit or explicit. You must use a cast expression to perform an explicit conversion.

Implicit numeric conversions

The following table shows the predefined implicit conversions between the built-in numeric types:

[!NOTE] The implicit conversions from int , uint , long , ulong , nint , or nuint to float and from long , ulong , nint , or nuint to double may cause a loss of precision, but never a loss of an order of magnitude. The other implicit numeric conversions never lose any information.

There are no implicit conversions to the byte and sbyte types. There are no implicit conversions from the double and decimal types.

There are no implicit conversions between the decimal type and the float or double types.

A value of a constant expression of type int (for example, a value represented by an integer literal) can be implicitly converted to sbyte , byte , short , ushort , uint , ulong , nint , or nuint , if it’s within the range of the destination type:

As the preceding example shows, if the constant value is not within the range of the destination type, a compiler error CS0031 occurs.

Explicit numeric conversions

The following table shows the predefined explicit conversions between the built-in numeric types for which there is no implicit conversion:

[!NOTE] An explicit numeric conversion might result in data loss or throw an exception, typically an xref:System.OverflowException.

When you convert a value of an integral type to another integral type, the result depends on the overflow-checking context. In a checked context, the conversion succeeds if the source value is within the range of the destination type. Otherwise, an xref:System.OverflowException is thrown. In an unchecked context, the conversion always succeeds, and proceeds as follows:

If the source type is larger than the destination type, then the source value is truncated by discarding its «extra» most significant bits. The result is then treated as a value of the destination type.

If the source type is smaller than the destination type, then the source value is either sign-extended or zero-extended so that it’s of the same size as the destination type. Sign-extension is used if the source type is signed; zero-extension is used if the source type is unsigned. The result is then treated as a value of the destination type.

If the source type is the same size as the destination type, then the source value is treated as a value of the destination type.

When you convert a decimal value to an integral type, this value is rounded towards zero to the nearest integral value. If the resulting integral value is outside the range of the destination type, an xref:System.OverflowException is thrown.

When you convert a double or float value to an integral type, this value is rounded towards zero to the nearest integral value. If the resulting integral value is outside the range of the destination type, the result depends on the overflow-checking context. In a checked context, an xref:System.OverflowException is thrown, while in an unchecked context, the result is an unspecified value of the destination type.

When you convert double to float , the double value is rounded to the nearest float value. If the double value is too small or too large to fit into the float type, the result is zero or infinity.

When you convert float or double to decimal , the source value is converted to decimal representation and rounded to the nearest number after the 28th decimal place if necessary. Depending on the value of the source value, one of the following results may occur:

If the source value is too small to be represented as a decimal , the result becomes zero.

If the source value is NaN (not a number), infinity, or too large to be represented as a decimal , an xref:System.OverflowException is thrown.

When you convert decimal to float or double , the source value is rounded to the nearest float or double value, respectively.

C# language specification

For more information, see the following sections of the C# language specification: