multiplying assignment operator c

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Assignment Operators in C

In C language, the assignment operator stores a certain value in an already declared variable. A variable in C can be assigned the value in the form of a literal, another variable, or an expression.

The value to be assigned forms the right-hand operand, whereas the variable to be assigned should be the operand to the left of the " = " symbol, which is defined as a simple assignment operator in C.

In addition, C has several augmented assignment operators.

The following table lists the assignment operators supported by the C language −

Simple Assignment Operator (=)

The = operator is one of the most frequently used operators in C. As per the ANSI C standard, all the variables must be declared in the beginning. Variable declaration after the first processing statement is not allowed.

You can declare a variable to be assigned a value later in the code, or you can initialize it at the time of declaration.

You can use a literal, another variable, or an expression in the assignment statement.

Once a variable of a certain type is declared, it cannot be assigned a value of any other type. In such a case the C compiler reports a type mismatch error.

In C, the expressions that refer to a memory location are called "lvalue" expressions. A lvalue may appear as either the left-hand or right-hand side of an assignment.

On the other hand, the term rvalue refers to a data value that is stored at some address in memory. A rvalue is an expression that cannot have a value assigned to it which means an rvalue may appear on the right-hand side but not on the left-hand side of an assignment.

Variables are lvalues and so they may appear on the left-hand side of an assignment. Numeric literals are rvalues and so they may not be assigned and cannot appear on the left-hand side. Take a look at the following valid and invalid statements −

Augmented Assignment Operators

In addition to the = operator, C allows you to combine arithmetic and bitwise operators with the = symbol to form augmented or compound assignment operator. The augmented operators offer a convenient shortcut for combining arithmetic or bitwise operation with assignment.

For example, the expression "a += b" has the same effect of performing "a + b" first and then assigning the result back to the variable "a".

Run the code and check its output −

Similarly, the expression "a <<= b" has the same effect of performing "a << b" first and then assigning the result back to the variable "a".

Here is a C program that demonstrates the use of assignment operators in C −

When you compile and execute the above program, it will produce the following result −

cppreference.com

Assignment operators.

Assignment and compound assignment operators are binary operators that modify the variable to their left using the value to their right.

[ edit ] Simple assignment

The simple assignment operator expressions have the form

Assignment performs implicit conversion from the value of rhs to the type of lhs and then replaces the value in the object designated by lhs with the converted value of rhs .

Assignment also returns the same value as what was stored in lhs (so that expressions such as a = b = c are possible). The value category of the assignment operator is non-lvalue (so that expressions such as ( a = b ) = c are invalid).

rhs and lhs must satisfy one of the following:

• both lhs and rhs have compatible struct or union type, or..
• rhs must be implicitly convertible to lhs , which implies
• both lhs and rhs have arithmetic types , in which case lhs may be volatile -qualified or atomic (since C11)
• both lhs and rhs have pointer to compatible (ignoring qualifiers) types, or one of the pointers is a pointer to void, and the conversion would not add qualifiers to the pointed-to type. lhs may be volatile or restrict (since C99) -qualified or atomic (since C11) .
• lhs is a (possibly qualified or atomic (since C11) ) pointer and rhs is a null pointer constant such as NULL or a nullptr_t value (since C23)

[ edit ] Notes

If rhs and lhs overlap in memory (e.g. they are members of the same union), the behavior is undefined unless the overlap is exact and the types are compatible .

Although arrays are not assignable, an array wrapped in a struct is assignable to another object of the same (or compatible) struct type.

The side effect of updating lhs is sequenced after the value computations, but not the side effects of lhs and rhs themselves and the evaluations of the operands are, as usual, unsequenced relative to each other (so the expressions such as i = ++ i ; are undefined)

Assignment strips extra range and precision from floating-point expressions (see FLT_EVAL_METHOD ).

In C++, assignment operators are lvalue expressions, not so in C.

[ edit ] Compound assignment

The compound assignment operator expressions have the form

The expression lhs @= rhs is exactly the same as lhs = lhs @ ( rhs ) , except that lhs is evaluated only once.

[ edit ] References

• C17 standard (ISO/IEC 9899:2018):
• 6.5.16 Assignment operators (p: 72-73)
• C11 standard (ISO/IEC 9899:2011):
• 6.5.16 Assignment operators (p: 101-104)
• C99 standard (ISO/IEC 9899:1999):
• 6.5.16 Assignment operators (p: 91-93)
• C89/C90 standard (ISO/IEC 9899:1990):
• 3.3.16 Assignment operators

[ edit ] See Also

Operator precedence

[ edit ] See also

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Home » Learn C Programming from Scratch » C Assignment Operators

C Assignment Operators

Summary : in this tutorial, you’ll learn about the C assignment operators and how to use them effectively.

Introduction to the C assignment operators

An assignment operator assigns the vale of the right-hand operand to the left-hand operand. The following example uses the assignment operator (=) to assign 1 to the counter variable:

After the assignmment, the counter variable holds the number 1.

The following example adds 1 to the counter and assign the result to the counter:

The = assignment operator is called a simple assignment operator. It assigns the value of the left operand to the right operand.

Besides the simple assignment operator, C supports compound assignment operators. A compound assignment operator performs the operation specified by the additional operator and then assigns the result to the left operand.

The following example uses a compound-assignment operator (+=):

The expression:

is equivalent to the following expression:

The following table illustrates the compound-assignment operators in C:

• A simple assignment operator assigns the value of the left operand to the right operand.
• A compound assignment operator performs the operation specified by the additional operator and then assigns the result to the left operand.

Print Cheatsheet

Mathmatical Symbols in C

C is able to perform basic mathematical operations on variables and values using the following symbols:

• Addition: +
• Subtraction: -
• Division: /
• Multiplication: *
• Incrementing: ++
• Decrementing: --

Assignment Operations in C

C can assign values to variables and perform basic mathematical operations using shorthand operators:

• Assignment: =
• Addition then assignment: +=
• Subtraction then assignment: -=
• Multiplication then assignment: *=
• Division then assignment: /=
• Modulo then assignment: %=

Comparing values in C

C can compare two values and/or variables against each other to return true or false. The operators are as follows:

• Do both sides have the same value? ==
• Do the two sides have different values? !=
• Is the left side a lower value than the right side? <
• Is the left side a lower or equal value to the right side? <=
• Is the left side a greater value than the right side? >
• Is the left side a greater or equal value to the right side? >=

Logical Operators in C

C can perform logical operations using the following operators:

• and: && (Are both sides true?)
• or: || (Is at least one side true?)
• not: ! (True becomes false and false becomes true.)

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Assignment operators

• 8 contributors

expression assignment-operator expression

assignment-operator : one of   =   *=   /=   %=   +=   -=   <<=   >>=   &=   ^=   |=

Assignment operators store a value in the object specified by the left operand. There are two kinds of assignment operations:

simple assignment , in which the value of the second operand is stored in the object specified by the first operand.

compound assignment , in which an arithmetic, shift, or bitwise operation is performed before storing the result.

All assignment operators in the following table except the = operator are compound assignment operators.

Assignment operators table

Operator keywords

Three of the compound assignment operators have keyword equivalents. They are:

C++ specifies these operator keywords as alternative spellings for the compound assignment operators. In C, the alternative spellings are provided as macros in the <iso646.h> header. In C++, the alternative spellings are keywords; use of <iso646.h> or the C++ equivalent <ciso646> is deprecated. In Microsoft C++, the /permissive- or /Za compiler option is required to enable the alternative spelling.

Simple assignment

The simple assignment operator ( = ) causes the value of the second operand to be stored in the object specified by the first operand. If both objects are of arithmetic types, the right operand is converted to the type of the left, before storing the value.

Objects of const and volatile types can be assigned to l-values of types that are only volatile , or that aren't const or volatile .

Assignment to objects of class type ( struct , union , and class types) is performed by a function named operator= . The default behavior of this operator function is to perform a member-wise copy assignment of the object's non-static data members and direct base classes; however, this behavior can be modified using overloaded operators. For more information, see Operator overloading . Class types can also have copy assignment and move assignment operators. For more information, see Copy constructors and copy assignment operators and Move constructors and move assignment operators .

An object of any unambiguously derived class from a given base class can be assigned to an object of the base class. The reverse isn't true because there's an implicit conversion from derived class to base class, but not from base class to derived class. For example:

Assignments to reference types behave as if the assignment were being made to the object to which the reference points.

For class-type objects, assignment is different from initialization. To illustrate how different assignment and initialization can be, consider the code

The preceding code shows an initializer; it calls the constructor for UserType2 that takes an argument of type UserType1 . Given the code

the assignment statement

can have one of the following effects:

Call the function operator= for UserType2 , provided operator= is provided with a UserType1 argument.

Call the explicit conversion function UserType1::operator UserType2 , if such a function exists.

Call a constructor UserType2::UserType2 , provided such a constructor exists, that takes a UserType1 argument and copies the result.

Compound assignment

The compound assignment operators are shown in the Assignment operators table . These operators have the form e1 op = e2 , where e1 is a non- const modifiable l-value and e2 is:

an arithmetic type

a pointer, if op is + or -

a type for which there exists a matching operator *op*= overload for the type of e1

The built-in e1 op = e2 form behaves as e1 = e1 op e2 , but e1 is evaluated only once.

Compound assignment to an enumerated type generates an error message. If the left operand is of a pointer type, the right operand must be of a pointer type, or it must be a constant expression that evaluates to 0. When the left operand is of an integral type, the right operand must not be of a pointer type.

Result of built-in assignment operators

The built-in assignment operators return the value of the object specified by the left operand after the assignment (and the arithmetic/logical operation in the case of compound assignment operators). The resultant type is the type of the left operand. The result of an assignment expression is always an l-value. These operators have right-to-left associativity. The left operand must be a modifiable l-value.

In ANSI C, the result of an assignment expression isn't an l-value. That means the legal C++ expression (a += b) += c isn't allowed in C.

Expressions with binary operators C++ built-in operators, precedence, and associativity C assignment operators

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• ❯ Multiplication Assignment

C++ Multiplication Assignment (*=) Operator

In this C++ tutorial, you shall learn about Multiplication Assignment operator, its syntax, and how to use this operator, with examples.

C++ Multiplication Assignment

In C++, Multiplication Assignment Operator is used to find the product of the value (right operand) and this variable (left operand) and assign the result back to this variable (left operand).

The syntax to find the product of a value 2 with variable x and assign the result to x using Multiplication Assignment Operator is

In the following example, we take a variable x with an initial value of 5 , multiply it with a value of 2 and assign the result back to x , using Multiplication Assignment Operator.

In this C++ Tutorial , we learned about Multiplication Assignment Operator in C++, with examples.

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Left Shift and Right Shift Operators in C/C++

In C/C++, left shift (<<) and right shift (>>) operators are binary bitwise operators that are used to shift the bits either left or right of the first operand by the number of positions specified by the second operand allowing efficient data manipulation. In this article, we will learn about the left shift and right shift operators.

Left Shift (<<) Operators

The left shift(<<) is a binary operator that takes two numbers, left shifts the bits of the first operand, and the second operand decides the number of places to shift. In other words, left-shifting an integer “ a ” with an integer “ b ” denoted as ‘ (a<<b)’ is equivalent to multiplying a with 2^b (2 raised to power b).  

• a is the integer value to be shifted.
• b specifies how many positions to shift the bits.

Example: Let’s take a=5 ; which is 101 in Binary Form. Now, if “ a is left-shifted by 2 ” i.e a=a<<2 then a will become a=a*(2^2) . Thus, a=5*(2^2)=20 which can be written as 10100.

Example of Left Shift Operator

Applications of left shift operator.

• Multiplication by Powers of Two : Left shifting a number by n positions is equivalent to multiplying it by 2^n and is much faster than normal multiplication
• Efficient Calculations : Used in performance-critical applications where arithmetic operations need to be fast.
• Bit Manipulation : Common in low-level programming, such as embedded systems and hardware interfacing.

Right Shift(>>) Operators

Right Shift(>>) is a binary operator that takes two numbers, right shifts the bits of the first operand, and the second operand decides the number of places to shift. In other words, right-shifting an integer “ a ” with an integer “ b ” denoted as ‘ (a>>b) ‘ is equivalent to dividing a with 2^b. 

Example: let’s take a=5 ; which is 101 in Binary Form. Now, if “ a is right-shifted by 2 ” i.e a=a>>2 then a will become a=a/(2^2) . Thus, a=a/(2^2)=1 which can be written as 01 .

Example of Right Shift Operator

Applications of right shift operators.

• Division by Powers of Two : Right shifting a number by n positions is equivalent to dividing it by 2^n and it is very fast.
• Efficient Calculations : Used in performance-critical applications for fast division operations.
• Bit Manipulation : Useful in extracting specific bits from data, common in data compression and cryptography.

Important Points of Shift Operators

1. The left-shift and right-shift operators should not be used for negative numbers. The result of is undefined behavior if any of the operands is a negative number. For example, results of both 1 >> -1 and 1 << -1 is undefined.

2. If the number is shifted more than the size of the integer, the behavior is undefined. For example, 1 << 33 is undefined if integers are stored using 32 bits. For bit shift of larger values 1ULL<<62   ULL is used for Unsigned Long Long which is defined using 64 bits that can store large values.

3. The left-shift by 1 and right-shift by 1 are equivalent to the product of the first term and 2 to the power given element(1<<3 = 1*pow(2,3)) and division of the first term and second term raised to power 2 (1>>3 = 1/pow(2,3)) respectively. 

Must Read: Bitwise Operators in C/C++

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Multiplication Assignment operator doesn't honor mathematical precedence

For example:

The first one I would have hoped would have worked correctly, and the compiler would have done the * first, but it didn’t. This is expected, because the * is on the left side to the assignment operator. Is there any way I can create another operator that operates like the ++ and – operators, kinda like this…

Operation and =* that would have the * on the right hand side, thus being a clear indicator that the * before the assignment operator is reached. I am just not sure how to break the rhs variable into pieces BEFORE it is evaluated.

Any ideas would be helpful.

• I am pretty sure I can do it in .NET Assemby, but that will just get overwritten each time I recompile. –  Bugs Bunny Commented Dec 14, 2018 at 1:15
• interesting question, on the docs it is stated the operator has higher precedence than the next section and lower precedence than the previous section (right-to-left). which means the addition will be done before the multiplication followed by the assignment - i = i * (15 + 12) . –  Bagus Tesa Commented Dec 14, 2018 at 1:26
• Yes, but the right hand side is evaluated with correct rules of math operator precedence, then that result is multiplied by the value of the left and side and then assigned to the value, which is not correct. Multiplication takes higher precedence than addition. It’s a flaw, but only kind of a flaw. If they created behavior like =* then it would make a lot more sense. Kind of like the difference between ++variable and variable++ (aka early increment or late increment). –  Bugs Bunny Commented Dec 14, 2018 at 1:31
• You are making an incorrect application of operator precedence: the opreator * is applied before the operator + in an expression, but the operator = and the operator *= are applied after the basic math operators + , - , * , / , hence the rhs is evaluated, then *= uses that. That is the definition in C# - operator evaluation is not a textual substitution (like a C pre-processor macro). Note that operator *= and operator * are not the same operator. –  NetMage Commented Dec 14, 2018 at 21:43
• If *= had higher precedence, then the expression is (value *= 15) + 12; so what do you think the +12 would mean? –  NetMage Commented Dec 14, 2018 at 21:45

2 Answers 2

Your code doesn't make sense if you're hoping to change the precendence of the *= operator.

If I started with this (ignoring the fact that it's not valid C#):

... and then had this code:

... the call to =* would set lhs as 10 , but it would have to set rhs as the result of 15 + 12 .

And that would simply be the same behaviour of the built-in *= operator.

• It’s pseudo code, and I wrote in it an email to Jon Skeet, not studio. I think you might understand what I want to do, but only maybe. I tested it in studio but wanted to express the issue more than copy my code from studio. –  Bugs Bunny Commented Dec 14, 2018 at 1:38
• 1 @BugsBunny - Please show real code whenever possible. Expressing ideas with pseudo code is often misleading. –  Enigmativity Commented Dec 14, 2018 at 1:39
• While completely correct, this might be better as a comment? :) –  Josh Noe Commented Dec 14, 2018 at 1:52
• 1 @Enigmativity Because this doesn't answer the question. It only comments on a mistake made in the question. I could be misunderstanding. –  Josh Noe Commented Dec 14, 2018 at 14:39
• 1 @JoshNoe - I was trying to illustrate that to pass a single value (i.e. Int32 rhs ) that the compiler must evaluate before calling the overload. There's no way for the compiler to get around this. You'd have to modify the compiler to get this functionality to work the way that the OP requested it. –  Enigmativity Commented Dec 14, 2018 at 22:17

Edit : Based on further comments by OP in the comment section of question, the operator was probably created to make it easier to write the aggregate operations : sum = sum + value , and so does not follow the normal mathematical rules, it is probably more useful then what you propose.

C# you cannot create new operator. The existing operators are built into the language, so to add any new one's at a minimum you would also want to implement some parts in compiler to parse them.

C# does allow operator overloading, but then it leads to defining your own integers and I don't think that is a happy road to go down.

There are languages other then C# which do allow this (notably functional languages where operators are just normal methods).

From C# language specifications

Certain operators can be overloaded. Operator overloading permits user-defined operator implementations to be specified for operations where one or both of the operands are of a user-defined class or struct type (Operator overloading).

• I can do it in .NET Assembly I am sure, but fix would be overwritten each compile. This is a perfect reason why all .NET Framework class should be labeled partial. LOL –  Bugs Bunny Commented Dec 14, 2018 at 1:35
• That's the reason I mentioned that it would involve rewrite of some parts of compiler –  peeyush singh Commented Dec 14, 2018 at 1:37
• Of course they can. I want to create a NEW operator that evaluates correctly with the rules of math operator precedence, not overload an existing one. –  Bugs Bunny Commented Dec 14, 2018 at 1:39
• Agreed, and I can do that, but was hoping there was an easier way that one of you could think of. –  Bugs Bunny Commented Dec 14, 2018 at 1:40
• Maybe I can create a post compile event that adds then assembly into the IL. That’s all I can think. –  Bugs Bunny Commented Dec 14, 2018 at 1:52

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