CODEWITHSUNDEEP
c++
westiti
westiti

Reputation: 35

Using assignment operator in the parameter of a function call

I am a beginner at c++ can anyone explain me this code:

#include <iostream>

void display(int b)
{
    std::cout << b << std::endl;
}

int main()
{
    int a;

    display(a=10);//display 10

    std::cout << a << std::endl;//also display 10

    return 0;
}

I know we can use = operator to set default values for a function parameters, but here it's in the function call, apparently "disply(a=10)" pass the value 10 to the function and store it in the variable "a" at the sametime.

is this correct coding in c++ and can anyone explain the assignment part?

Upvotes: 2

Views: 1867

Answers (5)

Cheers and hth. - Alf
Cheers and hth. - Alf

Reputation: 145269

The built-in assignment operator =

  • is right-associative
    which means it groups to right, e.g. a = b = c means a = (b = c),

  • is an lvalue expression that refers to the left hand side.

Note that in C an assignment produces a pure value, while in C++ it produces a reference (in the common language meaning of referring).

This means that you can assign a value to multiple variables:

a = b = c = 12345;

which is parsed as

a = (b = (c = 12345));

which first copies 12345 to c, then copies c to b, then copies b to a.

And it means that you can assign to the result of an assignment:

((a = b) = c) = 12345;

which first copies the b value to a, then copies the c value to a, then copies 12345 to a, leaving b and c unchanged…

In your case, the code

display(a=10);

is equivalent to

a = 10;  display( a );

Since the display function takes an int by value, this is equivalent to

display( 10 )

but if display had a reference argument then it could not be rewritten this way.

A common pitfall is to write

if( x = 12345 )

when one means to do a comparison,

if( x == 12345 )

Many compilers will warn about the first if the warning level is upped, as it should be.

More guaranteed ways to detect it include

  • Using const everywhere it can be used.
    x can’t be assigned to when it’s const. This is my preferred solution.

  • Writing if( 12345 == x ).
    Some people prefer this, but I find it hard to read, and as opposed to const it only helps to avoid the mis-typing when the writer is already, at that very point, very aware of the problem.

  • Defining a custom if construct via a macro.
    Technically this works, also for other constructs that use boolean conditions, but in order to be useful such a macro should be short, and this runs the risk of name collision. It's also hard on maintainers who are unfamiliar with the (effectively) custom language.

In C++03 the standard library required that any container element type should be assignable, and the assignable criterion required that a custom assignment operator T::operator= should return T& (C++03 §23.1/4) – which is also a requirement on the built-in assignment operator.

Until I learned that I used to define assignment operators with result type void, since I saw no point in supporting coding of expressions with side-effects (which is generally a bad practice) at the cost of both efficiency and verbosity.

Unfortunately this is a case where in C++ you pay for what you don’t use and generally should not use.

Upvotes: 1

ikh
ikh

Reputation: 10417

You need to know about = operator more. Not only is it assign rhs (right hand side) value to lhs (left hand side), but also it refers to the lhs.

Suppose this code:

a = b = c;

is exactly equal to

a = (b = c);

because = is right-associative.

If c is 10, the code assign 10 into b, and assign the value of b into a. So now a == b == c == 10.

Upvotes: 2

Mine
Mine

Reputation: 4241

The line

display(a=10);//display 10

equals to:

a = 10;
display(a);

This is because the value of the clause a = 10; is a.

I think this answers your question.

Upvotes: 3

prince
prince

Reputation: 1149

It is correct. 

 display(a=10); //It assigns 10 to a and 10 is passed as the parameter to function.

Upvotes: 0

Jan Spurny
Jan Spurny

Reputation: 5527

The assignment <variable> = <value> in C, C++ is and expression which means it have a value and this value is, of course, the <value> you've just assigned.

That's the reason why you can assign a value to multiple variables like this:

a = b = c = 1;

because internally it works something like this

a = value of (b = value of (c = 1));

and since the assignment does indeed have a value, the value of (c = 1) is 1, value of (b = (c = 1)) is 1 and therefore we get a = 1. And as a

If the assignment wouldn't be an expression and didn't have a value, we would get an error, because value of (c = 1) would not exist and we would get a syntax error.

So in your code, display(a=10); means: *set value a to 10 and pass the resulting value (which would be 10) as an argument to the function display.

Upvotes: 0

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