'How does the Comma Operator work

How does the comma operator work in C++?

For instance, if I do:

a = b, c;

Does a end up equaling b or c?

(Yes, I know this is easy to test - just documenting on here for someone to find the answer quickly.)

Update: This question has exposed a nuance when using the comma operator. Just to document this:

a = b, c;    // a is set to the value of b!

a = (b, c);  // a is set to the value of c!

This question was actually inspired by a typo in code. What was intended to be

a = b;
c = d;

Turned into

a = b,    //  <-  Note comma typo!
c = d;
c++comma-operator


Solution 1:[1]

It would be equal to b.

The comma operator has a lower precedence than assignment.

Solution 2:[2]

Take care to notice that the comma operator may be overloaded in C++. The actual behaviour may thus be very different from the one expected.

As an example, Boost.Spirit uses the comma operator quite cleverly to implement list initializers for symbol tables. Thus, it makes the following syntax possible and meaningful:

keywords = "and", "or", "not", "xor";

Notice that due to operator precedence, the code is (intentionally!) identical to

(((keywords = "and"), "or"), "not"), "xor";

That is, the first operator called is keywords.operator =("and") which returns a proxy object on which the remaining operator,s are invoked:

keywords.operator =("and").operator ,("or").operator ,("not").operator ,("xor");

Solution 3:[3]

The comma operator has the lowest precedence of all C/C++ operators. Therefore it's always the last one to bind to an expression, meaning this:

a = b, c;

is equivalent to:

(a = b), c;

Another interesting fact is that the comma operator introduces a sequence point. This means that the expression:

a+b, c(), d

is guaranteed to have its three subexpressions (a+b, c() and d) evaluated in order. This is significant if they have side-effects. Normally compilers are allowed to evaluate subexpressions in whatever order they find fit; for example, in a function call:

someFunc(arg1, arg2, arg3)

arguments can be evaluated in an arbitrary order. Note that the commas in the function call are not operators; they are separators.

Solution 4:[4]

The comma operator:

  • has the lowest precedence
  • is left-associative

A default version of comma operator is defined for all types (built-in and custom), and it works as follows - given exprA , exprB:

  • exprA is evaluated
  • the result of exprA is ignored
  • exprB is evaluated
  • the result of exprB is returned as the result of the whole expression

With most operators, the compiler is allowed to choose the order of execution and it is even required to skip the execution whatsoever if it does not affect the final result (e.g. false && foo() will skip the call to foo). This is however not the case for comma operator and the above steps will always happen*.

In practice, the default comma operator works almost the same way as a semicolon. The difference is that two expressions separated by a semicolon form two separate statements, while comma-separation keeps all as a single expression. This is why comma operator is sometimes used in the following scenarios:

  • C syntax requires an single expression, not a statement. e.g. in if( HERE )
  • C syntax requires a single statement, not more, e.g. in the initialization of the for loop for ( HERE ; ; )
  • When you want to skip curly braces and keep a single statement: if (foo) HERE ; (please don't do that, it's really ugly!)

When a statement is not an expression, semicolon cannot be replaced by a comma. For example these are disallowed:

  • (foo, if (foo) bar) (if is not an expression)
  • int x, int y (variable declaration is not an expression)

In your case we have:

  • a=b, c;, equivalent to a=b; c;, assuming that a is of type that does not overload the comma operator.
  • a = b, c = d; equivalent to a=b; c=d;, assuming that a is of type that does not overload the comma operator.

Do note that not every comma is actually a comma operator. Some commas which have a completely different meaning:

  • int a, b; --- variable declaration list is comma separated, but these are not comma operators
  • int a=5, b=3; --- this is also a comma separated variable declaration list
  • foo(x,y) --- comma-separated argument list. In fact, x and y can be evaluated in any order!
  • FOO(x,y) --- comma-separated macro argument list
  • foo<a,b> --- comma-separated template argument list
  • int foo(int a, int b) --- comma-separated parameter list
  • Foo::Foo() : a(5), b(3) {} --- comma-separated initializer list in a class constructor

* This is not entirely true if you apply optimizations. If the compiler recognizes that certain piece of code has absolutely no impact on the rest, it will remove the unnecessary statements.

Further reading: http://en.wikipedia.org/wiki/Comma_operator

Solution 5:[5]

The value of a will be b, but the value of the expression will be c. That is, in

d = (a = b, c);

a would be equal to b, and d would be equal to c.

Solution 6:[6]

b's value will be assigned to a. Nothing will happen to c

Solution 7:[7]

Yes Comma operator has low precedence than Assignment operator

#include<stdio.h>
int main()
{
          int i;
          i = (1,2,3);
          printf("i:%d\n",i);
          return 0;
}

Output : i=3
Because comma operator always return rightmost value.
In case of comma operator with Assignment Operator:

 int main()
{
      int i;
      i = 1,2,3;
      printf("i:%d\n",i);
      return 0;
}

Ouput: i=1
As we know comma operator has lower precedence than assignment.....

Solution 8:[8]

The value of a will be equal to b, since the comma operator has a lower precedence than the assignment operator.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 msanford
Solution 2
Solution 3 efotinis
Solution 4
Solution 5 H.S.
Solution 6 prakash
Solution 7 Roopam
Solution 8 Lightness Races in Orbit

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