commas in C - can you pass the puzzle ?

In a recent Deep-C training course in Bangalore I was discussing sequence points with some C programmers. I explained that the comma operator is one the very few operators that creates a sequence point. I like the comma. It's the name of a beautiful butterfly. K&R's famous Hello World program had a comma between hello and world. I should really put a comma into my blog picture!

During several cyber-dojos it became clear to me that many of the programmers (despite having programmed in C for several years), did not understand that in C, not all commas are the same. I've created a small piece of C code to try help C programmers understand the humble comma...

Have a look at the following 5 lines of code. Do you know what each line does?

int x = (1,2,3);
int x =  1,2,3;

   x =   1,2,3;
   x =  (1,2,3);
   x = f(1,2,3); 

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Scroll down for my answers once you've decided...
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int x = (1,2,3);

This declares an int called x and initializes it to the result of the expression (1,2,3). The commas inside this expression are operators. 1 is evaluated and its value (1) discarded, then a comma provides a sequence point, then 2 is evaluated and its value (2) discarded, then a comma provides a sequence point, then 3 is evaluted and its value is the value of the expression (1,2,3). So x is initialized to 3. You'll probably get warnings saying there are no side-effects in the expressions 1 and 2.

int x =  1,2,3;

This is different. If this compiled it would declare an int called x and initialize it to 1 and then declare two more ints called 2 and 3. It has the same structure as int x = 1,y,z; which declares three ints called x, y, and z. The commas are not operators, they are punctuators/separators. You can't declare variables called 2 or 3. It does not compile.

   x =   1,2,3;

In this fragment x is assumed to have already been declared. It is not a declaration. The commas are operators again. Assignment has higher precedence than the comma operator so this binds as (x = 1),2,3;. So 1 is assigned to x, and the result of this assignment expression (1) is discarded, then there is a sequence point, then 2 is evaluated and its value (2) is discarded, then there is a sequence point, then 3 is evaluated and its value (3) is discarded. You'll probaby get warnings saying there are no side-effects in the expressions 2 and 3.

   x =  (1,2,3);

Again, x is assumed to have already been declared. It is not a declaration. The commas are operators again. This is the same as the first fragment except it is not a declaration. x is assigned the value of the expression (1,2,3) which is 3. Again you'll probably get warnings saying there are no side-effects in the expressions 1 and 2.

   x = f(1,2,3);

Again, x is assumed to have already been declared. As has a function called f which accepts three int arguments. These commas are not operators. They are punctuators/separators. They separate the three expressions forming the three arguments to f. But they do not introduce any sequence points.

How did you do?