Can someone explain output stepwise?

Question

#include <stdio.h>

int main() {

int a=(1, 2,3);
int b=(++a, ++a, ++a) ;
int c=(b++, b++, b++) ;
printf ("%d, %d, %d", a, b, c) ; /* output is 6,9,8 */
return 0;
}

Answer 1

**Comma operator**
The comma operator expression has the form

lhs , rhs
where

lhs - any expression
rhs - any expression other than another comma operator (in other words, comma operator's associativity is left-to-right)
First, the left operand, lhs, is evaluated and its result value is discarded.

Then, a sequence point takes place, so that all side effects of lhs are complete.

Then, the right operand, rhs, is evaluated and its result is returned by the comma operator as a non-lvalue.
~~~~~~~~

int a = ((1, 2),3); // a = 3
int b = ((++a, ++a), ++a) ; // a = 6, b = 6
int c = ((b++, b++), b++) ; // b = 9, c = 8

____
https://en.cppreference.com/w/c/language/operator_other#Comma_operator

Answer 2

"Else the assignment operator (=) would take precedence"

In the case of a declarator the compiler issues an error

int n = 2,3; // error, comma assumed to begin the next declarator

Answer 3

C++ Soldier (Babak) true! Assignment takes precedence, and that's why it throws the error. We won't get any value. I was careless. :)

Answer 4

A variable name cannot start with a number lol

Answer 5

Also note that the parentheses are important. Else the assignment operator (=) would take precedence, and the values would be different.