How does the comma acts here as an operator or as a separator?

[radha gogia]

plz explain the working of below code ,here why both the printf functions are getting executed when comma is here acting like an operator and not like a separator.
‪

#‎include‬<stdio.h>
int main()
{ 
int i;
int g= i=8?(printf("hello"),printf("life")):printf("kill");
printf("%d",g);
return 0;
}

[weaknessforcats]

You have stumbled into the C "ternary operator" ? , :

It works like this:

? (condition), do if true : do if false.

So let's say we will test x and y. If x is less than y, print "LESS". If x is not less than y print "NOT LESS".

You could write:

if (x < y)
{
   printf("LESS");
}
else
{
   printf("NOT LESS");
}

Or the same logic could use the ternary operator:

?(x < y),  printf("LESS"): printf("NOT LESS");

[radha gogia]

Sir,I am asking about actually how the comma is acting here as an operator or as a separator,because if I write int a=(7,8) a gets the value 8 ,so then why here both the printf statements are getting printed.
Even When i write
if(10>9)
{
printf("kill"),printf("heloo");
}

so basically how this code works?

[radha gogia]

When comma used inside a paranthesis then the last argument gets executes ,so the last printf statement should execute so then why here both the printf statements get executed

[donbock]

A common way to write a for loop is

for (i=0; i<10; i++)
  dosomething();

A not uncommon idiom is

for (i=0,j=0; i<10; i++)
   dosomething();

Note the comma operator between i=0 and j=0. The effect of this operator is "first do the thing to my left, then do the thing to my right"; so both i and j get set to 0 before the loop is entered.

The comma operator isn't used much, which leads to confusion when it is used, which leads people who don't want their code to be confusing to avoid using it, and so it goes.

The comma operator introduces a sequence point. This means that the compiler is obligated to execute the arguments from left to right. The commas in a function call are separators, so not sequence points, so no similar assurance of the order that function arguments are evaluated. For

dosomething(j++, abc(j,k), --k)

you can't know what order the increment, abc() call, or decrement take place. Even if you instrument your code so that you think you know the order, you can't count on the order being the same the next time this snippet executes.

If this seems confusing to you then you're very perceptive.

[donbock]

Line 5 of the OP set i to 8, compares 8 to 0, finds that. 8 does not equal 0 (that is, that 8 is true), and prints "hello" and "world".

It only prints "kill" if 8 is equal to 0 (that is, if 8 is false ).

[weaknessforcats]

In both C and C++ the same symbol is used for multiple purposes. This is called operator overloading and what is it means is that you must use context to determine what the meaning is.

For example:

&

1) this is the address-of operator
2) this is the bitwise AND operator
3) in C++ this is the reference operator
4) this s part of the logical AND operator

These uses are mutually exclusive. So it requires the compile to see how you use the operator in order to determine how to compile it.

The same is true with the comma.

1) this is the sequence operator
2) this is the TRUE portion of the ternary operator.

So this example:

if(10>9)
{
printf("kill"),printf("heloo");
}
 

is examined by the compiler and it does not find the ? of the ternary operator making the comma the sequence operator. Hence, the printf calls are done one after the other.