Need a little help in c

Question

I have 3 programs..

First one:

#include<stdio.h>
int main() {
int i = (5, 6, 7);
printf("%d", i);
return 0;
}

Answer to this question is 7

Second one:

#include<stdio.h>
int main() {
int i = 5, 6, 7;
printf("%d", i);
return 0;
}

In this one compilation fails...

Third one:

#include<stdio.h>
int main() {
int i;
i = 5, 6, 7;
printf("%d", i);
return 0;
}

In this answer is 5...

Why?? All three of these are almost identical, yet different results..??

Answer 1

Normal usage of comma would be:
int a=1,b=2,c=3; //defines 3 integers a, b, and c giving values of 1, 2, and 3 respectively
second usage is:
for (a=1,b=2;a<b;a++) f(a);//assigns 1 to a and 2 to b looping once

In program 1, the parentheses separate the 3 expressions from the assignment. The result is the last expression of 7 which gets assigned to i. You could have used any valid expressions such as function calls or assignments.
int i = (f(5),a=6,7); //also legal when f and a are defined prior to this statement

In program 2, you are attempting to declare 6 and 7 as integers, which is illegal. Besides example above, you could use:
int i=5,j,k=6,l,m=7; //Creates 5 integers

In program 3, you have 3 valid expressions i=5 does your assignment. The values of the other two aren't used.

Answer 2

2nd
you declared i=5 then you tried to declare 6 and 7 as variable and according to the rule you cannot declare number or special character as a variable

1st and 3rd i'm not sure what exactly happened