How it's happening

Question

#include<stdio.h>
int main()
{
int a;
a=(20,30,40);
printf( "%d",a);
}
Output 40
----------------
Vs
#include<stdio.h>
int main()
{
int a;
a=20,30,40;
printf( "%d",a);
}
Output 20
------------------
And other case

#include<stdio.h>
int main()
{

int a=(20,30,40);
printf( "%d",a);
}
Output: error
----------------
Can anybody clearly explain.

Answer 1

Your first and last Question looking same it have no changes in first Question
Int a;
a=(20,30,40); alway last value will be print if u put bracket around the values because it behave like stack it all values will be assign to a means a=20 first then a=30 then a=40 and c language is procedural language so last value will be 40 .
But in second Question u wrote
Int a;
int a=20,30,40; here u assign 20 to a variable but all value won't be assign in a here u should assign like this
Int a=20,b=30,c=40;

int a=20,30,40; this is wrong way to define random values to single variable .May be Some Compilers will give error or warnings if you will write this . But some Compiler will give output as a 20 its depend on Compiler .

But if you dont want to use more variables then you can use array in this case .
Hope you understood
........Thankyou......

Answer 2

The error received when running the code expalins the problem (after fixing a other case errors, remember that C and C++ are case sensitive):

40

./Playground/file0.cpp: In function 'int main()':
./Playground/file0.cpp:7:4: warning: left operand of comma operator has no effect [-Wunused-value]
7 | a=(20,30,40);
| ^~
./Playground/file0.cpp:7:10: warning: right operand of comma operator has no effect [-Wunused-value]
7 | a=(20,30,40);
| ^~

Answer 3

When you assign like that it is same as a=20, a=30, a=40.. So 20,30 will be replaced by 40..
So 20,30 will have no effect.
So final value it contains, a = 40.