doubt in c : Declaring variables in one line?

Question

why should we not declare a variables like below?

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

[ int a, b, c = a+b; ]

float salary = 56.23;
char letter = 'Z';
a = 8;
b = 34;

printf("%d \n", c);
printf("%f \n", salary);
printf("%c \n", letter);

return 0;
}

Output gives wrong ans like:

0 [ though ans is 42 ]
56.2300
Z

Correct code:

#include <stdio.h>
int main() {
int a, b;
float salary = 56.23;
char letter = 'Z';
a = 8;
b = 34;
int c = a+b;

printf("%d \n", c);
printf("%f \n", salary);
printf("%c \n", letter);

return 0;
}

Output:
42
56.2300
Z

Answer 1

This is because at the time of assigning the values of "a+b" to variable "c", there are no values(thus a garbage value) in variables "a" and "b" in the first code

whereas in the second one values of "a" and "b" were initialised before the statement "c=a+b"

Answer 2

Arsenic is right 👍

Answer 3

When you decleared variable c=a+b here you have not intilize the value of a and b so by default a and b are garbage or 0 and u adding both Values so result is 0 after that you initializing a=8 and b=34 it wont be add u already assigned a+b in c and c is procedural language so it wont be pass in a and b .

Answer 4

You mean about the line
int a, b, c =a+b;
If yes, then a, b are not yet assigned any values,
(it is not works by references here, it is by values)..
So default values a=0,b=0 are assigning c=a+b=0,.. But you may get garbage values aslo...
Assignment happening to a, b after c=a+b;
If you add this after a, b assignment then c value change as you want as you doing it in 2nd program...

C evaluation of expressions are top to bottom... Line by line.. Sequential....
Hope it help...