Why the value of X is not incremented while assignment - c

Question

I'm new to C language and I'm not entirely sure why on Line 4 when post-increment is done the value of x isn't change what I mean is that
x = printf("%d",x++);
The value for x was 12 so the printf would print 12 and then x should be assigned 2 and while ++ was there x should be later changed with 2+1 and on line 6 pre-increment is done so output shouldn't be 124.
Why x on line 4 isn't added?
Please help.
#include <stdio.h>
int main(){
int x = 12;
x = printf("%d", x++);
printf("%d", ++x);
return 0;
}

Answer 1

Make yourself aware of sequence point. From this [emphasis mine]:
There is a sequence point after the evaluation of all function arguments and of the function designator, and before the actual function call.
From this [emphasis mine]:
Increment operators initiate the side-effect of adding the value 1 of appropriate type to the operand. Decrement operators initiate the side-effect of subtracting the value 1 of appropriate type from the operand. As with any other side-effects, these operations complete at or before the next sequence point.
Looks at this statement:
x = printf("%d", x++);
The post increment operator increase the value of operand by 1 but the value of the expression is the operand's original value prior to the increment operation.
So, the value of x passed to printf() will be its original value which is 12 and due to sequence point, before calling printf() the value of x will be incremented by 1. The return value of printf() will be assigned to x which overwrites the last value of x which is the incremented value due to post ++ operator. Hence, after this statement the value of x is 2.

Answer 2

The value for x was 12 so the printf would print 12 and then x should be assigned 2 and while ++ was there x should be later changed with 2+1 and on line 6 pre-increment is done so output shouldn't be 124.
no, the assignment is done after all concerning printf("%d", ++x);, your code is equivalent to that :
#include <stdio.h>
int main(){
int x = 12;
int y = printf("%d", x++);
x = y;
printf("%d", ++x);
return 0;
}
so x = printf("%d", ++x); does :
printf writes 12
then x is incremented to value 13
then x is assigned to the result of printf valuing 2
then you execute printf("%d", ++x); while x values 2 before, so x is incremented before to be given in argument, so 3 is written
and the final print result is 123
PS. As said by #H.S. in an other remark :There is a sequence point after the evaluation of all function arguments (x++ is an argument to printf()) and the pre/post increment/decrement operation complete at or before the next sequence point.

Answer 3

This will work for you:
++x is prefix increment and the x++ is postfix increment, and here is the difference on both of them:
The prefix increment returns the value of a variable after it has been incremented.
On the other hand, the more commonly used postfix increment returns the value of a variable before it has been incremented.
#include <stdio.h>
int main()
{
int x = 12;
x = printf("%d", ++x);
printf("%d", ++x);
return 0;
}