How does an arithmetic bit-shift affect a signed integer?

I am trying to figure out how exactly arithmetic bit-shift operators work in C, and how it will affect signed 32-bit integers.

To make things simple, let's say we work within one byte (8 bits):

x = 1101.0101
MSB[ 1101.0101 ]LSB

Reading other posts on Stack Overflow and some websites, I found that: << will shift toward MSB (to the left, in my case), and fill "empty" LSB bits with 0s.

And >> will shift toward LSB (to the right, in my case) and fill "empty" bits with MS bit

So, x = x << 7 will result in moving LSB to MSB, and setting everything to 0s.

1000.0000

Now, let's say I would >> 7, last result. This would result in [0000.0010]? Am I right?

Am I right about my assumptions about shift operators?

I just tested on my machine, **

int x = 1;   //000000000......01

x = x << 31; //100000000......00

x = x >> 31; //111111111......11 (Everything is filled with 1s !!!!!) 

Why?