Mechanism of the reaction of an alkene with excess HI in presence of CCl4
According to [1], 1,2-diiodopropane reacts with hydrogen iodide to give 2-iodopropane.

So the reaction path might be

$$\ce{CH2=CH\bond{-}CH2\bond{-}I + HI -&gt; CH3\bond{-}CHI\bond{-}CH2\bond{-}I}$$

$$\ce{CH3\bond{-}CHI\bond{-}CH2\bond{-}I + HI -&gt; CH3\bond{-}CHI\bond{-}CH3 + I2}$$

The mechanism of the second part is unclear to me. It may be elimination of $\ce{I2}$ followed by addition of $\ce{HI}$.

1 Th. M. Schmitt, Analysis of Surfactants, Second Edition, p. 67