The following program does not always lead to deadlock:

$/*$ thread$\_$one runs in this function

void $*$do$\_$work$\_$one$($void $*$param$)\{$

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

$/*$ do some work $*/$

pthread$\_$mutex$\_$unlock$($&second$\_$mutex$)$;

pthread$\_$mutex$\_$unlock$($&first$\_$mutex$)$;

$\}$

$/*$ thread$\_$two runs in this function $*/$

void $*$do$\_$work$\_$two$($void $*$param$)\{$

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

$/*$ do some work $*/$

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

pthread$\_$exit$()$;

$\}$

In which scenario would deadlock occur?

The following program does not always lead to deadlock:

$/*$ thread$\_$one runs in this function

void $*$do$\_$work$\_$one$($void $*$param$)\{$

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

$/*$ do some work $*/$

pthread$\_$mutex$\_$unlock$($&second$\_$mutex$)$;

pthread$\_$mutex$\_$unlock$($&first$\_$mutex$)$;

$\}$

$/*$ thread$\_$two runs in this function $*/$

void $*$do$\_$work$\_$two$($void $*$param$)\{$

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

$/*$ do some work $*/$

pthread$\_$mutex$\_$lock$($&first$\_$mutex$)$;

pthread$\_$mutex$\_$lock$($&second$\_$mutex$)$;

pthread$\_$exit$()$;

$\}$

In which scenario would deadlock occur?

If thread$\_$one acquires both locks before thread$\_$two.

If thread$\_$two acquire both locks before thread$\_$one.

If one thread is able to acquire only one lock before the other thread acquires the second lock.

All of the above