Diferencia entre revisiones de «Solución ejercicio 4»
De Wiki de Sistemas Operativos
(Página nueva: p1:<source lang="c"> for (int i=0; i<2; i++) { #1 send(msg); #2 recv(msg); #3 }</source> P2:<source lang="c"> for (int i=0; i<2; i++) { #1 recv...) |
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| + | Otra posible solución: | ||
| + | |||
| + | X = El proceso pasa a estado bloqueado. | ||
| + | / = El proceso pasa a estado preparado. | ||
| + | > = Fin de su ejecucion | ||
| + | |||
| + | |||
| + | | #1| #2| | | | #3| #1| #2| | | | #3| #1|FIN| | | | | | | | ||
| + | p1|---|---X | / |---|---|---X | | |---|---> | | | | | | | | ||
| + | | | | | | | | | | | | | | | | | | | | | | | ||
| + | | | | #1| #2| #3| | | | #1| #2| #3| | | #1|FIN| | | | | | | ||
| + | p2| | |---|---|---X / | |---|---|---X / |---> | | | | | | | ||
| + | | | | | | | | | | | | | | | | | | | | | | | ||
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Revisión del 17:11 9 may 2011
p1: for (int i=0; i<2; i++) { #1
send(msg); #2
recv(msg); #3
} for (int i=0; i<2; i++) { #1
recv(msg); #2
send(msg); #3
} /: bloqueado, @: preparado, >: fin del programa, X: se ejecuta el planificador
|i1=0 |i2=0 | i1=1| i2=1| i1=2|i2=2|
| #1| #2| | | #3| | #1| #2| | | #3| | #1| |
P1 |---|---/ | @---/ @---|---/ | @---/ @---> |
| | | | | | | | | | | | | | |
| | | #1| #2| | #3| | | #1| #2| | #3| | #1|
P2 | | |---|---/ @---/ | @---|---/ @---/ |--->
planif X | X | X X X X X | X X X X X
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Otra posible solución:
X = El proceso pasa a estado bloqueado.
/ = El proceso pasa a estado preparado.
> = Fin de su ejecucion
| #1| #2| | | | #3| #1| #2| | | | #3| #1|FIN| | | | | | |
p1|---|---X | / |---|---|---X | | |---|---> | | | | | | |
| | | | | | | | | | | | | | | | | | | | |
| | | #1| #2| #3| | | | #1| #2| #3| | | #1|FIN| | | | | |
p2| | |---|---|---X / | |---|---|---X / |---> | | | | | |
| | | | | | | | | | | | | | | | | | | | |
|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
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