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2 Z, c! f4 s7 r) C# ^- aprocess(q(14))
8 l; L T6 K- J) ^& ~4 Y begin& w5 x/ z+ {4 J* U# d1 B1 [7 }
if(q(14)'event and q(14)='1') then
8 Z! l) z3 X6 L/ b1 y if second<"111011" then3 e* u: X2 J, o
second <= second+1;/ V0 Q4 |' S3 y" ?) r
else( H* n8 ~2 I- I6 @8 s/ y
second <="000000";
1 b* i+ C$ b a: u: F& }9 ]/ A- M if minl < "1001" then
) ^0 M7 D! }; t7 z( W s minl <= minl+1;
! U; y4 |3 _% S* h0 x" M else) s7 p4 c( J: X. e1 C7 ?: J/ S
minl <= "0000";
9 u3 v- U0 B8 b, ^1 P/ J if minh < "101" then
& N0 h; K8 {; u9 I/ A6 p- B& | minh <= minh+1;3 h* f* S2 }/ y! f# ~
else
. j; L7 j0 b' q2 a7 u minh <="000";# x u% e2 }6 Y B* H- \
if hourh <"10" then
9 r4 L' u. g) o# ^6 F: T if hourl < "1001" then
& T& X/ {4 S4 Z) j9 J6 X hourl <= hourl+1;
1 v! x5 n6 z; } else ( o/ \% C: L+ M4 _6 I8 V
hourl <= "0000";1 {: `. o" J6 t) k
hourh <= hourh+1;! L! D' d% B" E0 h4 T3 s. J. ~
end if;
0 w# {: E; w7 K# {, A else if hourh = "10" then
7 O. U; R4 C* B) [( Y if hourl <"0011" then) a0 I2 ?& V& Z6 T) I8 ~: B3 g
hourl <= hourl+1;
! L2 e4 P4 f! [" F$ z7 p* V0 C+ d else/ A$ v. N7 ?2 _. c
hourl <= "0000";( D9 v! u6 H! [$ _+ A/ h
hourh <= "00";& i% E0 I6 h! l* A' C0 B; o6 g
end if;; R3 L' b% T& y; q; R
end if;1 h& M( k8 v, B) Z6 G5 Z) ?, j2 T& o
end if;2 H4 }: H; P2 f V
end if;+ w, c4 e, h8 p
end if;4 }" b8 N" s" u+ S
end if;; i% u; P" }2 W8 I T, V+ \& G- ~
end if;
4 J4 g' \* |: `) n- { end process;' `6 ]; F* b7 e4 Y8 h
我想问一下怎样理解上面的if嵌套语句,后面结尾end if需要多少个,是怎样来的,哪位大神对这样的嵌套语句有好的理解方式的请不吝赐教?非常感谢! |
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