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* C- Z. @. x& s$ m) G `
process(q(14))
+ C8 q; R: w& k begin5 G9 I7 d2 `8 o+ }9 e, n6 r9 `! s
if(q(14)'event and q(14)='1') then
( _; k6 q0 \ H+ Z; Y if second<"111011" then3 h& [4 w; ?* Z" n3 D
second <= second+1;
/ a6 M% P- w) `2 E else
/ Y( p6 R9 V! P* A# V second <="000000";, \) h. S4 o6 }0 {! n8 w4 }
if minl < "1001" then4 {* M9 ^* D4 B! ?: V
minl <= minl+1; Y6 o# a! C. y4 z5 i
else
* n2 `% {9 {! V' u/ Q minl <= "0000";
1 z+ _5 A# K; t% w* P; k6 k if minh < "101" then
+ t+ S) |' L* q/ Z% h3 e s minh <= minh+1; y1 Q5 V4 k U% c( [
else2 n" p) N8 Z7 @# W0 B+ V
minh <="000";
* g- I8 X V# A) H if hourh <"10" then/ J. |3 ]5 H! v) G/ N0 ?# o- Q
if hourl < "1001" then
: o, `; m+ y6 B hourl <= hourl+1;$ h# q+ N+ n- f1 ~2 _, h
else
2 w& O1 P' n; Z4 A1 A0 r hourl <= "0000";
( G* a1 D( Y6 k hourh <= hourh+1;
$ }1 Y/ \2 C7 B4 {+ d. Q$ ^1 L end if;
1 X. ], j8 f8 G7 `& \ else if hourh = "10" then " K5 p; L& h' \0 \: o3 P" [
if hourl <"0011" then
2 e: K* M6 _2 q9 z$ d; F4 O hourl <= hourl+1;1 J3 \; ?- }+ z+ P( L3 Z
else
0 g0 d+ g9 C# D3 z hourl <= "0000";* d- I6 Y }3 _% a+ s x" k
hourh <= "00";8 @3 o3 D! D* k
end if;7 ]* o, q! ~0 U
end if;" H- K) X v( r* b# t& A8 h
end if;
7 d* p+ f J1 R0 x/ N# c end if;
% x2 ?! q% g' J, {! t9 U: r$ c0 I end if;7 X! K- a/ s; s0 Z- \
end if;: ?& \# p" M3 b* D- |' d
end if;
0 Z" a8 Y/ a# t end process;: a9 ]. ?6 b2 ^+ O& y5 ^$ d
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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发表于 2019-4-9 10:53
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淘帖
支持！
反对！