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Re: 結構型態&共用型態的問題？
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初級會員
 
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還是不懂，像我寫了一個
union remote_re { unsigned int remote; struct { unsigned red_on_off:1; unsigned red_select:2; unsigned red_temperate:4; unsigned red_sleep:1; unsigned red_revolve:1; unsigned red_wind:3; unsigned red_mode:3; unsigned red_set:1; }; struct { unsigned :1; unsigned red_clock_minute_unit:4; unsigned red_clock_minute_ten:3; unsigned :1; unsigned red_clock_numbered_unit:4; unsigned red_clock_numbered_ten:2; unsigned :1; }; }remote_red; 這不是一樣位元可以共用嗎？
發表於: 2005/10/31 17:26
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Re: 結構型態&共用型態的問題？
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版主
 
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第一種型態是 Count 與 B<0:5> 共用一個 BYTE 的RAM，所以B0~B5 的位元變化會直接改變 Count 的值。至於其位址是由 LINKER 來決定的。
第二種型態與第一種相似，RB0 位元與 INT0 位元佔相同之位元。但 PORTB 與PORTBits 是個別獨立的宣告，其位址由別人定義 (extern)，在 C18 的例子裡位址會由 PIC18F452.asm 宣告。
發表於: 2005/10/31 12:27
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結構型態&共用型態的問題？
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初級會員
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看WAP002 Workshop看了很久還是看不懂@@
以下兩種宣告方式差別在那？怎麼覺的用起來是一樣的？ ------------------------------------------------------------------ near union { unsigned char Count; struct { unsigned B0:1; unsigned B1:1; unsigned B2:1; unsigned B3:1; unsigned B4:1; unsigned B5:1; }; }Bz=0; _____________________________________________ extern volatile near unsigned char PORTB; extern volatile near union{ struct { unsigned RB0:1; unsigned RB1:1; unsigned RB2:1; unsigned RB3:1; unsigned RB4:1; unsigned RB5:1; unsigned RB6:1; unsigned RB7:1; } ; struct { unsigned INT0:1; unsigned INT1:1; unsigned INT2:1; unsigned CCP2:1; } ; } PORTBbits ;
發表於: 2005/10/25 17:35
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