有如下程序段:MOV 31H, #24HMOV A, 31HSWAP AANL A, #0F0H执行结果是()
A: (A)=24H
B: (A)=42H
C: (A)=40H
D: (A)=00H
A: (A)=24H
B: (A)=42H
C: (A)=40H
D: (A)=00H
举一反三
- 有如下程序段:MOV31H,#24H;MOVA,31H;SWAPA;ANLA,#0F0H;执行结果是() A: B: =24H C: D: =42H E: F: =40H G: H: =00H
- 已知:(30H)=23H,(31H)=13H,(40H)=72H,(41H)=08H 执行: MOV A,30H ADD A,40H DA A MOV 30H,A MOV A,31H ADDC A,41H DA A MOV 31H,A 则结果为:(30H)=______H,(31H)=______H,(40H)=______H,(CY)=______。
- 执行如下三条指令后,30H单元的内容是( ) A: 、0 B: H C: H D、F D: H E: O F: O G: O H: R1,#30H I: 40H,#0 J: @R1,40H <br/>A、40H B、30H
- 根据下列程序段运行情况,将运行结果和指令分别补充完整。 (1)( );(A)=10H ( );(B)=10H ( );(A)=20H MOV 23H,A;(23H)=20H ( );(A)=21H MOV 24H,A;(24H)=21H INC A ;(A)=22H ( );(25H)= 22H (2)MOV A,#50H;(A)=( )H MOV R0,A;(R0)=( )H MOV @R0,#30H;(50H)=( )H ADD A,@R0;(A)=( )H MOVX @R0,A;(50H)=( )H
- 以下程序的输出结果是( )。 struct HAR { int x, y; struct HAR *p;} h[2]; main(){ h[0].x=1; h[0].y=2; h[1].x=3; h[1].y=4; h[0].p=&h[1]; h[1].p=h; printf("%d %d\n",(h[0].p)->x,(h[1].p)->y); }