#pragma NOAREGS char noaregfunc(char i) using 1 { return i+1; } void main(void) { char ch = noaregfunc(2); }
C_STARTUP: C:0x0000 020011 LJMP C:0011 2: char noaregfunc (char i) using 1 3: { C:0x0003 C0D0 PUSH PSW(0xD0) C:0x0005 75D008 MOV PSW(0xD0),#0x08 C:0x0008 8F10 MOV 0x10,R7 4: return i+1; C:0x000A E510 MOV A,0x10 C:0x000C 04 INC A C:0x000D FF MOV R7,A 5: } 6: C:0x000E D0D0 POP PSW(0xD0) C:0x0010 22 RET C:0x0011 787F MOV R0,#0x7F C:0x0013 E4 CLR A C:0x0014 F6 MOV @R0,A C:0x0015 D8FD DJNZ R0,C:0014 C:0x0017 758111 MOV SP(0x81),#0x11 C:0x001A 02001D LJMP main(C:001D) 7: void main(void) 8: { 9: char ch = noaregfunc(2); C:0x001D 7F02 MOV R7,#0x02 C:0x001F 120003 LCALL noaregfunc(C:0003) C:0x0022 8F11 MOV 0x11,R7 10: } C:0x0024 22 RET
I can't get the right answer after calling noaregfunc.why?#pragma NOAREGS doesn't works? This program works EXACTLY as it was programmed to work. main uses register bank 0 and noaregfunc uses registers in register bank 1. The reason you don't get the results you expect is because the return value from noaregfunc is stored in R7 (in register bank 1). Since the main function uses register bank 0, that's where it looks for the return value (in R7). And surprise, surprise. The correct value isn't there. It's in register bank 1! I'm not sure I understand what you are trying to accomplish with this program. There is no good reason to use register banks the way you do in this example. Register banks are really designed to be used with interrupt service routines to save time and so that the ISR doesn't have to push and pop all 8 registers. But, selecting a new register bank is not required. In fact, most 8051 programs only use register bank 0. You may want to take a careful look at the 8051 data sheet and focus your attention on the architecture section. Jon
Dear Mike, May I know why at beggining of code there is a jump to 0x011 in code space where C:0x0011 787F MOV R0,#0x7F C:0x0013 E4 CLR A C:0x0014 F6 MOV @R0,A C:0x0015 D8FD DJNZ R0,C:0014 C:0x0017 758111 MOV SP(0x81),#0x11 C:0x001A 02001D LJMP main(C:001D) it sets 0x7F of data space to 00h, then sets SP before returning. Isn't 7Fh just below the SFRs? Why does keil need to set this to 00h? Regards, KC
May I know why at beggining of code there is a jump to 0x011 in code space where This is the startup code. If you look at it carefully, you'll see that it clears the contents of DATA memory to 0. In the C programming language, all global variables which are not specifically initialized have a value of 0. Jon