hello all! *.h void Baudrate(unsigned int *ptr,unsigned char Uart_num) large; *.c void main(void) { ... *ptr = 1152; Baudrate(ptr,1); ... } In the routine ,the var Baud( Baud = *ptr; )is not 1152 but 0x00.When changing the defintion : void Baudrate(unsigned int pp,unsigned char Uart_num) large;,and be called : Baudrate(1152,1);,the routine run well. What is the wrong?
"unsigned int xdata *ptr; xdata unsigned int *ptr;" I believe the second example is obsolete syntax supported for backward compatibility. The new way of doing things is: char xdata * data ptr; Which is a pointer located in data space pointing to a char in xdata space. I think this makes things a bit easier as you only have to remember that the qualifier to the left of the '*' specifies the location of the object being pointed to and the qualifier to the right of the '*' specifies the location of the pointer. "Finally, to confuse you completely, answer what does next line do yourself: idata unsigned int xdata *ptr;" Yes, well, quite.
hi, I just compiled next example:
#include <reg51.h> void main(void) { idata char xdata *ptr; ptr = 0x11; *ptr = 0x22; while (1); }
; ptr = 0x11; MOV R0,#0x08 MOV @R0,#0x00 INC R0 MOV @R0,#0x11 ; *ptr = 0x22; DEC R0 MOV A,@R0 MOV R6,A INC R0 MOV A,@R0 MOV DPL,A MOV DPH,R6 MOV A,#0x22 MOVX @DPTR,A
"This shows that pointer is placed inside idata memory and points to xdata memory." Sorry, I should have explained that. I didn't realise it was a question. The correct syntax would be: char xdata * idata ptr;