LPC2132學習中遇到的錯誤1
const int8 NoCheckBusy = 0;
const int8 CheckBusy = 1;
const int32 LCD_RS = 1<<4;
const int32 LCD_RW = 1<<5;
const int32 LCD_EN = 1<<6;
const int32 LCD_DATA = 0xff<<7;
在c文件lcd.c和main.c中都要用到,但是在編譯的時候出現(xiàn)如下的問題:
Error: L6200E: Symbol NoCheckBusy multiply defined (by main.o and lcd.o).
Error: L6200E: Symbol LCD_RS multiply defined (by main.o and lcd.o).
上網(wǎng)站arm.com/help/index.jsp?topic=/com.arm.doc.dui0435a/index.html" target="_blank">http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dui0435a/index.html上去查找,有如下結果:
L6200E:
There are two common examples where this occurs:
This error is reported when functions that use semihosting SWIs are linked in from the C library, in the presence of the
To resolve this, you must provide your own implementations of these C library functions.
The ADS 1.2 Examplesembedded directory contains examples of how to re-implement some of the more common SWI-using functions - see the file
To identify which SWI-using functions are being linked-in from the C libraries:
1. Link with armlink -verbose -errors err.txt
2. Search err.txt for occurrences of __I_use_semihosting_swi
For example:
:
Loading member sys_exit.o from c_a__un.l.
:
This shows that the SWI-using function _sys_exit is being linked-in from the C library.
This means that there are two conflicting definitions of
stdio.o
To identify why stdio.o is being linked-in, you must link with the linkers "verbose" switch, e.g.:
Then study
To move forward, the user may have to either:
- Eliminate the calls like
- Re-implement the
具體是什么意思看的也不是很懂,后來看了別人的解決方法是:在頭文件中僅聲明變量,而把變量的定義都放到c文件中去,問題就解決了。
即把
const int8 NoCheckBusy = 0;
const int8 CheckBusy = 1;
const int32 LCD_RS = 1<<4;
const int32 LCD_RW = 1<<5;
const int32 LCD_EN = 1<<6;
const int32 LCD_DATA = 0xff<<7;
都放到lcd.c中就可避免該問題,時間不多,具體原因還有待研究。
評論