BREW AMR 编译错误 L6248E

本文探讨了全局变量及指针在位置独立代码构建中引发的链接错误,并提供了两种解决方案:一是使用全局数组替代全局指针;二是采用局部指针代替全局指针。对于包含多个元素的列表,建议使用二维数组进行声明。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

       主要是由于使用全局变量造成的, 如果要使用全局数组,如 int a[ ] = {1,2,3 };

请在前面加上const , 如下const static int a[ ] = {1,2,3 }; 请勿使用全局指针.

This linker error can occur when trying to build "Position Independent" code. Consider a small example like:

#include <stdio.h>
char *str = "test";
int main(void)
{
printf ("%s",str);
}

when compiled and linked with:

armcc -c -apcs /ropi pi.c
armlink -ropi pi.o

the linker will report a message of the form:

Error: L6248E: pi.o(.data) in ABSOLUTE region 'ER_RW' cannot have address/offset type
relocation to .constdata in PI region 'ER_RO'.

For the code above, the compiler generates a global pointer "str" to the char string "test". The global pointer "str" will need to be initialized to the address of the char string "test" in the .constdata section. However, absolute addresses cannot be used in a PI system, so the link step fails, because of the ABS32 relocations to (position independent) .constdata.

To resolve this, you must re-write the code to avoid the explicit pointer. Two possible ways are shown below:

1) Use a global array instead of a global pointer:

#include <stdio.h>
const char str[] = "test";
int main(void)
{
printf ("%s",str);
}

2) Use a local pointer instead of a global pointer:

#include <stdio.h>
int main(void)
{
char *str = "test";
printf ("%s",str);
}

Please note that if you are using a list with multiple elements, such as:

char * list[] = {"zero", "one", "two"};

You will get a separate link error for each element in the array. In this case, the recommended solution is:

char list[3][5] = {"zero", "one", "two"};

with the print instruction being (for example):

printf("%s", list[1]);

Note that you will need to declare a two dimensional array for the list, with the first dimension as the number of elements in the array, and the second dimension as the maximum size for an element in the array.

 

转载于:https://www.cnblogs.com/secbook/archive/2006/10/12/2655510.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值