为什么我会得到错误“错误：表达式必须具有常量值” [英] Why do i get error "error: expression must have a constant value"
问题描述
嗨！
我有以下代码，我在嵌入式系统中使用，
c-compiler ..但是我看到了同样的GCC的问题也是..
我需要一个地址指针的最后10位，在链接时完全知道
，但是会在编译时间。我使用
以下代码进行此初始化..
uint32 data_ptr_mask =（（uint32）data）& 0x3ff;
为了说明目的，我还包括了
uint32 data_ptr_off =（（uint32）数据）+ 0x3ff;
当我编译这段代码时（参见下面的例子例程）我得到
跟随gcc的错误
" error：expression必须有一个常量值"
有人可以告诉我为什么位操作符，*和/不工作
而+和 - 做。
TIA
-P.B. Srinivas
/ ******代码开头*** /
#include< stdio.h>
typedef int int32;
typedef unsigned uint32;
/ *全局变量声明** /
int32 data [64] ;
/ **这个工作** /
uint32 data_ptr =（uint32）数据;
/ **这也有用** /
uint32 data_ptr_off =（（uint32）数据）+ 0x3ff;
/ **这不起作用，只有+/-运营商** /
/ **似乎有效，为什么会这样？ **** /
uint32 data_ptr_mask =（（const uint32）data）& 0x3ff;
int main（）
{
printf（" Pointer Addition％x％x \ n"， data_ptr，data_ptr_off）;
printf（Pointer Masking％x％x \ n，data_ptr，data_ptr_mask）;
}
/ ***********代码结束************* /
Hi !
I have the following code, which I am using in an Embedded systems,
c-compiler.. However I see the same problem with GCC too..
I need the last 10 bits of an address pointer, which is completly know
at link time, but will be symbols at the compile time. I used the
following code for this initalization..
uint32 data_ptr_mask = ((uint32) data) & 0x3ff;
for illustration purpose, I have also included
uint32 data_ptr_off = ((uint32 ) data) +0x3ff;
When I compile this code (see the example routine below) I get the
following error with gcc
"error: expression must have a constant value"
Can somebody tell me why the bit operators, "*" and "/" don''t work
while "+" and "-" do.
TIA
-P.B. Srinivas
/****** Start of CODE ***/
#include <stdio.h>
typedef int int32;
typedef unsigned uint32;
/* Global variable declarations **/
int32 data[64];
/** this works **/
uint32 data_ptr = (uint32 ) data;
/** this also works **/
uint32 data_ptr_off = ((uint32 ) data) +0x3ff;
/** this doesn''t work , only +/- operators **/
/** seem to work, why is this ? ****/
uint32 data_ptr_mask = ((const uint32) data) & 0x3ff;
int main()
{
printf("Pointer Addition %x %x\n",data_ptr, data_ptr_off);
printf("Pointer Masking %x %x\n",data_ptr, data_ptr_mask);
}
/*********** End of code *************/
推荐答案
在你的typedef第二个typedef语句中，你没有指定
内置类型的标识符名称。你只为这种数据类型指定了修饰符
。
换句话说，''unsigned''是什么？
In your typedef second typedef statement you have not specified the
built in-type''s identifier name. You have only specified the modifier
for that data type.
In other words, an ''unsigned'' what?
2006-03-15，Albert< al ***************** @ gmail.com>写道：
On 2006-03-15, Albert <al*****************@gmail.com> wrote:
在你的typedef第二个typedef语句中，你没有指定
内置类型的标识符名称。你只为这种数据类型指定了修饰符。
换句话说，''unsigned''是什么？
In your typedef second typedef statement you have not specified the
built in-type''s identifier name. You have only specified the modifier
for that data type.
In other words, an ''unsigned'' what?
int，还有什么？
[嘿，你不会说'a''长''什么'，对吗？]
int, what else?
[hey, you don''t say "a ''long'' what", do you?]
2006年3月15日星期三06:13，Jordan Abel认为（
< sl ****************** ****@random.yi.org>）：
On Wednesday 15 March 2006 06:13, Jordan Abel opined (in
<sl**********************@random.yi.org>):
2006-03-15，Albert< al ************** ***@gmail.com>写道：
On 2006-03-15, Albert <al*****************@gmail.com> wrote:
在你的typedef第二个typedef语句中，你没有指定
内置类型的标识符名称。你只为这种数据类型指定了修饰符。
换句话说，一个''unsigned''是什么？
In your typedef second typedef statement you have not specified the
built in-type''s identifier name. You have only specified the modifier
for that data type.
In other words, an ''unsigned'' what?
int，什么别的？
[嘿，你不要说'a''长''什么'，对吗？]
int, what else?
[hey, you don''t say "a ''long'' what", do you?]
他一定忘记了他指的是什么，因为他没有包含任何
背景。 ;-)
-
BR，弗拉基米尔
谋杀总是一个错误 - 永远不应该做任何事情都不能用晚餐后讨论。
- 奥斯卡王尔德，多里安格雷的照片
He must have forgot what he was referring to as he did not include any
context. ;-)
--
BR, Vladimir
Murder is always a mistake -- one should never do anything one cannot
talk about after dinner.
-- Oscar Wilde, "The Picture of Dorian Gray"
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