这个警告是什么意思，我该如何摆脱它？ [英] What is the meaning of this warning, and how do I get rid of it?
问题描述
我在以下
行中有一部分代码：
volatile unsigned char x;
unsigned int f （unsigned char * y）;
当我这样做时
unsigned int z = f（& x）;
编译器发出以下警告：
警告：传递`f'的arg 1'从指针目标类型中丢弃限定符
这究竟是什么意思？如何更改我的代码以便
编译器对此感到满意？
I have a portion of code along the following
lines:
volatile unsigned char x ;
unsigned int f(unsigned char *y) ;
When I do
unsigned int z = f(&x) ;
the compiler issues the following warning:
warning: passing arg 1 of `f'' discards qualifiers from pointer target type
What does this exactly mean? How do I change my code so that the
compiler is happy about it?
推荐答案
James H. Newman写道：
James H. Newman wrote:
我有以下代码的一部分
行：
volatile unsigned char x;
unsigned int f（unsigned char * y）;
当我这样做时
unsigned int z = f（& x）;
编译器发出以下警告：
警告：传递f丢弃的arg 1指针目标类型的限定符
这究竟是什么意思？
I have a portion of code along the following
lines:
volatile unsigned char x ;
unsigned int f(unsigned char *y) ;
When I do
unsigned int z = f(&x) ;
the compiler issues the following warning:
warning: passing arg 1 of `f'' discards qualifiers from pointer target type
What does this exactly mean?
这意味着你要丢弃易变信号。限定符。
如何更改我的代码，以便
It means you''re discarding the "volatile" qualifier.
How do I change my code so that the
编译器对此感到满意？
compiler is happy about it?
不要使用volatile？
-
Mark McIntyre
CLC FAQ< http://c-faq.com/>
CLC自述文件：< http://www.ungerhu.com/jxh/ clc.welcome.txt>
don''t use volatile?
--
Mark McIntyre
CLC FAQ <http://c-faq.com/>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt>
在文章< g5 ********** @ registered.motzarella.org>，
James H. Newman< Ne ****** @ exicite.comwrote：
In article <g5**********@registered.motzarella.org>,
James H. Newman <Ne******@exicite.comwrote:
我在以下
行中有一部分代码：
I have a portion of code along the following
lines:
volatile unsigned char x;
unsigned int f（unsigned char * y）;
volatile unsigned char x ;
unsigned int f(unsigned char *y) ;
>当我做
>When I do
unsigned int z = f （& x）;
unsigned int z = f(&x) ;
>编译器发出以下警告：
>the compiler issues the following warning:
> ;警告：传递`f'的arg 1'从指针目标类型中丢弃限定符
>warning: passing arg 1 of `f'' discards qualifiers from pointer target type
这究竟是什么意思？
What does this exactly mean?
x是易失性的，因此& x是指向volatile变量的指针。但是，
接收函数的原型是非易失性的，因此不会为了保持易失性语义而知道。
x is volatile, so &x is a pointer to a volatile variable. However,
the receiving function is prototyped for non-volatile and so is not
going to know to preserve volatile semantics.
如何更改我的代码以便
编译器对此感到满意？
How do I change my code so that the
compiler is happy about it?
你可以用y表示原型f，因为它是易变的。或者您可以创建一个新的非易失性变量，用x'的当前值初始化，
将其发送到f并将修改后的结果存储到f之后。但是，如果x碰巧是内存映射的I / O寄存器之类的话，那么
，或者发生由信号处理例程设置的
，然后可能赢了'$
给你你想要的结果。
-
当前的垃圾邮件：每天750-800封邮件（3月4日， 2008）
You could prototype f with y being volatile. Or you could create a
new non-volatile variable, initialize that with x''s current value,
send that to f and store the modified result after f into x. However,
if x happens to be a memory-mapped I/O register or the like, or happens
to be set by a signal processing routine, then that probably won''t
give you are result you want.
--
Current spam load: 750-800 messages per day (March 4, 2008)
James H. Newman写道：
James H. Newman wrote:
我有以下代码的一部分
行：
volatile unsigned char x;
unsigned int f（unsigned char * y）;
当我这样做时
unsigned int z = f（& x）;
编译器发出以下警告：
警告：从指针目标传递arg 1'f''丢弃限定符
类型
I have a portion of code along the following
lines:
volatile unsigned char x ;
unsigned int f(unsigned char *y) ;
When I do
unsigned int z = f(&x) ;
the compiler issues the following warning:
warning: passing arg 1 of `f'' discards qualifiers from pointer target
type
这意味着它所说的。由于''f''原型为取无符号
char *，因此'x''的挥发性限定符在被传递时被丢弃
到''f ''。另请注意，表达式& x实际上是volatile类型
unsigned char ** - 而不是您希望传递给''f'的类型。
It means what it says. Since ''f'' is prototyped as taking an unsigned
char *, the volatile qualifier of ''x'' is discarded when it is passed
to ''f''. Also note that the expression &x is actually of type volatile
unsigned char ** - not the type that you want to be passing to ''f''.
这究竟是什么意思？如何更改我的代码以便编译器对它感到高兴？
？
What does this exactly mean? How do I change my code so that
the compiler is happy about it?
原型并将''''定义为：
unsigned int f（volatile unsigned char * y）{/ * ... * /}
如果''f''需要更改调用者'的论点副本那么你
可能需要：
unsigned int f（volatile unsigned char **）{/ * ... * /}
Prototype and define ''f'' as:
unsigned int f(volatile unsigned char *y) { /* ... */ }
If ''f'' needs to change the caller''s copy of it''s argument then you
probably want:
unsigned int f(volatile unsigned char **y) { /* ... */ }
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