清除编译器警告的最佳方法是什么？ [英] Best way of clearing a compiler warning?
问题描述
您好，
请考虑以下代码片段：
fileptr = fopen（param-> ; filename，" r"）;
if（fileptr == NULL）
{
error（" digest.c：digest_file ：打开文件，错误）;
返回（-2）;
}
使用上面的代码，编译器（gcc）给出以下警告：
digest.c：121：警告：传递arg 1的'错误''丢弃限定符来自
指针目标类型。
函数错误声明如下：
void error（char * errmsg，int code）;
为什么会给出这个消息，我该如何让它静音？
谢谢
-
Daniel Rudy
电子邮件地址已经过base64编码以减少垃圾邮件
使用b64decode解码电子邮件地址或uudecode -m
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Daniel Rudy写道：考虑以下代码片段：
fileptr = fopen（param-> filename，" r"）;
if（fileptr == NULL）
{
错误（ digest.c：digest_file：打开文件，错误）;
返回（-2）;
}
使用上面的代码，编译器（gcc）给出以下警告：
digest.c：121：警告：传递arg 1的'错误''从指针目标类型中丢弃限定符。
函数错误声明如下：
void error（char * errmsg，int code）;
为什么会给出这条消息
因为你传递了一个字符串文字，这是const。 'error''是
声明为char *，这意味着它不承诺不修改其参数的内容。如果确实如此，那么糟糕的事情就会发生。
我该如何沉默呢？
如果'错误''真的没有修改`errmsg' '，改变原型和
函数来取代const char *。
如果它确实修改了'errmsg''，你就会有将消息复制到新的
分配的字符串。
S.
关于时间是11/6/2005 4:59 PM，Skarmander声明如下：Daniel Rudy写道：
请考虑以下代码片段：
fileptr = fopen（param-> filename，" r"）;
if（fileptr == NULL）
{
错误（" digest.c：digest_file ：打开文件，错误）;
返回（-2）;
}
使用上面的代码，编译器（gcc）给出以下警告：
digest.c：121：警告：传递arg 1的'错误''丢弃限定符来自
指针ta rget类型。
函数错误声明如下：
void error（char * errmsg，int code）;
为什么它给出这条消息
因为你传递的是字符串文字，即const。 'error''被声明为采用char *，这意味着它不承诺不修改其参数的内容。如果是这样，坏事就会发生。
我明白了。错误不会修改* errmsg的内容，只需使用syslog或fprintf将它打印到stderr。
我怎么让它沉默？
如果'错误''真的没有修改`errmsg''，改变原型和
函数来取一个const char *相反。
如果它确实修改了'errmsg''，你必须将消息复制到新分配的字符串。
S. / blockquote>
啊，谢谢。
-
Daniel Rudy
电子邮件地址已经过base64编码以减少垃圾邮件
解码电子邮件地址使用b64decode或uudecode -m
为什么极客喜欢电脑：看起来聊天日期触摸grep make unzip
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Skarmander写道：
Daniel Rudy写道：考虑以下代码片段：
fileptr = fopen（param-> filename，" r"）;
if（fileptr == NULL）
{
错误（" digest.c：digest_file：Open File"，errno）;
返回（-2）;
}
以上代码，编译器（gcc）给出以下警告：
digest.c：121：
警告：传递arg 1的`错误''丢弃来自
指针目标类型的限定符。
函数错误声明如下：
void error（char * errmsg，int code）;
为什么会给出这条消息
因为你传递了一个字符串文字，这是const。
一个字符串文字不是const合格。
-
pete
Hello,
Consider the following code fragment:
fileptr = fopen(param->filename, "r");
if (fileptr == NULL)
{
error("digest.c: digest_file: Open File ", errno);
return(-2);
}
With the above code, the compiler (gcc) gives the following warning:
digest.c:121: warning: passing arg 1 of `error'' discards qualifiers from
pointer target type.
The function error is declared as follows:
void error(char *errmsg, int code);
Why does it give this message and how do I silence it?
Thanks
--
Daniel Rudy
Email address has been base64 encoded to reduce spam
Decode email address using b64decode or uudecode -m
Why geeks like computers: look chat date touch grep make unzip
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解决方案Daniel Rudy wrote:Consider the following code fragment:
fileptr = fopen(param->filename, "r");
if (fileptr == NULL)
{
error("digest.c: digest_file: Open File ", errno);
return(-2);
}
With the above code, the compiler (gcc) gives the following warning:
digest.c:121: warning: passing arg 1 of `error'' discards qualifiers from
pointer target type.
The function error is declared as follows:
void error(char *errmsg, int code);
Why does it give this message
Because you''re passing a string literal, which is const. `error'' is
declared as taking a char*, meaning it does not promise not to modify
the contents of its argument. If it does, Bad Things will happen.
and how do I silence it?
If `error'' really doesn''t modify `errmsg'', change the prototype and the
function to take a const char* instead.
If it does modify `errmsg'', you''ll have to copy the message to a newly
allocated string.
S.
At about the time of 11/6/2005 4:59 PM, Skarmander stated the following:Daniel Rudy wrote:Consider the following code fragment:
fileptr = fopen(param->filename, "r");
if (fileptr == NULL)
{
error("digest.c: digest_file: Open File ", errno);
return(-2);
}
With the above code, the compiler (gcc) gives the following warning:
digest.c:121: warning: passing arg 1 of `error'' discards qualifiers from
pointer target type.
The function error is declared as follows:
void error(char *errmsg, int code);
Why does it give this message
Because you''re passing a string literal, which is const. `error'' is
declared as taking a char*, meaning it does not promise not to modify
the contents of its argument. If it does, Bad Things will happen.
I see. error does not modify the contents of *errmsg, just prints it
out using either syslog or fprintf to stderr.
and how do I silence it?
If `error'' really doesn''t modify `errmsg'', change the prototype and the
function to take a const char* instead.
If it does modify `errmsg'', you''ll have to copy the message to a newly
allocated string.
S.
Ah, thanks.
--
Daniel Rudy
Email address has been base64 encoded to reduce spam
Decode email address using b64decode or uudecode -m
Why geeks like computers: look chat date touch grep make unzip
strip view finger mount fcsk more fcsk yes spray umount sleep
Skarmander wrote:
Daniel Rudy wrote:Consider the following code fragment:
fileptr = fopen(param->filename, "r");
if (fileptr == NULL)
{
error("digest.c: digest_file: Open File ", errno);
return(-2);
}
With the above code, the compiler (gcc) gives the following warning:
digest.c:121:
warning: passing arg 1 of `error'' discards qualifiers from
pointer target type.
The function error is declared as follows:
void error(char *errmsg, int code);
Why does it give this message
Because you''re passing a string literal, which is const.
A string literal isn''t const qualified.
--
pete
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