将字符串常量赋给char * [英] assigning string constant to char *
问题描述
以下第一个函数编译，第二个给出了什么
我认为是一个虚假的错误：
无法转换`const char [ 5]''to char *''in assignment"
void foo（int m）{
char * str;
if（m< 4）str ="％01x";
else if（m< 9）str ="％02x";
else str ="％03x";
}
void foo（int m）{
char * str;
str = m< 4？ "％01X" ：m< 9？ "％02X" ：％03x;
}
任何想法为什么gcc不喜欢第二个版本？如果我编译
它作为C文件而不是C ++，它工作正常。
Steve
The first of the following functions compiles, the second gives what
I think is a spurious error:
"cannot convert `const char[5]'' to `char *'' in assignment".
void foo(int m) {
char *str;
if (m < 4) str = "%01x";
else if (m < 9) str = "%02x";
else str = "%03x";
}
void foo(int m) {
char *str;
str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
}
Any idea why gcc doesn''t like the second version? If I compile
it as a C file instead of C++, it works fine.
Steve
推荐答案
Steve Pope写道：
Steve Pope wrote:
以下第一个函数编译，第二个给出了什么
我认为是一个虚假的错误：
无法在赋值中将`const char [5]''转换为`char *''。
void foo（int m）{
char * str;
if（m< 4）str ="％01x";
否则if（m <9）str ="％02x";
else str ="％03x";
}
void foo（int m）{
char * str;
str = m< 4？ "％01X" ：m< 9？ "％02X" ：％03x;
}
任何想法为什么gcc不喜欢第二个版本？如果我将它编译成一个C文件而不是C ++，它可以正常工作。
The first of the following functions compiles, the second gives what
I think is a spurious error:
"cannot convert `const char[5]'' to `char *'' in assignment".
void foo(int m) {
char *str;
if (m < 4) str = "%01x";
else if (m < 9) str = "%02x";
else str = "%03x";
}
void foo(int m) {
char *str;
str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
}
Any idea why gcc doesn''t like the second version? If I compile
it as a C file instead of C++, it works fine.
这不是一个虚假的错误，只是一个不起眼的错误。字符串常量
（如％01x）实际上是const char [5]类型，它衰减为const
char *。由于历史原因（为了不打破很多现有的
代码），你可以为非const char *分配一个字符串常量，尽管
它会弯曲（如果不打破）const-correctness。在第一个例子中，你是
正在为char *分配一个字符串常量，并且这是允许的。在
第二个例子中，你不是指定一个字符串常量，而是指定一个
表达式给char *。这是不允许的 - 例外
到const-correctness有意保持狭窄。
要解决您的问题，请将您的第二个示例更改为以下内容：
void foo（int m）{
const char * str = m< 4？ "％01X" ：m< 9？ "％02X" ："％03x"
}
因为你现在要分配给一个const char *，而不是一个非const char *，
问题消失了。
祝你好运，
Tom
It is not a spurious error, just an obscure one. A string constant
(like "%01x") is actually of type const char[5], which decays to const
char*. For historical reasons (so as not to break a lot of existing
code), you can assign a string constant to a non-const char*, although
it bends (if not breaks) const-correctness. In the first example, you
are assigning a string constant to a char*, and that''s permitted. In
the second example, you are assigning not a string constant, but an
expression instead, to the char*. That''s not permitted - the exception
to const-correctness is intentionally kept narrow.
To solve your problem change your second example to the following:
void foo(int m) {
const char *str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
}
Since you''re now assigning to a const char*, not to a non-const char*,
the problem disappears.
Best regards,
Tom
Thomas Tutone< Th *********** @ yahoo.comwrote：
Thomas Tutone <Th***********@yahoo.comwrote:
> Steve Pope写道：
>Steve Pope wrote:
> void foo（int m）{
char * str;
str = m< 4？ "％01X" ：m< 9？ "％02X" ："％03x" ;;
}
>void foo(int m) {
char *str;
str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
}
>任何想法为什么gcc不喜欢第二个版本？如果我将它编译为C文件而不是C ++，它可以正常工作。
>Any idea why gcc doesn''t like the second version? If I compile
it as a C file instead of C++, it works fine.
>这不是一个虚假的错误，只是一个不起眼的错误。字符串常量
（如％01x）实际上是const char [5]类型，它衰减为const
char *。由于历史原因（为了不打破很多现有的代码），你可以为非const char *分配一个字符串常量，尽管它会弯曲（如果没有中断）const-correctness。在第一个例子中，你正在为char *分配一个字符串常量，并且这是允许的。在第二个例子中，你不是为char *指定一个字符串常量，而是指定
表达式。这是不允许的 - 对于const-correctness的例外有意保持狭窄。
>It is not a spurious error, just an obscure one. A string constant
(like "%01x") is actually of type const char[5], which decays to const
char*. For historical reasons (so as not to break a lot of existing
code), you can assign a string constant to a non-const char*, although
it bends (if not breaks) const-correctness. In the first example, you
are assigning a string constant to a char*, and that''s permitted. In
the second example, you are assigning not a string constant, but an
expression instead, to the char*. That''s not permitted - the exception
to const-correctness is intentionally kept narrow.
好的，这是有道理的。
Okay, that makes sense.
>要解决您的问题，请将您的第二个示例更改为以下：
>To solve your problem change your second example to the following:
> void foo（int m）{
const char * str = m< 4？ "％01X" ：m< 9？ "％02X" ："％03x" ;;
}
>void foo(int m) {
const char *str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
}
>因为你现在要分配给一个const char *，而不是一个非const char *，
问题消失了。
>Since you''re now assigning to a const char*, not to a non-const char*,
the problem disappears.
这样可行，但奇怪的是以下内容：
const char * str;
str = m< 4？ "％01X" ：m< 9？ "％02X" ：％03x;
if（m 12）str ="％04x";
似乎是const对于char *类型并不意味着和其他类型的
一样多。
Steve
That works, but oddly so does the following:
const char *str;
str = m < 4 ? "%01x" : m < 9 ? "%02x" : "%03x";
if (m 12) str = "%04x";
It seems the "const" does not mean as much for type char * as
it does for other types.
Steve
>
致以最诚挚的问候，
Tom
>
Best regards,
Tom
Steve Pope写道：
Steve Pope wrote:
似乎是const对于类型char *而言，并不意味着
对于其他类型。
It seems the "const" does not mean as much for type char * as
it does for other types.
这是一个与char类型无关的问题，而是与字符串文字相关的问题：
允许以下转换：
string literal -const char *
string literal -char * //已弃用，但有效
但当然，它's无效
const char * -char *
所以，你可以直接将字符串文字转换成char *，而不是通过
const char *（你尝试使用条件运算符'''''）。
再见，
Zeppe
It''s an issue related not to the char type, but to the string literals:
the following conversion are allowed:
string literal -const char*
string literal -char* // deprecated, but valid
but, of course, it''s not valid
const char* -char*
so, you can convert string literal to char* only directly, not through
const char* (as you tried with the conditional operator ''?'').
Bye,
Zeppe
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