使用gcc（？）识别字符串类型的类 [英] Class not recognising string type with gcc (?)

问题描述

我对c ++不是很有经验所以我确定这是很明显很容易解决的问题，但是很明显对于生活
我b $ b我似乎无法弄明白。我将非常感激

对此有一些了解。


我的班级职位有一个字符串（私人，公共 - 不

似乎很重要，我在一个成员构建

函数并希望从main（）返回调用。但是，虽然我已经将返回类型声明为string编译器（g ++

2.95和3.4）类型似乎没有认识到这一点。


为了说明问题，我创建了一个小类和

main（）说明了这个问题：


===================== ============================= ========

#include< string> ;


班级职位{


私人：


int p;


public：


Position :: Position（）;

string makeString（）;

string s;


};


职位::职位（）{

p = 5;

}


string Position :: makeString（）{

s [0] =''h'';

s [1] =''我';

返回s;

}

main（）{

位置p;

string myString;


myString = p.makeString;

}


========================================== ======== =======

在我的真实程序中，makeString是一个更复杂的函数

，创建的字符串基于Position的数据成员，

所以makeString需要是一个成员函数。


g ++ 2.95抱怨：


g ++ test7.cpp

test7.cpp：函数`int main （）''：

test7.cpp：34：不匹配`string& = {unknown type}''

/usr/include/g++/std/bastring.h:166：候选人是：class

basic_string< char，string_char_traits< char> ，__ defau lt_alloc_template< false，0>

&
basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0> :: operator =（const
basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0>&）
/usr/include/g++/std/bastring.h:232:class

basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0> &安培;
basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0> :: operator =（const char *）
/usr/include/g++/std/bastring.h:234:class, br />
basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0> &安培;
basic_string< char，string_char_traits< char>，__ defau lt_alloc_template< false，0> :: operator =（char）

g ++ 34 test7.cpp

test7.cpp：8：错误：`string''没有命名类型

test7.cpp：13：错误：`string''没有命名类型

test7.cpp：22：错误：`string''没有命名类型

test7.cpp：在函数`int main（）''中：

test7.cpp：32：错误：`string''未声明（首先使用此函数）

test7.cpp：32：错误:(每个未声明的标识符仅报告一次。）

test7.cpp：32：错误：预期`;''在'myString'之前'

test7。 cpp：34：错误：`myString''未声明（首先使用此函数）

test7.cpp：34：错误：''class Position''没有名为''makeString''的成员/>

如果有人知道这个问题，我会很感激。我已经用Google搜索了这个并且看起来没有喜悦的Stroustrup。我很可能

做一些搞笑的傻事:-)


TIA

DCH

解决方案

" David Carter-Hitchin" <哒*** @ carter-hitchin.clara.co.uk>写了...

我对c ++不是很有经验所以我确定这是显而易见的，很容易解决的，但对于我的生活来说我似乎无法弄明白。我将非常感谢您对此有一些了解。


获取一本关于C ++的最新书。您需要了解名称空间

，特别是关于''std''命名空间。

[...]
======== ========================================== ========
#include< string>

班级职位{

私人：

int p;

public：

Position :: Position（）;
string makeString（）;


制作


std :: string makeString（）;


here。

string s;


这里：


std :: string s;


以及其他任何地方。

};

[...]

Victor

在文章< 10 **************** @ doris.uk.clara.net>中，David Carter-Hitchin写道：

你好，
我的班级职位有一个字符串（私人，公共 - 似乎并不重要），我在一个成员
函数中构建并希望从main（）返回调用。但是，虽然我已经将返回类型声明为string编译器（g ++
2.95和3.4）类型似乎没有认识到这一点。


这不是字符串，它是std :: string

为了说明问题，我创建了一个小班并且
main（）说明了这个问题：

=============================== ================== =========
#include< string>

班级职位{
int p;

公开：

位置:: Position（）;
字符串makeString （）;


std :: string makeString（）;

string s;


std :: string s;

};

位置::位置（）{
p = 5 ;
}


为什么不：

位置:: position（）：p（5）{}

？

string Position :: makeString（）{


std :: string Position :: makeString（）{


注意：当你来到这里时，s是空的。这意味着s [0]和

s [1]不存在。在非const字符串上，这是未定义的

行为。提示：改为使用s.append（''h''）。

s [0] =''h'';
s [1] =''i'';
返回s;
}

main（）{

位置p;
字符串myString;

myString = p.makeString;
}

================================= ================ ========
在我的真实程序中，makeString是一个更复杂的功能
并且创建的字符串是基于Position的的数据成员，
所以makeString需要成为一个成员函数。

-

Robert Bauck Hamar
blockquote>

Victor Bazarov写道：

" David Carter-Hitchin" <哒*** @ carter-hitchin.clara.co.uk>写了...

我对c ++不是很有经验所以我确定这是显而易见的，很容易解决的，但对于我的生活来说我似乎无法弄明白。我非常感谢你对这方面的一些了解。

获取一本关于C ++的最新书。你需要了解名称空间
，特别是关于''std''命名空间。

他说他正在使用Stroustrup''。它省略了using指令，只是

使用不合格的名称。我认为它在

介绍中提到它。


- Pete


< snip>

Hi,

I''m not very experienced with c++ so I''m sure this is
something obvious that is easily solved, but for the life
of me I can''t seem to figure it out. I''d be very grateful
for some knowledgeable insight into this.

My class Position has a string (private, public - doesn''t
seem to matter which), which I construct in a member
function and wish to return calls from main(). However, although
I''ve declared the return type as "string" the compiler (g++
2.95 and 3.4) the type does not seem to recognise this.

To illustrate the problem, I''ve created a small class and
main() that illustrates this problem:

================================================== ========
#include <string>

class Position {

private:

int p;

public:

Position::Position();
string makeString();
string s;

};

Position::Position() {
p = 5;
}

string Position::makeString () {
s[0] = ''h'';
s[1] = ''i'';
return s;
}
main () {

Position p;
string myString;

myString = p.makeString;
}

================================================== =======
In my real program makeString is a more complicated function
and the string created is based on Position''s data members,
so makeString needs to be a member function.

g++ 2.95 complains:

g++ test7.cpp
test7.cpp: In function `int main()'':
test7.cpp:34: no match for `string & = {unknown type}''
/usr/include/g++/std/bastring.h:166: candidates are: class
basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0>

& basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0>::operator =(const basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0> &) /usr/include/g++/std/bastring.h:232: class
basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0> & basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0>::operator =(const char *) /usr/include/g++/std/bastring.h:234: class
basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0> & basic_string<char,string_char_traits<char>,__defau lt_alloc_template<false,0>::operator =(char)

g++34 test7.cpp
test7.cpp:8: error: `string'' does not name a type
test7.cpp:13: error: `string'' does not name a type
test7.cpp:22: error: `string'' does not name a type
test7.cpp: In function `int main()'':
test7.cpp:32: error: `string'' undeclared (first use this function)
test7.cpp:32: error: (Each undeclared identifier is reported only once
for each function it appears in.)
test7.cpp:32: error: expected `;'' before "myString"
test7.cpp:34: error: `myString'' undeclared (first use this function)
test7.cpp:34: error: ''class Position'' has no member named ''makeString''

If someone knows the problem, I''d appreciate a hint. I''ve
googled for this and looked a Stroustrup without joy. I''m probably
doing something hilariously stupid :-)

TIA
DCH

解决方案

"David Carter-Hitchin" <da***@carter-hitchin.clara.co.uk> wrote...

I''m not very experienced with c++ so I''m sure this is
something obvious that is easily solved, but for the life
of me I can''t seem to figure it out. I''d be very grateful
for some knowledgeable insight into this.
Get a more recent book on C++. You need to learn about namespaces
and especially about the ''std'' namespace.
[...]
================================================== ========
#include <string>

class Position {

private:

int p;

public:

Position::Position();
string makeString();
Make it

std::string makeString();

here.
string s;
And here:

std::string s;

And everywhere else.

};

[...]

Victor

In article <10****************@doris.uk.clara.net>, David Carter-Hitchin wrote:

Hi,
My class Position has a string (private, public - doesn''t
seem to matter which), which I construct in a member
function and wish to return calls from main(). However, although
I''ve declared the return type as "string" the compiler (g++
2.95 and 3.4) the type does not seem to recognise this.
It''s not string, it''s std::string
To illustrate the problem, I''ve created a small class and
main() that illustrates this problem:

================================================= =========
#include <string>

class Position {

private:

int p;

public:

Position::Position();
string makeString();
std::string makeString();
string s;
std::string s;

};

Position::Position() {
p = 5;
}
Why not:
Position::position() : p(5) {}
?
string Position::makeString () {
std::string Position::makeString() {

Note: by the time you come here, s is empty. That means that s[0] and
s[1] doesn''t exist. On a non-const string, this is undefined
behaviour. Hint: use s.append(''h'') instead.
s[0] = ''h'';
s[1] = ''i'';
return s;
}
main () {

Position p;
string myString;

myString = p.makeString;
}

================================================= ========
In my real program makeString is a more complicated function
and the string created is based on Position''s data members,
so makeString needs to be a member function.

--
Robert Bauck Hamar

Victor Bazarov wrote:

"David Carter-Hitchin" <da***@carter-hitchin.clara.co.uk> wrote...

I''m not very experienced with c++ so I''m sure this is
something obvious that is easily solved, but for the life
of me I can''t seem to figure it out. I''d be very grateful
for some knowledgeable insight into this.

Get a more recent book on C++. You need to learn about namespaces
and especially about the ''std'' namespace.

He said he was using Stroustrup''s. It omits the using directives, and just
uses unqualified names. I think it mentions it somewhere in the
introduction.

- Pete

<snip>

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