将两根弦连在一起 [英] Attaching two strings together

问题描述

大家好，


我在这里他妈的疯了。我想要杀死所有人的电脑周围的所有电脑。好到这一点......我;我想做的事非常非常简单（至少它应该很简单）我;我想把两根弦放在一起

基本上。


我有一个名为currentstr的公共字符串的源代码。 Currentstr是一个

字符串; s在源代码中无处不在;


Currentstr = Currentstr +" - "

printf（"％s"，CurrentStr.c_str）;


我已尝试过其他一些事情，比如


试一试：


string a;

string b;

string c;

a = currentstr;

b =" - >"

c + = a

c + = b

printf（"％s"，c.c_str）;


试试两个


string a;

string b;

string c;

a = currentstr;

b =" - >"

c = a + b;

printf（"％s"，c.c_str）;


现在好了...当我做printf时只有Currentstr打印时没有 - >

任何人都知道如何解决这个问题。我已经搜索了互联网。但是我可以;没想来。

了吧


谢谢!!!!!

解决方案

Joah Senegal写道：

大家好，


我在这里他妈的疯了。我想要杀死所有人的电脑周围的所有电脑。好到这一点......我;我想做的事非常非常简单（至少它应该很简单）我;我想把两根弦放在一起

基本上。


我有一个名为currentstr的公共字符串的源代码。 Currentstr是一个

字符串; s在源代码中无处不在;


Currentstr = Currentstr +" - "

printf（"％s"，CurrentStr.c_str）;

首先，你的问题是std :: string :: c_str（）是一个函数，而不是

数据成员。你需要括号来调用它。


其次，你应该更喜欢iostreams到printf，因为它是类型安全的。

参见 http://www.parashift.com/c++-faq-lit .... html＃faq-15.1 。


干杯！ --M

Joah Senegal写道：

我想要做一些非常非常的事情

简单（至少它应该很简单）我;我试图将两个字符串放在一起

基本上。


我有一个名为currentstr的公共字符串的源代码。 Currentstr是一个

字符串; s在源代码中无处不在;


Currentstr = Currentstr +" - "

printf（"％s"，CurrentStr.c_str）;


我已尝试过其他一些事情，比如


试一试：


string a;

string b;

string c;

a = currentstr;

b =" - >"

c + = a

c + = b

printf（"％s"，c.c_str）;


试试两个


string a;

string b;

string c;

a = currentstr;

b =" - >"

c = a + b;

printf（"％s"，c.c_str）;


现在好了...当我做printf时只有Currentstr打印时没有 - >

任何人都知道如何解决这个问题。我已经搜索了互联网。但是我可以;无论如何形象

out out

请复制并粘贴以下程序，然后编译它并

执行它。请报告输出结果。


#include< string>

#include< iostream>

＃包括< ostream>


int main（）

{

使用命名空间std;

string currentstr（Joah）;

cout<< currenttr<< endl;

currentstr = currentstr +" - " ;;

cout<< currenttr<< endl;

string a;

string b;

string c;

a = currentstr;

b =" - >" ;;

c + = a;

c + = b;

cout<< c<<结束;

}


祝你好运，


Tom

Joah Senegal写道：

大家好，

$ b $我在这儿他妈的疯了。我想要杀死所有人的电脑周围的所有电脑。好到这一点......我;我想做的事非常非常简单（至少它应该很简单）我;我想把两根弦放在一起

基本上。


我有一个名为currentstr的公共字符串的源代码。 Currentstr是一个

字符串; s在源代码中无处不在;


Currentstr = Currentstr +" - "

printf（"％s"，CurrentStr.c_str）;


我已尝试过其他一些事情，比如


试一试：


string a;

string b;

string c;

a = currentstr;

b =" - >"

c + = a

c + = b

printf（"％s"，c.c_str）;


试试两个


string a;

string b;

string c;

a = currentstr;

b =" - >"

c = a + b;

printf（"％s"，c.c_str）;


现在好了...当我做printf时只有Currentstr打印时没有 - >

任何人都知道如何解决这个问题。我已经搜索了互联网。但是我可以;无论如何，我需要支付b

请不要在你的信息中重新输入代码。它引入了错误（上面的b / b
是不可编译的C ++）并且意味着人们必须猜测你的问题可能是什么。直接从您的编辑器复制和粘贴

遵循
http://www.parashift.com/c++-faq-lit...t.html#faq-5.8


那说，这段代码对你有用吗？如果没有，请按照常见问题解答中的指导原则解释错误

，你应该得到很多帮助。


#include< string>

#include< stdio.h>

使用std :: string;


int main（）

{

string a =" a string" ;;

string b =" - >" ;;

string c = a + b;

字符串d;

d + = a;

d + = b;


printf（"％s \ n"，c.c_str（））;

printf（"％s \ n"，d.c_str（））;

}


输出结果是

a string->

a string->


Gavin Deane

Hello all,

I'';m getting fucking crazy out here. I want to kill people an all the
computer around. Well to the point... I;m trying to do something very very
simple (atleast it should be simple) I;m trying to put two strings together
basically.

I have sourcecode with a public string called currentstr. Currentstr is a
string that;s called everywhere in the source;

Currentstr=Currentstr+" -"
printf ("%s", CurrentStr.c_str);

And I;ve tried some other things like

Try one:

string a;
string b;
string c;
a = currentstr;
b = "->"
c +=a
c+=b
printf ("%s", c.c_str);

Try two

string a;
string b;
string c;
a = currentstr;
b = "->"
c = a + b;
printf ("%s", c.c_str);

Well now... when I do printf only Currentstr is printed without "->"
Anyone know how to fix this.. I;ve searched the internet. but I can;t figure
it out

THANKS!!!!!

解决方案

Joah Senegal wrote:

Hello all,

I'';m getting fucking crazy out here. I want to kill people an all the
computer around. Well to the point... I;m trying to do something very very
simple (atleast it should be simple) I;m trying to put two strings together
basically.

I have sourcecode with a public string called currentstr. Currentstr is a
string that;s called everywhere in the source;

Currentstr=Currentstr+" -"
printf ("%s", CurrentStr.c_str);

First, your problem is that std::string::c_str() is a function, not a
data member. You need parentheses to invoke it.

Second, you should prefer iostreams to printf because it is type-safe.
See http://www.parashift.com/c++-faq-lit....html#faq-15.1.

Cheers! --M

Joah Senegal wrote:

I;m trying to do something very very
simple (atleast it should be simple) I;m trying to put two strings together
basically.

I have sourcecode with a public string called currentstr. Currentstr is a
string that;s called everywhere in the source;

Currentstr=Currentstr+" -"
printf ("%s", CurrentStr.c_str);

And I;ve tried some other things like

Try one:

string a;
string b;
string c;
a = currentstr;
b = "->"
c +=a
c+=b
printf ("%s", c.c_str);

Try two

string a;
string b;
string c;
a = currentstr;
b = "->"
c = a + b;
printf ("%s", c.c_str);

Well now... when I do printf only Currentstr is printed without "->"
Anyone know how to fix this.. I;ve searched the internet. but I can;t figure
it out

Please copy and paste the following program, then compile it and
execute it. Please report back on what the output is.

#include <string>
#include <iostream>
#include <ostream>

int main()
{
using namespace std;
string currentstr("Joah");
cout << currentstr << endl;
currentstr = currentstr + " -";
cout << currentstr << endl;
string a;
string b;
string c;
a = currentstr;
b = "->";
c += a;
c += b;
cout << c << endl;
}

Best regards,

Tom

Joah Senegal wrote:

Hello all,

I'';m getting fucking crazy out here. I want to kill people an all the
computer around. Well to the point... I;m trying to do something very very
simple (atleast it should be simple) I;m trying to put two strings together
basically.

I have sourcecode with a public string called currentstr. Currentstr is a
string that;s called everywhere in the source;

Currentstr=Currentstr+" -"
printf ("%s", CurrentStr.c_str);

And I;ve tried some other things like

Try one:

string a;
string b;
string c;
a = currentstr;
b = "->"
c +=a
c+=b
printf ("%s", c.c_str);

Try two

string a;
string b;
string c;
a = currentstr;
b = "->"
c = a + b;
printf ("%s", c.c_str);

Well now... when I do printf only Currentstr is printed without "->"
Anyone know how to fix this.. I;ve searched the internet. but I can;t figure
it out

Please don''t retype code into your message. It introduces errors (bits
of the above are not compileable C++) and means people have to guess
what your problem might be. Copy and paste directly from your editor
following the guidelines in
http://www.parashift.com/c++-faq-lit...t.html#faq-5.8

That said, does this code work for you? If not, explain waht''s wrong
following the guidelines in that FAQ and you should get plenty of help.

#include <string>
#include <stdio.h>
using std::string;

int main()
{
string a = "a string";
string b = "->";
string c = a + b;
string d;
d += a;
d += b;

printf("%s\n", c.c_str());
printf("%s\n", d.c_str());
}

The output is
a string->
a string->

Gavin Deane

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