将2个字符串连接在一起的程序

[Gzd2003]

要求编写程序将2个字符串连接在一起，程序如下：
#include <iostream>
#include <string>
using namespace std;
void main()
{
string s1="abc",s2="def";
copy(s2,s2.size(),s1+s1.size());
cout<<s1<<endl;
}
编译出错，信息如下：
--------------------Configuration: Test - Win32 Debug--------------------
Compiling...
CopyTest.cpp
D:\Test\CopyTest.cpp(7) : error C2782: 'class std::basic_string<_E,_Tr,_A> __cdecl std::operator +(const class std::basic_string<_E,_Tr,_A> &,const _E)' : template parameter '_E' is ambiguous
could be 'unsigned int'
or 'char'
D:\Test\CopyTest.cpp(7) : error C2784: 'class std::basic_string<_E,_Tr,_A> __cdecl std::operator +(const class std::basic_string<_E,_Tr,_A> &,const _E *)' : could not deduce template argument for 'const class std::basic_string<_E,_Tr,_A> &' from 'cl
ass std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >'
D:\Test\CopyTest.cpp(7) : error C2784: 'class std::basic_string<_E,_Tr,_A> __cdecl std::operator +(const _E,const class std::basic_string<_E,_Tr,_A> &)' : could not deduce template argument for 'const ' from 'class std::basic_string<char,struct std:
:char_traits<char>,class std::allocator<char> >'
D:\Test\CopyTest.cpp(7) : error C2784: 'class std::basic_string<_E,_Tr,_A> __cdecl std::operator +(const _E *,const class std::basic_string<_E,_Tr,_A> &)' : could not deduce template argument for 'const *' from 'class std::basic_string<char,struct
std::char_traits<char>,class std::allocator<char> >'
D:\Test\CopyTest.cpp(7) : error C2784: 'class std::basic_string<_E,_Tr,_A> __cdecl std::operator +(const class std::basic_string<_E,_Tr,_A> &,const class std::basic_string<_E,_Tr,_A> &)' : could not deduce template argument for 'const class std::bas
ic_string<_E,_Tr,_A> &' from 'class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >'
D:\Test\CopyTest.cpp(7) : error C2784: 'class std::reverse_iterator<_RI,_Ty,_Rt,_Pt,_D> __cdecl std::operator +(_D,const class std::reverse_iterator<_RI,_Ty,_Rt,_Pt,_D> &)' : could not deduce template argument for '' from 'class std::basic_string<ch
ar,struct std::char_traits<char>,class std::allocator<char> >'
D:\Test\CopyTest.cpp(7) : error C2676: binary '+' : 'class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >' does not define this operator or a conversion to a type acceptable to the predefined operator
D:\Test\CopyTest.cpp(7) : error C2782: '_OI __cdecl std::copy(_II,_II,_OI)' : template parameter '_II' is ambiguous
could be 'unsigned int'
or 'class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >'
Error executing cl.exe.

Test.exe - 8 error(s), 0 warning(s)

错在哪里？请指教。谢谢。

[yinghui130]

支持6楼



引用 2 楼 lann64 的回复:
s1=s1+s2;

UP.
搞COPY的测试应该是这样:

C/C++ code

#include <iostream> #include <string> #include <iterator> using namespace std; int main(){ string s1="DN",s2="CS"; copy(s1.begin(),s1.end(),back_inserter(s2)); cout << s2 << endl; return 0; }

[zhanshen2891]

自己看一下书！

[Cricketol]

string已经重载了+操作符了，直接s1 += s2;就得了

[xianyuxiaoqiang]

既然用string，那就直接s1+=s2;
否则,可以自己实现之:

#include "string.h"
char * stradd(char **dest,const char *src)
{
    char *temp = NULL;
    int len_dest = 0;
    int len_src = 0;
    int index = 0;
    int i = 0;
	if(dest == NULL)return NULL;
    if(src == NULL || *src == '\0')
    {
        return *dest;
    }
    len_src = strlen(src);
    if(*dest == NULL)
    {
        *dest = new char[len_src + 1];
        strcpy(*dest,src);
    }
    else
    {
        len_dest = strlen(*dest);
        temp = new char[len_src + len_dest +1];
        while((*dest)[i] != '\0')
        {
            temp[index] = (*dest)[i];
            i++;
            index++;
        }
        i = 0;
        while(src[i] != '\0')
        {
            temp[index] = src[i];
            i++;
            index++;
        }
        temp[index] = '\0';
		delete [](*dest);
        *dest = temp;
     }
	return *dest;
}
int main(void) 
{ 
	char *a = new char[10];
        a[0]='a';
	a[1]='\0';
	char *b = "bcd";
        stradd(&a,b);
	printf(a);
}

[General1982]

[Quote=引用 10 楼 Cricketol 的回复:]
string已经重载了+操作符了，直接s1 += s2;就得了
[/Quote]

正解！！

[ugrg]

s1+s1.size() 这个有些问题吧

[猫已经找不回了]

哎。用C++写还这么费劲干吗，重载的运算符，用了省事。

[xxgamexx]

我也写了一个

#include <iostream>
#include <string.h>
#include <assert.h>
using namespace std;
void stradd(char *head,const char *tail,int size)
{  
       assert(head!=NULL||tail!=NULL);
       while((size--)!=0)*head++;
       while((*head++=*tail++)!='\0');
}    
int main()
{
   
 char a[20]="hello";
  char b[10]=" world!";
  stradd(a,b,strlen(a));
  cout<<a;
return 0;
}

[s79947171]

楼上的最后应该手动加上/0吧？

[xxgamexx]

[Quote=引用 16 楼 s79947171 的回复:]
楼上的最后应该手动加上/0吧？
[/Quote]

不用了 已经赋值了

[lgccaa]

用strncpy吧,不要用copy,除非是自己写的实现功能的函数

[lann64]

s1=s1+s2;

[Super.Jiju]

void main()
{
string s1="abc",s2="def";
copy(s2,s2.size(),s1+s1.size());
cout < <s1 < <endl;
}

要么就s1+=s2;
要么就重新申请一个空间

[BaihowFF]

既然是string这种C++的东西 直接用加号就可以了啊...

[cyj626]

string 可以直接++的

[ChamPagneZ]

[Quote=引用 2 楼 lann64 的回复:]
s1=s1+s2;
[/Quote]
UP.
搞COPY的测试应该是这样:

#include <iostream>
#include <string>
#include <iterator>
using namespace std;
int main(){
	string s1="DN",s2="CS";
	copy(s1.begin(),s1.end(),back_inserter(s2));
	cout << s2 << endl;
	return 0;
}

[zxianrong]

s1=s1+s2;

LS的那样可以么？？