C ++中的连接运算符？ [英] Concatenation operator in C++?

问题描述

我有一个应用程序，我需要在一个变量中组合字符串，如：

  int int_arr [4] 
 int_arr [1] = 123; 
 int_arr [2] = 456; 
 int_arr [3] = 789; 
 int_arr [4] = 10; 
 std :: string _string =Text+ int_arr [1] +Text+ int_arr [2] +Text+ int_arr [3] +Text+ int_arr [4] 
  

它给我的编译错误

 错误C2210：'+'操作符不能在表达式的第二个字符串上添加指针。
  

至于我可以告诉我组合字符串字面量和整数，而不是指针。

有没有另一个连接运算符，我应该使用？

我使用Visual Studio 2010

解决方案

您可以在 Java 中执行此操作，因为它使用 toString（）每个部分。

如果你想在C ++中使用相同的方式，你必须显式转换这些整数到字符串，以使这个工作。 p>

如下：

  #include< iostream> 
 #include< sstream> 
 
 std :: string intToStr（int i）{
 std :: ostringstream s; 
 s<<一世; 
 return s.str（）; 
} 
 
 int main（void）{
 int var = 7; 
 std :: string s =Var is'+ intToStr（var）+'; 
 std :: cout<< s<< std :: endl; 
 return 0; 
} 
  

当然，您可以使用：

  std :: ostringstream os; 
 os<< Var is'< var< '; 
 std :: string s = os.str（）; 
  

这很容易。

I have an application in which I need to combine strings within a variable like so:

int int_arr[4];
int_arr[1] = 123;
int_arr[2] = 456;
int_arr[3] = 789;
int_arr[4] = 10;
std::string _string = "Text " + int_arr[1] + " Text " + int_arr[2] + " Text " + int_arr[3] + " Text " + int_arr[4];

It gives me the compile error

Error C2210: '+' Operator cannot add pointers" on the second string of the expression.

As far as I can tell I am combining string literals and integers, not pointers.

Is there another concatenation operator that I should be using? Or is the expression just completely wrong and should figure out another way to implement this?

BTW I am using Visual Studio 2010

解决方案

You can do this in Java since it uses the toString() method automatically on each part.

If you want to do it the same way in C++, you'll have to explicitly convert those integer to strings in order for this to work.

Something like:

#include <iostream>
#include <sstream>

std::string intToStr (int i) {
    std::ostringstream s;
    s << i;
    return s.str();
}

int main (void) {
    int var = 7;
    std::string s = "Var is '" + intToStr(var) + "'";
    std::cout << s << std::endl;
    return 0;
}

Of course, you can just use:

    std::ostringstream os;
    os << "Var is '" << var << "'";
    std::string s = os.str();

which is a lot easier.

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