C ++：如何将'const char *'转换为char [英] C++: How to convert 'const char*' to char

问题描述

我知道在StackOverflow上有很多问题，但我没有找到任何帮助解决我的情况。每当我尝试这样做：

  // str =一些字符串或字符数组
 
 //一些魔法来绕过精确的错误
 stringstream convert; 
 convert<< str; 
 //捕获流的临时字符串
 const string store = convert.str（）; 
 //获取一个可管理的数组
 const char * temp = store.c_str（）; 
  

然后尝试执行 atoi（temp [0]），我继续得到经典的转换错误，char无法转换为const char。在 atoi 和许多其他函数的文档中，const char是必需的参数。如果只有一个字符，如何发送一个字符？检索特定位置的字符是否自动转换为字符？

解决方案

我不确定这是什么原因错误，但是atoi取其参数不是一个char，而是指向一个。所以代替
atoi（temp [0]）

请尝试此

 code> atoi（& temp [0]）
  

/ p>

I know there are a lot of questions like this out there on StackOverflow, but I haven't been able to find any that help resolve my case. Whenever I try to do something like this:

// str = some string or char array

// some magic to get around fpermissive errors
stringstream convert;
convert << str;
// capture the stream's temporary string
const string store = convert.str();
// get a manageable array
const char* temp = store.c_str();

and then try to do something like atoi(temp[0]), I keep getting the classic conversion error that char couldn't be converted to const char. In the documentation for atoi and many other functions, const char is a required parameter. How can a char be sent in if there's only a const one? Does retrieving a char at a specific position auto-cast to char?

解决方案

I'm not sure if this is what is causing the error, but atoi takes as its parameter not a char, but the pointer to one. So instead of atoi(temp[0])

try this

atoi(&temp[0])

as that is a pointer.

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