C++ 如何摆脱 - “警告：从 NULL 转换为非指针类型‘char’”？
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how to get rid of - "warning: converting to non-pointer type 'char' from NULL"?
提问by Owen
I have this block of code:
我有这个代码块：
int myFunc( std::string &value )
{
char buffer[fileSize];
....
buffer[bytesRead] = NULL;
value = buffer;
return 0;
}
The line - buffer[bytes] = NULL is giving me a warning: converting to non-pointer type 'char' from NULL. How do I get rid of this warning?
行 - buffer[bytes] = NULL 给了我一个警告：从 NULL 转换为非指针类型 'char'。我如何摆脱这个警告？
回答by Xeo
Don't use NULL? It's generally reserved for pointers, and you don't have a pointer, only a simple char. Just use \0(null-terminator) or a simple 0.
不使用NULL？它通常为指针保留，并且您没有指针，只有一个简单的char. 只需使用\0(null-terminator) 或简单的0.
回答by iammilind
buffer[bytesRead] = 0;// NULL is meant for pointers
buffer[bytesRead] = 0;// NULL 用于指针
As a suggestion, if you want to avoid copying and all then, below can be considered.
作为建议，如果您想避免复制等等，可以考虑以下内容。
int myFunc (std::string &value)
{
s.resize(fileSize);
char *buffer = const_cast<char*>(s.c_str());
//...
value[bytesRead] = 0;
return 0;
}
回答by ninjalj
NULL≠ NUL.
NULL≠ NUL。
NULLis a constant representing a null pointer in C and C++.
NULL是一个常量，表示 C 和 C++ 中的空指针。
NULis the ASCII NUL character, which in C and C++ terminates strings and is represented as \0.
NUL是 ASCII NUL 字符，它在 C 和 C++ 中终止字符串并表示为\0.
You can also use In C++, character constants are type 0, which is exactly the same as \0, since in C character literals have inttype.char.
您也可以使用在 C++ 中，字符常量是 type 0，它与 完全相同\0，因为在 C 字符文字中有int类型。char。
