C++ 错误：从“const char*”到“char”的无效转换[-fpermissive]
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error: invalid conversion from ‘const char*’ to ‘char’ [-fpermissive]
提问by VarunVyas
#include<stdio.h>
#include<iostream>
#include<fstream>
#include<string.h>
using namespace std;
class base {
public:
int lookup(char c);
}; // class base
int base::lookup(char c)
{
cout << c << endl;
} // base::lookup
int main() {
base b;
char c = "i";
b.lookup(c);
} // main
On Compiling above code I am getting below error :
在编译上面的代码时，我收到以下错误：
g++ -c test.cpp test.cpp: In function ‘int main()': test.cpp:20:10: error: invalid conversion from ‘const char*' to ‘char' [-fpermissive]
g++ -c test.cpp test.cpp：在函数“int main()”中：test.cpp:20:10：错误：从“const char*”到“char”的无效转换[-fpermissive]
回答by mathematician1975
Try replacing
尝试更换
char c = "i";
with
和
char c = 'i';
回答by paxdiablo
"i"is not a character, it's a character array that basically decays to a pointer to the first element.
"i"不是字符，它是一个字符数组，基本上衰减到指向第一个元素的指针。
You almost certainly want 'i'.
你几乎肯定想要'i'。
Alternatively, you may actually wanta lookup based on more than a single character, in which case you shouldbe using "i"but the type in that case is const char *rather than just char, both when defining cand in the base::lookup()method.
或者，您实际上可能需要基于多个字符的查找，在这种情况下，您应该使用，"i"但在这种情况下，在定义和方法中的类型都const char *不仅仅是。charcbase::lookup()
However, if that were the case, I'd give serious thought to using the C++ std::stringtype rather than const char *. It may not be necessary,but using C++ strings may make your life a lot easier, depending on how much you want to manipulate the actual values.
但是，如果是这种情况，我会认真考虑使用 C++std::string类型而不是const char *. 这可能不是必需的，但使用 C++ 字符串可能会让您的生活更轻松，这取决于您想要操纵实际值的程度。
回答by Mankarse
"i"is a string literal, you probably wanted a char literal: 'i'.
"i"是一个字符串文字，你可能想要一个字符文字：'i'。
String literals are null terminated arrays of const char(which are implicitly converted to char const*when used in that expression, hence the error).
字符串文字是以空终止的数组const char（char const*在该表达式中使用时隐式转换为，因此出现错误）。
char literals are just chars
字符文字只是chars
