C++ 字符出错？

Question

#include <iostream>

int main()
{
    char[] name = { "Nitish prajapati" }; 
    char* namePointer = &name ;

    std::cout << "\n name    = " << name;
    std::cout << "\n &name    = " << &name;
    std::cout << "\n &namePointer    = " << &namePointer;
    std::cout << "\n namePointer    = " << namePointer;

    return 0;
}

为什么这个程序报错：expected unqualified-id before '[' token？和解释你如何实际使用 char 以及引用和取消引用（即指针）

Answer 1

应该是：

const char name[] = {"Nitish prajapati"}; 
const char* namePointer = name ;

Answer 2

这两种说法都是错误的

char[] name = { "Nitish prajapati" }; 
char* namePointer = &name ;

在 C++ 中，数组的有效声明类似于

char name[] = { "Nitish prajapati" };

至于第二条语句，则没有从类型 char ( * )[17] 到 char * 的隐式转换。声明的初始值设定项的类型为 char ( * )[17]，而声明的指针的类型为 char *

你应该写任何一个

char* namePointer = name ;

或

char ( *namePointer)[17] = &name ;

C++ 字符出错？

C++ char getting errors?

c++

arrays

pointers

char