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;
}

为什么这个程序会出错：expected unqualified-id before '[' token？和解释你如何实际使用char以及引用和解除引用（即指针）

Answer 1

这两个陈述都是错误的

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

在C ++中，数组的有效声明类似于

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

至于第二个语句，则没有从类型char ( * )[17]到char *的隐式转换。声明的初始化器具有类型char ( * )[17]，而声明的指针具有类型char *

你应该写

char* namePointer = name ;

或

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

Answer 2

应该是：

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

C ++ char得到错误？

2 个答案: