c++ - 为什么数组名 [i] 在这种情况下在 C++ 中不等于 *(arrayname + i)

Question

对于下面的代码，为什么arrayname [i]在这种情况下不等于*(arrayname + i)并且输出可能很奇怪:

#include <iostream>
using namespace std;

struct fish
{
    char kind[10] = "abcd";
    int weight;
    float length;
};

int main()
{
    int numoffish;
    cout << "How many fishes?\n";
    cin >> numoffish;
    fish *pfish = new fish[numoffish];
    cout << pfish[0].kind << endl; //the output is "abcd"

    /*if the above code is changed to
    "cout << (*pfish.kind);"
     then compile error happens */

    /*and if the above code is changed to
    "cout << (*pfish->kind);"
    then the output is only an "a" instead of "abcd"*/

    delete [] pfish;
    return 0;
}

Answer 1

. 运算符和 -> 运算符的优先级高于一元 * 运算符。

在访问像

这样的成员之前，你必须添加括号来计算 *

cout << ((*pfish).kind);