C++结构体中const使用场景

冷不防 2023-02-20 07:52 9阅读 0赞

看如下代码：

#include<iostream>
    using namespace std;
    
    #include<string>
    
    
    //结构体
    struct Student {
    
    	string name;
    	int age;
    	int score;
    
    }st3;
    
    //值传递
    void printStufdents(struct Student st2) {
    	cout << "子函数" << endl;
    	st2.age = 150;
    	cout << "名字：" << st2.name << "	年龄：" << st2.age << "	分数：" << st2.score << endl;
    }
    
    //指针传递只占四个字节  节省内存空间  而且不会复制新的副本出来
    void printStufdents(struct Student *s) {
    	cout << "子函数" << endl;
    	s->age = 150;
    	cout << "名字：" << s->name << "	年龄：" << s->age << "	分数：" << s->score << endl;
    }
    
    
    
    int main() {
    
    	
    	struct Student st2 = { "张三丰",120,70 };
    	cout << "名字" << st2.name << "年龄" << st2.age << "分数" << st2.score<< endl;
    
    	printStufdents(st2);
    
    
    	system("pause");
    }

1，地址传递只占四个字节  节省内存空间  而且不会复制新的副本出来

2，地址传递有个问题，会改变地址的值，为了防止误操作，我们引入了const关键字修饰指针

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#include<iostream>
    using namespace std;
    
    #include<string>
    
    
    //结构体
    struct Student {
    	string name;
    	int age;
    	int score;
    }st3;
    
    void printStufdents(struct Student st2) {
    	cout << "子函数" << endl;
    	st2.age = 150;
    	cout << "名字：" << st2.name << "	年龄：" << st2.age << "	分数：" << st2.score << endl;
    }
    
    //指针传递只占四个字节  节省内存空间  而且不会复制新的副本出来
    void printStufdents(const struct Student *s) {
    	cout << "子函数" << endl;
    	s->age = 150;
    	cout << "名字：" << s->name << "	年龄：" << s->age << "	分数：" << s->score << endl;
    }
    
    int main() {
    
    	struct Student st2 = { "张三丰",120,70 };
    	cout << "名字" << st2.name << "年龄" << st2.age << "分数" << st2.score<< endl;
    
    	printStufdents(st2);
    	system("pause");
    }

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