可以通过三种方式将结构的值从一个函数传递到另一个函数。它们如下-
将单个成员作为函数的参数传递。
将整个结构作为参数传递给函数。
将结构的地址作为函数的参数传递。
现在,让我们了解如何将结构的地址作为函数的参数传递。
结构的地址作为参数传递给函数。
它收集在指向函数标头中结构的指针中。
将结构的地址作为参数传递给函数的优点如下:
无需浪费内存,因为无需再次创建副本。
无需返回值,因为该函数可以间接访问整个结构,然后对其进行处理。
以下程序显示如何将结构的地址作为参数传递给函数-
#include<stdio.h> struct date{ int day; char month[10]; int year; }; int main(){ struct date d; printf("enter the day,month and year:"); scanf("%d%s%d",&d.day,d.month,&d.year); display(&d); return 0; } void display(struct date *p){ printf("day=%d\n",p->day); printf("month=%s\n",p->month); printf("year=%d\n",p->year); }输出结果
执行以上程序后,将产生以下结果-
enter the day, month and year:20 MAR 2021 day=20 month=MAR year=2021
下面给出的C程序通过调用整个函数作为参数来演示结构和函数。由于这种调用函数的方法,因此不会浪费内存,因为我们不需要再次复制并返回值。
#include<stdio.h> //Declaring structure// struct student{ char Name[100]; int Age; float Level; char Grade[50]; char temp; }s[5]; //Declaring and returning Function// void show(struct student *p){ //Declaring variable for For loop within the function// int i; //For loop for printing O/p// for(i=1;i<3;i++){ printf("The Name of student %d is : %s\n",i,p->Name); printf("The Age of student %d is : %d\n",i,p->Age); printf("The Level of student %d is : %f\n",i,p->Level); printf("The Grade of student %d is : %s\n",i,p->Grade); p++; } } void main(){ //Declaring variable for for loop// int i; //Declaring structure with pointer// struct student *p; //Reading User I/p// for(i=0;i<2;i++){ printf("Enter the Name of student %d : ",i+1); gets(s[i].Name); printf("Enter the Age of student %d : ",i+1); scanf("%d",&s[i].Age); printf("Enter the Level of student %d :",i+1); scanf("%f",&s[i].Level); scanf("%c",&s[i].temp);//Clearing Buffer// printf("Enter the Grade of student %d :",i+1); gets(s[i].Grade); } //Assigning pointer to structure// p=&s; //Calling function// show(&s); }输出结果
执行以上程序后,将产生以下结果-
Enter the Name of student 1 : Lucky Enter the Age of student 1 : 27 Enter the Level of student 1 :2 Enter the Grade of student 1 :A Enter the Name of student 2 : Pinky Enter the Age of student 2 : 29 Enter the Level of student 2 :1 Enter the Grade of student 2 :B The Name of student 1 is : Lucky The Age of student 1 is : 27 The Level of student 1 is : 2.000000 The Grade of student 1 is : A The Name of student 2 is : Pinky The Age of student 2 is : 29 The Level of student 2 is : 1.000000 The Grade of student 2 is : B