有三种方法可以将结构的值从一个函数传递到另一个函数。它们如下所示:
将单个成员作为参数传递给函数。
将整个结构作为参数传递给函数。
将结构的地址作为参数传递给函数。
现在让我们看看如何将整个结构作为参数传递给函数。
在函数调用中给出结构变量的名称作为参数。
在函数头中将其收集到另一个结构变量中。
缺点是浪费内存,会再次创建整个结构的副本。
以下程序演示了如何将整个结构作为参数传递给函数。
演示
#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);//passing entire structure as an argument to function return 0; } void display(struct date d){ printf("day=%d</p><p>",d.day); printf("month=%s</p><p>",d.month); printf("year=%d</p><p>",d.year); }
当上述程序被执行时,它产生以下结果 −
enter the day, month and year:18 JAN 2021 day=18 month=JAN year=2021
考虑另一个例子,在这个例子中,解释了一个C程序如何将整个结构体作为参数传递给函数。
在线演示
#include<stdio.h> //Declaring structure// struct add{ int var1; int var2; }a; //Declaring and returning Function// void show(struct add a){ //Declaring sum variable// int sum; //Arithmetic Operation// sum=a.var1+a.var2; //Printing O/p// printf("Added value is %d",sum); } void main(){ //Declaring structure// struct add a; //Reading User I/p// printf("Enter variable 1 = "); scanf("%d",&a.var1); printf("Enter variable 2 = "); scanf("%d",&a.var2); //Calling function// show(a); }
当上述程序被执行时,它产生以下结果 −
Enter variable 1 = 20 Enter variable 2 = 50 Added value is 70
这是另一个C程序,演示了将整个结构作为参数传递给函数的方法,其中解释了声明结构、声明和返回函数等内容。
在线演示
#include<stdio.h> //Declaring structure// struct student{ int s1,s2,s3; }s[5]; //Declaring and returning Function// void addition(struct student s[]){ //Declaring sum variable and For loop variable// int i,sum; //Arithmetic Operation// for(i=1;i<4;i++){ sum=s[i].s1+s[i].s2+s[i].s3; printf("Student %d scored total of %d</p><p>",i,sum); } } void main(){ //Declaring variable for For loop// int i; //Reading User I/p through For loop// for(i=1;i<4;i++){ printf("Enter marks for student %d in subject 1 = ",i); scanf("%d",&s[i].s1); printf("Enter marks for student %d in subject 2 = ",i); scanf("%d",&s[i].s2); printf("Enter marks for student %d in subject 3 = ",i); scanf("%d",&s[i].s3); } //Calling function// addition(s); }
当上述程序被执行时,它产生以下结果 −
Enter marks for student 1 in subject 1 = 25 Enter marks for student 1 in subject 2 = 89 Enter marks for student 1 in subject 3 = 45 Enter marks for student 2 in subject 1 = 12 Enter marks for student 2 in subject 2 = 45 Enter marks for student 2 in subject 3 = 89 Enter marks for student 3 in subject 1 = 12 Enter marks for student 3 in subject 2 = 78 Enter marks for student 3 in subject 3 = 12 Student 1 scored total of 159 Student 2 scored total of 146 Student 3 scored total of 102
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