如何在C語言中將結構體的地址作為引數傳遞給函式?
將結構體的值從一個函式傳遞到另一個函式,共有三種方法:
將單個成員作為引數傳遞給函式。
將整個結構體作為引數傳遞給函式。
將結構體的地址作為引數傳遞給函式。
現在,讓我們瞭解如何將結構體的地址作為引數傳遞給函式。
將結構體的地址作為引數傳遞給函式。
它在函式頭中被收集到一個指向結構體的指標中。
優點
將結構體的地址作為引數傳遞給函式的優點如下:
不浪費記憶體,因為不需要再次建立副本。
不需要返回值,因為函式可以間接訪問整個結構體並對其進行操作。
示例
下面的程式演示瞭如何將結構體的地址作為引數傳遞給函式:
#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
",p->day); printf("month=%s
",p->month); printf("year=%d
",p->year); }
輸出
執行上述程式時,會產生以下結果:
enter the day, month and year:20 MAR 2021 day=20 month=MAR year=2021
示例
下面是一個C程式,透過將整個函式作為引數來演示結構體和函式。由於這種函式呼叫方法,不會浪費記憶體,因為我們不需要再次複製和返回值。
#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
",i,p->Name); printf("The Age of student %d is : %d
",i,p->Age); printf("The Level of student %d is : %f
",i,p->Level); printf("The Grade of student %d is : %s
",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); }
輸出
執行上述程式時,會產生以下結果:
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
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