Question

我试图在一个函数中声明一个变量（结构类型），并从其他函数操纵它（读/写）。但是，当我尝试在未声明该变量的任何函数中使用该变量时，该变量仅包含垃圾。

这是我的代码：

    #include <stdio.h>
    #include <stdlib.h>

    typedef struct
      char name[25];
      int roll;
      float marks;  
    }Student;

    void getInfo(Student student);
    void display(Student student);

    int main(int argc, char *argv[]) {
      Student student;
      getInfo(student);
      display(student);
      return 0;
    }

    void getInfo(Student student){
      printf("Enter student's name: \n");
      scanf("%s", &student.name);
      printf("Enter student's roll: \n");
      scanf("%d", &student.roll);
      printf("Enter student's grade: \n");
      scanf("%f", &student.marks);
    }

    void display(Student student){
      printf("NAME: %s\n", student.name);
      printf("ROLL: %d\n", student.roll);
      printf("GRADE: %.2f\n", student.marks);
    }

Answer 1

您应该通过引用（＆运算符）传递您的结构

#include <stdio.h>
#include <stdlib.h>

typedef struct{
    char name[25];
    int roll;
    float marks;    
} Student;

void getInfo(Student *student);
void display(Student *student);

int main(int argc, char *argv[]) {
    Student student;
    getInfo( &student );
    display( &student );
    return 0;
}

void getInfo(Student *student){
    printf("Enter student's name:");
    scanf("%s", student->name);
    printf("Enter student's roll:");
    scanf("%d", &student->roll);
    printf("Enter student's grade:");
    scanf("%f", &student->marks);
}

void display(Student *student){
    printf("NAME: %s\n", student->name);
    printf("ROLL: %d\n", student->roll);
    printf("GRADE: %.2f\n", student->marks);
}

演示：https://repl.it/repls/LowNonstopWeb

C-从不同的函数访问结构变量

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