Question

你对c编程很新。我有以下代码，代码基本上通过终端从用户获取值并打印出值。所以我有一个get函数和print函数。不知何故，当我输入学生ID后，程序停止并直接显示注册选项提示并打印出名称和ID。我尝试重新排列选项，然后它工作。伙计们好吗？提前致谢

Answer 1

请参阅代码中的更正。 http://ideone.com/BQjS1A

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

#define MAX_LEN 80

// ****  Add a constant for the size of name
#define MAX_NAME 100

struct Student {
     char name[MAX_NAME];    // ** Save the hassle using malloc
     int id;                   // student number
     char enroll;    // *** Just need a character - not a character pointer
};

struct Student student1;

void getStudent(struct Student *s)
{
    printf("Type the name of the student: ");
// *** Do not need malloc as using an array
//    s->name = malloc(100);   // assume name has less than 100 letters 
    fgets(s->name, MAX_NAME, stdin);

    printf("\nType the student number: ");  // ** Corrected a typo
    scanf("%d", &(s->id));  // *** You should check the return value here and take appropriate action - I leave that you the reader

    printf("\nType the student enrollment option (D or X): ");
   // *** Added a space in front of %c to read white space
    scanf(" %c", &(s->enroll)); //  scanf requires a charcter pointer here
//    return;   This is not needed
}

void printStudent(struct Student s)
{
// *** THIS CODE DOES NOT MAKE ANY SENSE
//    char name[MAX_LEN];
//    char enroll[MAX_LEN];
//    int id;

//    s.id = id;
//    s.name = name;
//    s.enroll = enroll;
// *** Need %c to print enroll
    printf("Student Details: %d %s %c \n", s.id, s.name, s.enroll );
 ///   return;  Not needed
}

int main(int argc, char *argv[]){
    getStudent(&student1);
    printStudent(student1);
    return 0;
}

无法在C中打印值

1 个答案: