访问struct数组元素

时间:2013-04-19 17:37:38

标签: c arrays struct

该程序应该捕获用户输入到结构中的输入,然后打印出给定信息的直方图。到目前为止一切正常,除了当我尝试打印直方图时,无论学生的成绩如何,所有'*'字符都属于F级。我认为正在发生的是学生的数组索引被传递而不是实际变量本身,但我很困惑,因为在直方图打印之前的输出中它显示正确的值。有什么建议吗?

#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 28
#define MAXGRADE 100

struct studentData{
    char studentID[MAXSIZE];
    char studentName[MAXSIZE];
    int examPercent;
} studentRecords[MAXSIZE];

// function prototype for histogram
void displayHist(struct studentData *records, int classSize);

int main()
{
    int i, students = -1;
    //struct studentData *studentRecords[MAXSIZE];
    while(students < 0 || students > MAXSIZE)
    {
        printf("Please enter the number of students in your class:\n");
        scanf("%d", &students);

        if(students > MAXSIZE || students <= 0)
        {
            printf("Try again..\n");
            scanf("%d", &students);
        }

    }

 for(i=0;i<students;i++) {

        printf("Please enter the student #%d's lastname:\n", i+1);
        scanf("%s", &studentRecords[i].studentID);


        printf("Please enter the student #%d's ID#:\n", i+1);
        scanf("%s", &studentRecords[i].studentName);

        printf("Please enter the student's exam percent:\n");
        scanf("%d", &studentRecords[i].examPercent);

 }

 //This is just here to view the input...
 for(i=0;i<students;i++) {

        printf("Student #%d's name is %s\n", i+1, studentRecords[i].studentName);
        printf("student #%d's ID#:%s\n", i+1, studentRecords[i].studentID);
        printf("student #%d's grade was %d\n", i+1, studentRecords[i].examPercent);

  }

    displayHist(&studentRecords[students], students);

    return 0;
}

void displayHist(struct studentData *records, int classSize)
{
    int i;


       printf("A:");
    for(i=0;i<classSize;i++)
    {

        if(records[i].examPercent >=90)
        {
            printf("*");
        }

    }


       printf("\n");
       printf("B:");
    for(i=0;i<classSize;i++)
    {
        if(records[i].examPercent< 90 && records[i].examPercent >= 80)
        {
            printf("*");
        }
    }

       printf("\n");
       printf("C:");
     for(i=0;i<classSize;i++)
     {
        if(records[i].examPercent < 80 && records[i].examPercent >= 70)
        {
            printf("*");
        }

    }

   printf("\n");
   printf("D:");
   for(i=0;i<classSize;i++)
    {
        if(records[i].examPercent< 70 && records[i].examPercent >= 60)
        {
            printf("*");
        }

    }

   printf("\n");
   printf("F:");
   for(i=0;i<classSize;i++)
    {
        if(records[i].examPercent < 60)
        {
            printf("*");
        }

    }
}

2 个答案:

答案 0 :(得分:2)

displayHist(&studentRecords[students], students);

&studentRecords[students]是数组studentRecords之后的地址。在displayHists中,对records[i]的访问将尝试取消引用studentRecords[students+i],这超出了数组的范围。

正确的电话可能是:

displayHist(&studentRecords[0], students);

相当于:

displayHist(studentRecords, students);

顺便说一下,& scanf中的char *char (*)[]无关,因为char *和{{1}}可能有不同的内存表示形式。

答案 1 :(得分:0)

scanf("%s", &studentRecords[i].studentID);

scanf("%s", &studentRecords[i].studentName);

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[28]’ [-Wformat]

当您使用address-of,即&时,它变为char **,这不是scanf所期望的。

所以尝试使用这种方式。

scanf("%s", &(*studentRecords[i].studentID));

displayHist(studentRecords, students);