错误：表达式在 c 中必须具有结构或联合类型

Question

我一直在研究一种代码来存储学生的姓名和成绩，然后在输入学生姓名时回忆起成绩。

这是我的代码：

#include <stdio.h>

#define N 10
#define M 2

struct a
{
     char name[50];
     int grade;
};

int main()
{
    int i;
    int j;
    struct a A[N][M];

    for(i=0;i<N;i++)
    {
         printf("Please Enter Students' Names:/n");
            scanf("%s", &A[i].name);
    }
    for(j=0;j<M;j++)
    {
        printf("Please Enter Students' Grades:/n");
            scanf("%d", &A[j].grade);
    }
    printf("Which Student's Grades Would You Like To View?/n");
        if(scanf("%s", *A[i].name))
        {
            printf("Their Grade Is:%d/n", *A[j].grade);
        }
    return 0;
}

我遇到了这些错误：

hw2problem2.c(21): error: expression must have struct or union type
                        scanf("%s", &A[i].name);
                                     ^
hw2problem2.c(26): error: expression must have struct or union type
                        scanf("%d", &A[j].grade);
                                     ^
hw2problem2.c(29): error: expression must have struct or union type
                if(scanf("%s", *A[i].name))
                                ^
hw2problem2.c(31): error: expression must have struct or union type
                        printf("Their Grade Is:%d/n", *A[j].grade);
                                                       ^
compilation aborted for hw2problem2.c (code 2)

对于错误或一般程序的任何帮助将不胜感激。 谢谢。

Answer 1

您将struct a A 定义为二维数组，并且仅指定 scanf("%s", &A[i].name); 和 scanf("%d", &A[j].grade); 中的一维。

您还有其他一些问题，例如 scanf("%s", &A[i].name);... & 是不必要的。

Answer 2

你的程序应该是这样的

for(i=0;i<N;i++)
{
   for(j=0;j<M;j++)
   {
    printf("Please Enter Students' Grades:/n");
        scanf("%d", &A[i][j].grade);
    printf("Please Enter Students' Names:/n");
        scanf("%s", &A[i][j].name);

    }
}

因为A[i] 的类型是struct a*，而不是struct a。应该是 A[i][j]

从逻辑上讲，您的数组应该是一维的。因此，循环应该是：

struct a A[N];

for(i=0;i<N;i++)
{
     printf("Please Enter Students' Names:/n");
        scanf("%s", &A[i].name);
}
for(j=0;j<N;j++)
{
    printf("Please Enter Students' Grades:/n");
        scanf("%d", &A[j].grade);
}

如果它是主题明智的，那么它应该是 2D 并使用如图所示的嵌套循环。

Answer 3

无需使用2D 数组。试试这个..

#include <stdio.h>
#define N 3

struct a
{
 char name[50];
 int grade;
};

int main()
{
int i;
struct a A[N];
    char sname[50];
for(i=0;i<N;i++)
{
     printf("Please Enter Students' Names:/n");
        scanf("%s", A[i].name);
}
for(i=0;i<N;i++)
{
    printf("Please Enter Students' Grades:/n");
        scanf("%d",&A[i].grade);
}

     printf("Which Student's Grades Would You Like To View?/n\n");
     printf("Enter the name...\n");
     scanf("%s",sname);
  for(i=0;i<N;i++)
  {
    if(strcmp(A[i].name,sname)==0)
    {
    printf("%s grade is %d\n",A[i].name,A[i].grade);
    break;
    }
    else
    {
    if(i==N-1)
    printf("No such a name in your list...\n");
    }
  }
 return 0;
 }