C - 无法将结构指针传递给函数

Question

当我编译这个时，我得到很多错误，说我的 struct 的每个部分都是无效的，按照以下行:

sort.c:16: 请求成员“last”不是结构或 union

对于我使用 strcpy 的情况，错误如下:

sort.c:18: 警告:从不兼容的指针类型传递 'strcpy' 的 arg 2

所以我一定是误用了指针...但我不确定为什么。

我在 DBrecord.h 中定义了 struct:

typedef struct{
    int DBrecordID;         //ID for each entry, range 0-319
    char *last;             //student last name
    char *first;            //student first name
    char studentID[8];      //student ID
    int age;                //student age
    class year;             //year in school
    float gpa;              //GPA
    int expGradYear;        //expected graduation year
}DBrecord;


#include <stdio.h>
#include <string.h>
#include "DBrecord.h"

void bubbleSort(DBrecord **record, int numEntries, int sortChoice) {
int i, j;
char temp[100];

for(i=0; i<numEntries; i++)
    for(j = 0; j < numEntries-1; j++)
        switch(sortChoice){
            //sort by last name
            case  1 : if(strcmp(record->last[j], record->last[j+1]) > 0){
                    //swap the two elements
                    strcpy(temp, record[j]);
                    strcpy(record[j], record[j+1]);
                    strcpy(record[j+1], temp);
                  }
            //sort by first name
            case  2 : if(strcmp(record->first[j], record->first[j+1]) > 0){
                    //swap the two elements
                    strcpy(temp, record[j]);
                    strcpy(record[j], record[j+1]);
                    strcpy(record[j+1], temp);
                  }
            //sort by student ID
            case  3 : if(atoi(record->studentID[j]) > atoi(record->studentID[j+1])){
                    //swap the two elements
                    temp = record[j];
                    record[j] = record[j+1];
                    record[j+1] = temp;
                  }
            //sort by age
            case  4 : if(atoi(record->age[j]) > atoi(record->age[j+1])){
                    //swap the two elements
                    temp = record[j];
                    record[j] = record[j+1];
                    record[j+1] = temp;
                  }
            //sort by class
            case  5 : if(record->class[j] > record->class[j+1]){
                    //swap the two elements
                    temp = record[j];
                    record[j] = record[j+1];
                    record[j+1] = temp;
                  }
            //sort by gpa
            case  6 : if(atoi(record->gpa[j]) > atoi(record->gpa[j+1])){
                    //swap the two elements
                    temp = record[j];
                    record[j] = record[j+1];
                    record[j+1] = temp;
                  }
            //sort by expected graduation year
            case  7 : if(atoi(record->expGradYear[j]) > atoi(record->expGradYear[j+1])){
                    //swap the two elements
                    temp = record[j];
                    record[j] = record[j+1];
                    record[j+1] = temp;
                  }
            default : break;
}

Answer 1

因为 record 是一个指向 DBrecord 的指针，record->last[j] 不会产生任何感觉。 record 处的对象(指向指针的指针)没有名为 last 的字段；它根本没有字段，它只是一个指针。

您的意思可能是 record[j]->last，以获取第 j 个 DBrecord 的 last 成员。等等。