结构数组和malloc [C]

Question

I have a problem with this code. I want to code a program that has various personal info in an array. And I want 15 Arrays to be set up in one place in the memory (malloc). Also the programm should output (printf) the personal info of one person on request (angestellter[0 - 14]).

The Code Errors I recive are the following:

 gcc ANGDB.c ANGDB.c: In function 'print_angestellter': ANGDB.c:14:18: error: subscripted value is neither array nor pointer nor vector nu = angestellter[x].nummer; ^ ANGDB.c:15:18: error: subscripted value is neither array nor pointer nor vector vn = angestellter[x].vorname; ^ ANGDB.c:16:18: error: subscripted value is neither array nor pointer nor vector nn = angestellter[x].nachname; ^ ANGDB.c: In function 'main': ANGDB.c:25:13: error: subscripted value is neither array nor pointer nor vector angestellter[0] -> nummer = 1; ^ ANGDB.c:26:13: error: subscripted value is neither array nor pointer nor vector angestellter[0] -> vorname = "George"; ^ ANGDB.c:27:13: error: subscripted value is neither array nor pointer nor vector angestellter[0] -> nachname = "Washington"; 

This is my code:

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

struct angestellter{
    int nummer;
    char vorname[50];
    char nachname[50];
}angestellter;

void print_angestellter(int x){
    int nu;
    char vn[50];
    char nn[50];
    nu = angestellter[x].nummer;
    vn = angestellter[x].vorname;
    nn = angestellter[x].nachname;
    printf("%d, %s, %s\n", nu, vn, nn);
}

int main(){
    struct angestellter **db = malloc(sizeof(angestellter)*15);
    angestellter[0] -> nummer = 1;
    angestellter[0] -> vorname = "George";
    angestellter[0] -> nachname = "Washington";
    print_angestellter(0);
}

Answer 1

Where you're using angestellter , which is a single instance of struct angestellter , you should be using db , which is your dynamically allocated array. You should also declare it as struct angestellter * instead of struct angestellter ** . This will also need to be passed to print_angestellter .

You also need to use strcpy to copy strings. You can't directly assign a string to a character array.

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

struct angestellter{
    int nummer;
    char vorname[50];
    char nachname[50];
};

void print_angestellter(struct angestellter *db, int x){
    int nu;
    char vn[50];
    char nn[50];
    nu = db[x].nummer;
    strcpy(vn, db[x].vorname);     // use strcpy to copy strings
    strcpy(nn, db[x].nachname);
    printf("%d, %s, %s\n", nu, vn, nn);
}

int main(){
    struct angestellter *db = malloc(sizeof(struct angestellter)*15);
    db[0].nummer = 1;
    strcpy(db[0].vorname, "George");
    strcpy(db[0].nachname, "Washington");
    print_angestellter(db, 0);
}

Answer 2

You seem to have two issues. First of all, in your print_angestellter() function, you use the global angestellter struct, which is perfectly fine. However, it is a struct , not an array or pointer to memory. Therefore you cannot use the operator [] , hence the errors. It seems that you will have to redesign print_angestellter() or make angestellter an array of struct angestellter s. You also made a similar mistake in main() , doing angestellter[0] instead of db[0] . To fix this, I suggest adding a pointer argument to print_angestellter() , say making it print_angestellter(int, struct *angestellter a) and replacing angestellter[0] with a[0] . You also use = to copy strings, but that copies the pointer addresses, not the contents of the string. Use strcpy() instead.

Answer 3

您写了“ struct angestellter ** db = malloc（sizeof（angestellter）* 15）;”表示数组的指针，可以，但是您应该使用“ db [0]-> number”代替“ angestellter [0]-> nummer“；函数” void print_angestellter（）“，应在使用” malloc“时添加形式参数” dp“，不要忘记使用” free“释放您申请的内存；