array of structs and 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);
}

angestellter is a single variable of type struct angestellter; you cannot use subscripting on it. — Jonathan Leffler
– Jonathan Leffler, Commented Jul 28, 2016 at 23:34
It looks like you are allocating memory and assigning it to a variable named db. Shouldn't you use db instead of angestellter? — bruceg
– bruceg, Commented Jul 28, 2016 at 23:47

dbush · Accepted Answer · 2016-07-28 23:48:28Z

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);
}

answered Jul 28, 2016 at 23:48

dbush

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1 Comment

chqrlie Over a year ago

print_angestellter could pass the struct members directly: printf("%d, %s, %s\n", db[x].nummer, db[x].vorname, db[x].nachname);

DeepCoder · Accepted Answer · 2016-07-28 23:47:20Z

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 angestellteran array of struct angestellters. 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.

wenwei · Accepted Answer · 2016-07-29 05:32:55Z

0

you wrote "struct angestellter **db = malloc(sizeof(angestellter)*15);"means pointer of array,its ok,but you should use "db[0]->number" to replace "angestellter[0] -> nummer";The function "void print_angestellter()",you should add formal parameter"dp",when you use "malloc",don't forget use "free" to free the memory you apply；

answered Jul 29, 2016 at 5:32

wenwei

1

1 Comment

David C. Rankin Over a year ago

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