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);
}
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);
}
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.
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;
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