Array of struct with pointer

Question

typedef struct line {
  int a;
  int b;
} line;

int main() {
  line *v;
  int c, d, j;
  char a;
  int i;
  scanf("%d", &n)
  for (i = 0; i < n; i++) {
    scanf("%c", &a);
    v = (line*) malloc(n * sizeof(line));
    if (a == '+') {
      scanf("%d %d", &v[j]->a, &v[j]->b);
    }
  }

I want to make an array that holds a struct info, and then use that info in my function . But I am getting some errors and I don't know if I used well the pointer.

I have tried with v[j].a but it didn't work. (I used this because I am more familiar from linked-list.)

Bobo, Code is missing declaration of n. Suggest posting compilable code. — chux
– chux, Commented Apr 19, 2018 at 22:29
Rather than just say, "I am getting some errors", post the exact text of the error. That helps all of us. "it didn't work." is vague. Post how you determined or the evidence of how it "did not work". — chux
– chux, Commented Apr 19, 2018 at 22:30
Common error: scanf("%c", &a); --> scanf(" %c", &a); (Add space). — chux
– chux, Commented Apr 19, 2018 at 22:32
@P.Antoniadis , ty for pointing me the mistake at j , but i still get an error invalid type argument of ‘->’ (have ‘line {aka struct line}’) — Bobo Lemnaru
– Bobo Lemnaru, Commented Apr 19, 2018 at 22:33
regarding: scanf("%d", &n) This statement is missing the trailing ; The posted code does not compile! amongst other things it is missing the #include statement for stdio.h — user3629249
– user3629249, Commented Apr 21, 2018 at 17:24

Panos Antoniadis · Accepted Answer · 2018-04-19 22:50:37Z

1

Use &(v[j].a) instead, which is equal to &((v+j)->a). The name of an array is a pointer to the first element. So v[j] is an element of type struct line. In order to get one of its fields you use .a.

Also, check your code for some other errors because some variebles are not initialized.

edited Apr 19, 2018 at 22:50

answered Apr 19, 2018 at 22:26

Panos Antoniadis

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2 Comments

0___________ Over a year ago

@P.Antoniadis &(v[j].a) == &v[j].a. In this program OP allocates the array of the structs not the array to the structs.

Panos Antoniadis Over a year ago

@PeterJ_01 can you explain this a little bit?

0___________ · Accepted Answer · 2018-04-19 22:39:06Z

0

First of all you malloc a new memory on every iteration and the previous one becomes unavailable.

Abstracting from the logic of the program

typedef struct line {
  int a;
  int b;
} line;


int main(void) {
  line *v;
  int c, d, j;
  char a;
  int i,n;

  scanf("%d", &n);
  v = malloc(n * sizeof(line));
  if(v == NULL) return -1;
  for (i = 0; i < n; i++) {
    scanf("%c", &a);
    if (a == '+') {
      scanf("%d %d", &v[j].a, &v[j].b);
    }
  }
}

answered Apr 19, 2018 at 22:39

0___________

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2 Comments

chux Over a year ago

scanf("%c", &a); --> Hmmm. Likely a will get the value of '\n'.

0___________ Over a year ago

@chux - as I wrote -> abstracting from the logic of the program. I an not here to teach C. My answer is to show how to malloc and access the fields of struct in the allocated table and their addresses

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