传递结构数组

Question

How can I pass an array of structures?

So far I have this which is global:

typedef struct _line
{
  float val1;
  int val2;
} line;

Then I read data from a file into this structure

struct _line* read_file()
{
  typedef struct _line *Lines
  Lines *array = malloc(num_lines * sizeof(Lines));
  //read values into structures here

Then I fill up the structures in the array with values. If I do printf("%d", (*array[1]).val1); I get the right value here in this particular method

Then I return the array like so

return *array

But when I do so, only the 0th structure reads correctly in the method I returned to. Reading the 1st element just prints random values. What am I doing incorrectly?

Answer 1

You should not dereference the array when you return it ¹ , it's actually of incompatible type with the function return type, just

return array;

also, check that array != NULL after malloc() before reading the values, and you don't really need the typedef it makes your code a bit confusing.

If your code compiled which I doubt, then you don't have warnings enabled in your compiler command, enable them so you can prevent this kind of issue.

---

(1) *array is equivalent to array[0] .

Answer 2

Expanding on my comments, your code (as you describe and show it) you have undefined behavior :

This is because you allocate an array of pointers , but you apparently do not allocate the pointers in that array. So when you dereference a pointer (which you haven't allocated and whose value is indeterminate and so will point to a seemingly random location) you have this undefined behavior.

Instead of using a type-alias like Line use the structure name, like

struct _line *array = malloc(num_lines * sizeof(*array));

That will allocate num_lines structures (instead of pointers), then you use it like a normal array, without the pointer dereferencing

array[x].val1 = something;

And you of course return that pointer array as-is:

return array;