c将结构传递给更新方法

Question

I'm having some issues with the following code snippet:

#include <stdio.h>

struct some_numbers 
{
int id;
char *somestring;
};

typedef struct some_numbers numb;

void print_numbers(numb *a)
{
printf("%d: %s\n", a->id, a->somestring);
}

void add_number(numb *a)
{

 // do someting magical
 // push the new result to the existing struct
 // put something into like:
  a->somestring[5] = "widdley";
}

int main(void)
{

// put some stuff in the struct
numb entries[50];
int x;
for(x=0; x < 4; x++)
{
    numb a = entries[x];
    a.id = x;
    a.somestring = "goats";
    print_numbers(&a);
}

add_numbers(&a);  // i want to call a method 

return 0;
}

I want to create an array of structs, pass the struct to a method, and pop more items into the array. Everything I've tried thus far has failed miserably, and I'm having a hard time thinking my way out of this conundrum. I can print the values without any issue:

> ./struct 
0: goats
1: goats
2: goats
3: goats
> 

I would like the output to look like:

> ./struct 
0: goats
1: goats
2: goats
3: goats
4: widdley
>

Please help. I'm not good at c, so be gentle!

edit: clarified the code example to take the focus off of the wrong areas.

Answer 1

Here:

a->somestring[5] = "widdley";

the type of somestring[5] is char , not char* . If you need a string array you need to define:

struct some_numbers {
  int id;
  char *somestring[20];  // 20 is an example
};

And somehow manage these strings depending on your actual goal.

If you want to add a new number into entries then define it with more than 4 items and keep track of valid locations:

numb entries[20]; // 20 is an example
int num_entries = 0;
entries[num_entries++] = new_entry(); // some function that returns an entry

or just use a dynamic array, which requires dynamic memory management (malloc/realloc);

#include <stdio.h>

#include <stdlib.h>
#include <string.h>
struct some_numbers 
{
  int id; 
  char *somestring;
};

typedef struct some_numbers numb;

void print_numbers(numb *a) {
  printf("%d: %s\n", a->id, a->somestring);
}

void add_entry(numb **list, int *n, int id, const char *str) {
  int cnt = *n; 
  *list = realloc(*list, sizeof(numb)*(cnt + 1));
  (*list)[cnt].id = id; 
  (*list)[cnt].somestring = malloc(strlen(str)+1);
  strcpy((*list)[cnt].somestring, str);
  *n = cnt + 1;
}

int main(void)
{

  // put some stuff in the struct
  numb *entries = 0;
  int x, num_entries=0;
  for(x=0; x < 4; x++)
  {
    add_entry(&entries, &num_entries, x, "goats");
  }

  for (x=0; x<num_entries; x++)
    print_numbers(&entries[x]);
  printf("\n\n");
  add_entry(&entries, &num_entries, 6, "widdley"); 
  for (x=0; x<num_entries; x++) 
    print_numbers(&entries[x]);

  return 0;
}

Answer 2

If you want to add more values to your array, you need to either allocate enough memory upfront to store the maximum possible number of structs, or allocate the memory dynamically. So:

numb entries[100];

Or

numb *entries = malloc(sizeof(numb)*100);

Then, you'll need to pass a variable to the add_number function to keep track of where your array ends:

void add_number(numb *a, int position) {
        a[position].somestring = "widdley";
}

Answer 3

You commented out the call to add_numbers() , so of course, the structs in your array won't change. I suspect you did this because you are getting a compiler error. a->somestring[5] = "widdley"; should be a->somestring = "widdley"; since you want to set the value of the whole string, not just one char in that string. In the future, please post any compiler errors that you get. After this one change to your code, you should print out the array after calling add_numbers() .