how to change a character array in typedef struct from a function?

Question

i'm trying to change array a inside set of typedef through function readset but it doesn't do anything to it and im trying to understand how to use a pointer so i would just send it to the function but its not working how to do that?

#include <stdlib.h>
#include <string.h>

typedef struct set {
    char a[128];
}set;

 set SETA={""};

void readset (set rset)
{
    char tmp[128];
    int i;
    i=0;
    while(i<128)
        rset.a[i++]='0';
    i=0;
    while(i<128)
        tmp[i++]='0';

        tmp[arr[1]-1]='1';
    strcpy(rset.a, tmp);
    printf("tmp in function is: %s\n",tmp);

}

int main()
{
    int i=0;

    while(i<128)
        SETA.a[i++]='0';
    printf("\n");
    printf("setA before: %s\n",SETA.a);
    readset(setPtr);
    printf("setA after: %s\n",SETA.a);

return 0;
} ```

Please make a [mre]. setPtr is undefined – Yunnosch Jan 11 '20 at 15:15 — Yunnosch, Jan 11 '20 at 15:15

ssd · Answer 1 · 2020-01-11T15:39:45.230

Either you need to define a separate pointer named setPtr and assign the address of SETA, just before calling the function readset, like;

set *setPtr = &SETA;
readset(setPtr);

or, just send the memory address of the SETA to readset;

readset(&SETA);

Btw, you may need to change the function decoration as well, like;

void readset (set *rset) {
    ...
        rset->a[i++]='0';
    ...
}

In addition, you may use a string terminating character (\0) at the end of your char array in order to get the printf & strcpy functions do their task properly.

score 1 · Accepted Answer · answered Jan 11 '20 at 15:40

Say you have a struct, defined like this:

typedef struct bar_s
{
    char data[128];
    int someInt;
} bar_t;

You would define your function like this:

void foo(bar_t* p) //Note the asterisk
{
    strcpy(p->data, "Hello!"); //copy a string into data field
    p->someInt = 0; //assign value to someInt field. Note the "->"
    (*p).someInt = 0; //Same as above, using a dereference operator.
}

Now how do you pass a pointer to this function? You would do something like this:

bar_t sample;
foo(&sample); // & = address-of operator.

Your readSet function needs a bit of work to get it functional, but that's how to work with pointers in C. In your case you would do something like this:

void readSet(set* rset)
{
    int i = 0;
    for(i = 0; i < 127; ++i)
    {
        rset->a[i] = '0';
    }

    //Note that all strings in C must be terminated with a null character.
    //So, let's put one at the last place in the array.
    rset->a[127] = '\0';
    ...
}

so my understanding is i must use a pointer from that kind of typedef in order to change a variable inside the struct And the mistake i did with the last bit of the string array of not putting null Thanks i got it ! — AdmDTR, Jan 12 '20 at 12:51

score 1 · Answer 3 · answered Jan 11 '20 at 16:14

structs and unions are passed by the value so you can: https://godbolt.org/z/3NUWHW

Pass it by value

struct str
{
  char a[100];
  int x;
  /*  ... */
};

struct str read_a(struct str s)
{ 
   strcpy(s.a, "Test");
   return s;
}

int main()
{
   struct str s;
   s.x = 5;
   s = read_a(s);

   printf("s.a=%s, s.x=%d\n",s.a,s.x);
}

Pass by pointer

struct str
{
  char a[100];
  int x;
  /*  ... */
};

struct str *read_a(struct str *s)
{ 
   strcpy(s->a, "Test1");
   return s;
}

int main()
{
   struct str s;
   s.x = 6;
   read_a(&s);

   printf("s.a=%s, s.x=%d\n",s.a,s.x);
}

how to change a character array in typedef struct from a function?

3 Answers3