I want to initialize a couple of arrays which are members of a struct which is passed to a function by reference.
Appreciate any help.
typedef struct Snake_pos
{
char field[10][10];
int Y[10];
int X[10];
}snake_pos;
int main()
{
snake_pos pos1;
pos_init(&pos1);
return 0;
}
void pos_init(snake_pos *pos1)
{
pos1->X={};
pos1->Y={};
pos1->field={};
}
You can only use the syntax = {} when defining the variable. So to zero every member you can either define it as
snake_pos pos1 = { 0 };
Or if it is passed to a function, like this
void pos_init(snake_pos *pos1)
{
memset(pos1, 0, sizeof *pos1);
}
You can access any array element inside of an array through a
structure pointer this way :
for 1 D array :
structure_pointer->array_valiable[index1]
for 2 D array :
structure_pointer->array_valiable[index1][index2]
so now, you can initialize each member of arrays X , Y and field of structure snake_pos using scanf() this way :
#include <stdio.h>
typedef struct Snake_pos
{
char field[10][10];
int Y[10];
int X[10];
}snake_pos;
void pos_init(snake_pos *pos1)
{
int i,j;
//initializing each member
//of array X
for(i=0;i<10;i++)
scanf("%d",&pos1->X[i]);
//of array Y
for(i=0;i<10;i++)
scanf("%d",&pos1->Y[i]);
//of the two dimensional array field
for(i=0;i<10;i++)
for(j=0;j<10;j++)
scanf(" %c",&pos1->field[i][j]);
//notice the space before %c is to consume white spaces returning from before scanf's
}
int main()
{
snake_pos pos1;
pos_init(&pos1);
return 0;
}
Note : if you want to initialize all members to a single fixed value,
then you can instead of using scanf() in for loop, assign the value
this way :
for(i=0;i<10;i++) //initializes all members with a value of 0
pos1->X[i]=0;
If you want to initialize with zero's, Why not memset the structure ?
void pos_init(snake_pos *pos1)
{
memset(pos1, 0, sizeof(snake_pos));
}
Related
I just start to learn pointers to structures and I'm confused.I have to create a type of data ARRAY (which is associated with an array which contains integers.) like a structure which contains: numbers of array's elements and the array's elements stored in a part of memory(heap), dynamically allocated.
So I wrote:
typedef struct ARRAY
{
int nrElem; // number of elements
int *v[100];
};
Now I need to create 2 functions, one for reading an array from keyboard and the second one to display it using the structure I declared.
I tried but I get stuck.
void arrayDisplay(ARRAY *ps)
{
int i;
for(i=0;i<pd->nrElem;++i)
{
printf("%d",)
}
}
void readArray(ARRAY *ps)
{
int i;
for(i=0;i<pd->nrElem;++i)
{
printf("%d",)
scanf("%d",&);
}
}
How to continue?
Instead of an array of pointers int *v[100]; you need an array of ints int v[100]; in your data structure.
See code below:
#include <stdio.h>
#include <stdlib.h>
typedef struct ARRAY
{
int nrElem; // number of elements
int v[100];
} ARRAY;
void arrayDisplay(ARRAY *ps)
{
int i;
for(i=0;i<ps->nrElem;++i)
{
printf("%d\n", ps->v[i]);
}
}
void readArray(ARRAY *ps)
{
int i;
for(i=0;i<ps->nrElem;++i)
{
printf("%d: ", i);
scanf("%d",&ps->v[i]);
}
}
int main()
{
ARRAY a;
a.nrElem = 5;
readArray(&a);
arrayDisplay(&a);
return 0;
}
If you really want to use an array of int pointers you need to allocate the array first. And a different level of redirection for printf and scanf. But I'm not sure why you want to allocate memory for an integer array like this.
typedef struct ARRAY
{
int nrElem; // number of elements
int *v[100];
} ARRAY;
void arrayDisplay(ARRAY *ps)
{
int i;
for(i=0;i<ps->nrElem;++i)
{
printf("%d\n", *ps->v[i]);
}
}
void readArray(ARRAY *ps)
{
int i;
for(i=0;i<ps->nrElem;++i)
{
printf("%d: ", i);
scanf("%d",ps->v[i]);
}
}
int main()
{
ARRAY a;
int i;
a.nrElem = 5;
for(i=0;i<a.nrElem;++i) {
a.v[i] = (int*)malloc(sizeof(a.v[i]));
}
readArray(&a);
arrayDisplay(&a);
return 0;
}
I have following struct defined which I can not change:
typedef struct _content {
int length;
char **lines;
} content_t;
I initialized it in the main function like that:
struct _content cont;
cont.length = 6;
cont.lines[cont.length-1][255];
I try to assign a value to a certain element of the multidimensional array and also
to print the value out- which doesn't work:
cont.lines[1][1] = "A";
printf("\n\n%c", cont.lines[1][1]);
What am I doing wrong here?
Many Thanks
int main(){
content_t cont;
cont.length = 6;
cont.lines=malloc(sizeof(char*)*cont.length);
for(int i=0;i<cont.length;i++) cont.lines[i]=malloc(255);
printf("\n\n%c", cont.lines[1][1]);
for(int i=0;i<cont.length;i++) free(cont.lines[i]);
free(cont.lines);
}
that's the right way to do it
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 100000
typedef struct {
int day;
int month;
int year;
} DATE;
typedef struct {
char name[100];
int age;
float hrlyWage;
float hrsWorked;
float regPay;
float otPay;
float totalPay;
DATE payDate;
} PAYRECORD;
int newRecord(struct PAYRECORD record[], int index){
//set name to \0 so it can work as string
record[index].name = {'\0'};
index++;
return index;
}
int main(){
char menuChoice = 'X';
struct PAYRECORD record[SIZE];
int index = 0;
while (menuChoice != 'Q'){
system("pause");
system("cls");
menuChoice = runMenu();
switch (menuChoice){
case 'A':
index = newRecord(record, index);
}
}
}
main sets up an array of structs the gets passed into newRecord, and the goal is to make it so that I can input the data here and then return the new index to keep track of my array of structs. However something is going wrong where my program doesn't seem to be recognizing newRecord as a function, which ends up throwing the whole program off.
I get syntax errors for all the functions inside of newRecord, though I beleive it's because, as I mentioned, the program seems to be unable to recognize newRecord as a User defined Function.
Use of struct PAYRECORD is wrong since there is no such type. You only have a typedef named PAYRECORD.
If you want to be able to use struct PAYRECORD as well as just PAYRECORD, change the definition of the struct to:
typedef struct PAYRECORD {
char name[100];
int age;
float hrlyWage;
float hrsWorked;
float regPay;
float otPay;
float totalPay;
DATE payDate;
} PAYRECORD;
If that's not your goal, change the use of struct PAYRECORD by just PAYRECORD.
Also, the line:
record[index].name = {'\0'};
in newRecord is not correct. You cannot assign to an array like that. Change it to:
record[index].name[0] = '\0';
The struct PAYRECORD does not exist, the compiler has no idea how big that is.
Note that PAYRECORD is a typedef to an anonymous struct. So your function
should look like this:
int newRecord(PAYRECORD record[], int index){
//set name to \0 so it can work as string
record[index].name[0] = 0;
index++;
return index;
}
Also note that {'\0'}; works only when initializing a array when you declare
it:
char arr1[10] = { '\0' }; // OK
char arr2[10];
arr2 = { '\0' }; // NOT OK
// error: expected expression before ‘{’ token
// a = { '\0' };
// ^
And when writing functions that take arrays as an argument, you should also pass
the size of the array.
int newRecord(PAYRECORD record[], int index, size_t len){
if(record == NULL)
return -1; // error, NULL passed
if(index >= len)
return -1; // error, cannot access array
record[index].name[0] = 0;
index++;
return index;
}
And then you can call it from main like this:
PAYRECORD record[SIZE];
...
int index = 0;
if(newRecord(record, index, sizeof record / sizeof *record) != index)
{
// error handling
}
This makes the code more robust. You always have to check the array boundaries,
otherwise you might read/write out of bounds. And also check that NULL has not
been passed as well, if you dereference NULL, your program will crash with
segfault.
Also, the parameter to newRecord could be a PAYARRAY, not an array directly; based on declaring
typedef struct { } PAYRECORD, PAYARRAY[SIZE];
int newRecord(PAYARRAY record, int index) {...}
int main(){
...
PAYARRAY record;
...
case 'A':
index = newRecord(&record, index);
}
The compiler should be converting the PAYARRAY or PAYRECORD[] argument to a PAYRECORD * pointing to the first element, so use of the '&' is indicated for the function call.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct data{
char name[20];
char lastname[25];
int age;
}person;
void insert(person *p,int *num);
int main()
{
int num;
person p;
insert(&p,&num);
printf("Name: %s",p[0].nome); /* Here i would print the first struct by
my array, but: is not array or not pointer Why?? */
}
void insert(person *p, int *num)
{
int dim;
person *arr;
printf("Insert how many people do you want? "); /* How many index the
array should have */
scanf("%d",&dim);
arr = (person *) malloc(dim*sizeof(person)); /* I'm not sure for
this explicit cast. */
for(int i = 0; i < dim; i++)
{
printf("Insert name: ");
scanf("%s",arr[i].name);
printf("Insert lastname: ");
scanf("%s",arr[i].lastname);
printf("Insert age:': ");
scanf("%d",&arr[i].age);
}
*num = dim;
*p = *arr;
}
I've tried: `person *insert(int *num)
And it's works,but how can pass an array reference?`
This programm should ask how many person do you want to insert ( in function insert) and with a for, he should ask name,surname,age.
After the insert, he should print, but for quick, i would tried with first element (index) of array (structs).
You can't return entire array from the function, but you can return the base location of the array. For example you can do like this : person *insert(int *sz);. But i see in your code you're passing &p and the &num variable into the insert method, maybe you want to modify them within that function and manipulate it afterwards in your main(). For that i would have these recommendations:
Change Line 16 person p to person *p. Since p is supposed to hold base value of an array. Remember array name is nothing but a base address to the first element of the list.
Change your function definition to recieve person** rather than person*. Since you want to modify a pointer variable and therefore you'd need a pointer to a pointer variable. Change it like this:` void insert(person **p, int *num)
Free the memory after usage; add a free(p) at the end of main.
`
How should I parse an array of structures as parameter to a function?
For example, I have the following structure definition:
struct Town
{
char *TownName;
char **GiftList;
int *GiftCount;
int GiftTypes;
};
and a declaration of an array of such structures, in my main:
struct Town TownList[100];
struct Town AuxiliaryStructure;
I have written a custom sorting function for this array, in which I want to make use of each structure's fields, but I do not know how to provide the array TownList[100] to the sort function.
To pass an array of anything to a function, you can just pass a pointer to the first element plus an array length:
struct mystruct {
char* something;
/* ... */
}
struct mystruct myarray[100];
void do_something(struct mystruct* array, int length)
{
int i;
for (i=0; i<length; ++i)
{
array[i].something = ...
}
}
int main(void)
{
do_something(myarray, 100);
return 0;
}
Based on what you wrote, you would reference it in the caller and pass the result. For example:
void sort(char* str){
...
}
int main(){
for(int i=0; i<100; ++i){
sort(array[i].something);
}
}
Alternately you may want to pass the full array and handle it in another function, which would be like this:
void sort(struct mystruct * array){
...
}
int main(){
for(int i=0; i<100; ++i){
sort(array);
}
}
If, on the other hand, you meant that you want to sort the outer array of structures by the contents of something, then you would either have to pass the array as above and implement your own sort, or use a sorting function that takes a function pointer so you can write a comparator. The latter is available in the standard C library, and could be used something like this.
#include <stdlib.h>
int compare_mystruct_by_something(const void *a, const void *b){
return strcmp(((struct mystruct*)a)->something, ((struct mystruct *)b)->something);
}
int main(){
qsort(array, 100, sizeof(struct mystruct), compare_mystruct_by_something);
}