how to give integer value to array inside structure through pointer

Question

I have created a structure and I want to give it a value without creating an object (struct integer a). In string I can use strcpy for it but what can I do for integer values.

#include<stdio.h>
struct integer
{
   int x[5];
};

main()
{
  struct integer *pointer;
  pointer->x[0]=5;
  printf("pointer->x[0]");
}

Answer 1

The problem is you didn't make the poiter point to any object, so either use malloc or use it like this:

int main(){
    struct integer obj;
    struct integer *pointer = &obj;
    pointer->x[0]=5;
    printf("%d\n",pointer->x[0]);
}

Answer 2

As pointer is a pointer you need to allocate memory before using it.

do

pointer = malloc(sizeof(*pointer));
pointer->x[0]=5;

Answer 3

First make the pointer to point to some allocated space.

#include<stdio.h>
struct integer{
int x[5];
};

int main(){
struct integer *pointer = malloc(1*sizeof(struct integer)); //Allocated memory 

pointer->x[0]=5; //Initialize the element(s) of x

printf("%d",pointer->x[0]); //Use proper format specifier for printf 

free(pointer); //Always free memory after use.

return 0;

}

Answer 4

As you are living in the C world you need to use malloc/free.

i.e.

#include<stdio.h>
struct integer{
int x[5];
};
main(){
struct integer *pointer;
pointer = malloc(sizeof(struct integer));
pointer->x[0]=5;
printf("pointer->x[0] = %d", pointer->x[0]);
free(pointer);

}

Should have put in the indentation but could not be bothered to fire up my IDE

Answer 5

strcpy will not work for you if you have not allocated the corresponding memory beforehand.

There is also general function with similar properties to strcpy - memcpy it basically copies the memory pointed by one pointer to another.

#include<stdio.h>
#include <stdlib.h>
struct integer {
   int x[5];
};

main(){
   struct integer *pointer, source;
   source.x[0] = 5;
   pointer = malloc (sizeof(struct integer));
   memcpy(pointer, &source, sizeof(struct integer));
   printf("%d", pointer->x[0]); //will be 5
}

Answer 6

You cannot assign the value without creating the object , the code below can cause segmentation fault as it may be accessing memory not allocated to the program.

main(){
struct integer *pointer; //<-- 'pointer' is just some garbage address
pointer->x[0]=5; //<-- as pointer may contain memory which may not be accessible to current process which causes "Segmentation Fault"
printf("pointer->x[0]");
}

@Ed heal: thanks for correcting me

Either you use [lang: C++]struct integer *pointer=new integer(); [lang: C] pointer = (struct integer*) malloc(sizeof(struct integer)); to allocate memory or do not use pointer, like below

main(){
struct integer *pointer = (struct integer*)malloc(sizeof(struct integer)); //<-- 'pointer' is now pointing to memory allocated for struct 'integer'
pointer->x[0]=5;
printf("pointer->x[0]");
}

Answer 7

The problem is that your pointer didn't point to any object!!!!

You have 3 options:

use malloc like this

struct integer *pointer = (struct integer*)malloc(1*sizeof(struct integer)); 

OR

main()
{
struct integer nonPointer;
nonPointer.x[0]=5;
printf("pointer->x[0] %d",nonPointer.x[0]);
}

OR

 main()
 {
    struct integer integerObj;
    struct integer *pointer=&integerObj;
    pointer->x[0]=5;
    printf("pointer->x[0] %d",pointer->x[0]);
 }

Now about the copy, if you need to copy object (like strcpy with string ) you can use memcpy:

void *memcpy(void *str1, const void *str2, size_t n)

in your case you can use it like this:

struct integer *target, *source;
target=(struct integer*)malloc(1*sizeof(struct integer));
source=(struct integer*)malloc(1*sizeof(struct integer));
source->x[0]=5;

memcpy(target, source, sizeof(struct integer));
printf("target->x[0] %d",target->x[0]);