Array to store integers and string

I want to store 7 integers and 7 string in array

Here is program to store 7 integers

Code:

---

#include <stdio.h>

int main()
{
  int i;
 
  int date[8]= {1,2,3,4,5,6,7};
 
  for ( i = 0; i < 7; i++)
          printf ("Date  %d\n", date[i]);
 
  return 0;
}

---

Here is program to store 7 strings

Code:

---

#include <stdio.h>

int main()
{
  int i;
 
  char days[8][9] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};
 
  for ( i = 0; i < 7; i++)
          printf ("Day %s \n", days[i]);
 
  return 0;
}

---

I have few doubts

What is meaning of this line

Code:

---

int date[8]= {1,2,3,4,5,6,7};

---

I have declared and initialize array

Does it line mention that 7 integers has been stored ?

Code:

---

    char days[8][9] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};

---

Does it line mention that 7 string has been stored ?

If we want to store multiple variables then we use array. I think both lines are storing integers and strings

> Does it line mention that 7 integers has been stored ?
The rule is that if you have fewer initialisers than the declared size, the remainder will be filled with zero (or '\0' for chars, 0.0 for floats or NULL for pointers).

So

Code:

---

int date[8]= {1,2,3,4,5,6,7,0};
and
char days[8][9] = {
    "Monday\0\0\0",
    "Tuesday\0\0",
    "Wednesday",
    "Thursday\0",
    "Friday\0\0\0",
    "Saturday\0",
    "Sunday\0\0\0",
    "\0\0\0\0\0\0\0\0\0"};

---

Incidentally, you need 10 characters to store "Wednesday" as a proper string.

> If we want to store multiple variables then we use array.
> I think both lines are storing integers and strings
You would begin with a structure.

Code:

---

struct day {
  int dayNumber;
  char dayName[10];
};

// Now create an array of the structure and initialise
// with the data we want.
struct day days[7] = {
  { 1, "Monday" },
  { 2, "Tuesday" },
};

---

Quote:

---

Originally Posted by Salem

> Does it line mention that 7 integers has been stored ?
The rule is that if you have fewer initialisers than the declared size, the remainder will be filled with zero (or '\0' for chars, 0.0 for floats or NULL for pointers).

---

Please look at following programs

Code:

---

#include <stdio.h>
int main()
{

  int date[8]= {1,2,3,4,5,6,7};

  return 0;
}

---

Code:

---

#include <stdio.h>
int main()
{
  int i;

  int date[8]= {1,2,3,4,5,6,7};
 
  for ( i = 0; i < 7; i++)
      printf("%d \n "date[i]);


  return 0;
}

---

In first program I am not using printf statement and loop and second program i am using printf statement and loop

I know the printf statment is use to print on screen and for loop to repeat process 8 time

Is my first program store the 1,2,3,4,5,6,7 value ?

I think yes, because if we want to store integer or string we have to declare and initialize variable. I am doing both things here

> I know the printf statment is use to print on screen and for loop to repeat process 8 time
No, for ( i = 0; i < 7; i++) loops 7 times, not 8 times.

Try making the loop < 8, then you'll see that 0 is the last value.