unwanted chars are added at the end of the array in C - c

Question

i want to type a sequence of chars and save them with a temporary array. After that, i want to create the actual array with a certain size with the values of the temporary array. Here is the code:
#include <stdio.h>
int main()
{
char c;
char temp[100];
char array[24];
int i;
char *ptrtemp = temp;
// create a temporary array with getchar function
while(1) {
c = getchar();
if(c == '\n')
break;
*ptrtemp = c;
i++;
ptrtemp++;
}
// type wrong in case of wrong size
if(i != 24) {
printf("Data is wrong");
exit(0);
}
char *ptrarray = array;
char *ptrtemp2 = temp;
// create the actual array
for(i=0;i<24;i++) {
*ptrarray = *ptrtemp2;
if(i == 23)
break;
ptrarray++;
ptrtemp2++;
}
//printing the actual array
printf("\n%s\n", array);
}
However, I get interesting elements after the actual sequence. The array's size was stated as 24 but 25th, 26th, 27th etc. elements are being also printed.
Here is what I get
Every time I try, I see different extra chars. Can anyone explain what is happening here?

Answer 1

You are doing the stuff far too complicated.
First, as already denoted, i is not initialised. Second, you do not leave space for the terminating 0 in the array. Finally, you can easily write to your array directly:
char array[24 + 1]; // need space for the terminating 0 character
for(int i = 0; i < 24; ++i)
{
// write to array directly:
array[i] = getchar();
if(array[i] == '\n')
{
printf("Data is wrong");
return 0;
}
}
array[24] = 0; // this is important if using %s!
printf("\n%s\n", array);
Actually, you have an alternative, as you know you always want to print exactly 24 characters:
char array[24]; // skipping(!) space for the terminating 0 character
for(int i = 0; i < 24; ++i)
{
// just as above
}
// NOT now (would be UB, as array is too short):
// array[24] = 0;
// but now need to give precision to tell how many bytes to write
// (maximally; if a terminating 0 would be encountered before,
// printf would stop there)
printf("\n%.24s\n", array);

Answer 2

I modified the following:
Initialized i to 0 in the beginning.
memset(temp, 0, sizeof(temp));
Added following to make sure that it does not cross array boundary.
if (i == 99)
{
break;
}
Similarly added following for array variable.
memset(array, 0, sizeof(array));
Following check is done beginning of for loop to make sure it does not write to 24th element of array.
if(i == 23)
break;
#include <stdio.h>
int main()
{
char c;
char temp[100];
char array[24];
int i = 0; //Should be initialized to 0
char *ptrtemp = temp;
memset(temp, 0, sizeof(temp));
// create a temporary array with getchar function
while(1) {
c = getchar();
if(c == '\n')
break;
*ptrtemp = c;
i++;
ptrtemp++;
if (i == 99)
{
break;
}
}
char *ptrarray = array;
char *ptrtemp2 = temp;
memset(array, 0, sizeof(array));
// create the actual array
for(i=0;i<24;i++) {
if(i == 23)
break;
*ptrarray = *ptrtemp2;
ptrarray++;
ptrtemp2++;
}
//printing the actual array
printf("\n%s\n", array);
}
Let me know if you face any problem.

Answer 3

Two problems:
You are using i uninitialized in your code. (I don't think the issue you are seeing is because of that)
All C-strings need to be '\0' terminated. In order to ensure that it is the case, you can always do a:
memset(array, 0, sizeof(array));