printing values of dynamic array

Question

I have made a dynamic array of integers in C, here is my code

#include <stdio.h>

int main(){
   int count=0, i, input;
   int *myarr;
   myarr=(int*)malloc(4*sizeof(int));

   while(1){
     scanf("%d", &input);
     myarr[count]=input;
     count++;
     if (input == -1) break;
   }

   for (i=0; i<count; i++){
     printf("%d ", myarr[i]);
   }

   return 0;
}

From the code, I thought i clearly made an array of 4 integers only i.e myarr[0] up to myarr[3], how come when i insert even 10 integers, it still prints all of them, it doesn't print garbage as i thought it would after the fourth integer... Maybe i didn't understand the point of dynamic creating an array?? Make me straight please!

@Venkatesh sorry i didn't see that, thanks i fixed it already — user3564573, Nov 17 '14 at 15:19
possible duplication of http://stackoverflow.com/questions/24648635/c-dynamic-pointer-array-allocation-why-can-i-access-elements-out-of-range — Venkatesh, Nov 17 '14 at 15:28
There are many http://stackoverflow.com/questions/1239938/c-accesses-an-array-out-of-bounds-gives-no-error-why — Venkatesh, Nov 17 '14 at 15:31

score 0 · Answer 1 · answered Nov 17 '14 at 15:18

0

You should only access myarr[0] up to and including myarr[3].

Accessing any other index is undefined behaviour: it might work, it might not.

Also, myarr[count]==input looks like a typo. Did you mean myarr[count] = input? The way you have it is testing if myarr[count] equals input. Technically the way you have it is undefined behaviour for any element of myarr since you are making use of uninitialised data.

answered Nov 17 '14 at 15:18

Bathsheba

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So it depends on the compiler?? – user3564573 Nov 17 '14 at 15:20
It could depend on the compiler, the weather, or what you ate last night. – Bathsheba Nov 17 '14 at 15:21
About the assignment operator, i fixed it already.. typing mistake – user3564573 Nov 17 '14 at 15:22

printing values of dynamic array

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