Zero in int array treat as null not digit

Zero in int array treat as null not digit

1) how can i use as digit
2) when i remove zero then it print four digit in array and print some garbage value, how remove this remedies

Code:

---

#include<iostream>
using namespace std;

void check(int *t){
    while(*t != NULL){
    cout << *t++ << endl;
    t++;
  }
}

int main(){
        int a[] = {5,4,7,8,0};
        check(a);
}

---

print when int array does not contain zero

Code:

---

5
4
7
8
4636772
21473443
4009792
2293664
4198619
1
4013768
4009224
4718596
2293652
-1
2293648

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First, you should compare with 0 (it's an array of integers), not NULL (which is for pointers).

Plus, how did you get those results, when you're incrementing t TWICE in each loop?

++ has higher precedence than *, both t++ increment the pointer. Is this your intention

Indeed, this cannot be the real code. This code will only output 5 and 7. Remove the extra increment.
In your real code you need to pass the length of the array, obtained through using sizeof, into the check function.

that's right！ the size of the array should be included in the display function not compare with NULL， because the compile just take 0 as NULL， so when *t=0，the loop is over ，and you do twice increment in one loop while once is ok。
so you can write like this ：

Code:

---

#include<iostream>
using namespace std;

void check(int *t, size_t size){
        int i = 0;
        while(i++ < size){
                cout << *t++ << endl;
                //t++;
        }
}

int main(){
        int a[] = {5,4,7,8,0};
        check(a,sizeof(a)/sizeof(*a));
}

---

is this hopeful for you ？

Better yet, use std::vector.