What is meant by asking if the pointer that points to an array is true or false?

Question

I'm quite new to programming and have been looking at a few example codes, I would just like to know what is meant by the if(!pi) statement in this code.

// allocate and populate array
    int* pi = new int[N];
    if (!pi)
        return 1;
    for (int i = 0; i < N; i++)
        pi[i] = rand() % 100000;

Does this answer your question? [Checking for NULL pointer in C/C++](https://stackoverflow.com/questions/3825668/checking-for-null-pointer-in-c-c) — rsjaffe, Dec 10 '20 at 16:41

score 6 · Answer 1 · answered Dec 10 '20 at 15:58

6

Pointers are implicitly convertible to bool. If the pointer is null, then the result is false and otherwise true. operator ! is the logical NOT operator.

In this case the check is redundant, because new[] will never return a null pointer and therefore the condition is never true.

answered Dec 10 '20 at 15:58

eerorika

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For illustration, GCC eliminates the check and the corresponding branch under optimizations: https://godbolt.org/z/7T4hfb. – Daniel Langr Dec 10 '20 at 16:07
Thanks for the help. – Alex O'Neill Dec 15 '20 at 18:49

What is meant by asking if the pointer that points to an array is true or false?

1 Answers1