index shows NULL inside for loop

Question

I want to multiply matrixes. I create matrix by malloc() function. Index (variable named i) shows NULL, instead of number

int** matrix_a = (int**)malloc(lines * sizeof(int*));
for (int i = 0; i < lines; i++) {
    if (i == NULL) 
        printf("'i' is NULL!");
    matrix_a[i] = (int*)malloc(trans * sizeof(int*));
}

Output:

'i' is NULL!

Error:

Unhandled exception at 0x00007FF6003C35BB in MPI_C.exe: 0xC0000005: Access violation writing location 0xFFFFFFFFABA57A00.

`NULL` could be something like `0` which is the initial value of `i`. Read https://www.tutorialspoint.com/c_standard_library/c_macro_null.htm — J...S, Oct 08 '17 at 13:27
NULL is zero... not magical like in some higher level languages. — Grady Player, Oct 08 '17 at 13:30
@J...S That site looks crappy. Those are not the only possible values of `NULL`, so why mention `0L` specifically? (The C standard also allows `(42 - 42)`, `(1 / sizeof "A")`, etc.) — melpomene, Oct 08 '17 at 13:30
https://stackoverflow.com/a/18781550/5375464 Any integer constant expression with the value `0` or such an expression cast to type `(void *)`. — J...S, Oct 08 '17 at 13:32
Use effective 2D matrices instead of pointer-to-pointer pseudo matrices. — Jens Gustedt, Oct 08 '17 at 14:05
Just curious why all the down votes on this? The user stated what they want to accomplish and provided what they tried. I am not very familiar with C so maybe I am missing something. — Max Young, Oct 08 '17 at 14:43

score 1 · Answer 1 · answered Oct 08 '17 at 13:28

It doesn't make sense to compare non-pointers to NULL. The NULL macro expands to an integer constant 0, possibly cast to (void *).

In the former case, your code just checks if (i == 0), which is true in the first iteration of the loop. In the latter case, you're doing if (i == ((void *)0)), which your compiler should be complaining about because it's a type error.

Only use NULL for pointer values, not numbers.

index shows NULL inside for loop

1 Answers1