Hash define to the first element of an array causing a null point exception

Question

I have code that looks like this:

char* talk[516] = {(char*)1};
#define testValue (*(int*)talk[0])

I receive a null pointer exception when the following line of code is then called.

testValue = 0;

Why is that? Haven't all value of the talk[] been initialised?

EDIT

What I want to do is, there are 516 number values (floats and ints) which are stored in char* array. testValue should point to the int value that is stored in the first element of the array. The next value along might be #define testValue2(*(float*)talk[1]).

Because all the elements (i.e `char*`) of `talk` is initialized with value `0` (i.e. null pointer). — songyuanyao, Sep 21 '16 at 06:43
It's still ill-formed. What do want to do? `*(int*)talk[0]` is ill-formed too. — songyuanyao, Sep 21 '16 at 06:51
Someone else has written a file which gives names to all 516 elements in the array using the format above. I just want to store values into the correct element of talk[] using the names they have defined. — Shani de Leeuw, Sep 21 '16 at 06:55
The name in this case is 'testValue'. There are 516 number values (floats and ints) which are stored in char* array. testValue should point to the int value that is stored in the first element of the array. The next value along might be #define testValue2(*(float)*talk[1]) — Shani de Leeuw, Sep 21 '16 at 07:07

songyuanyao · Accepted Answer · 2016-09-21T07:21:52.860

For char* talk[516] = {0};, all the elements (i.e char*) of talk are initialized with value 0 (i.e. null pointer).

For char* talk[516] = {(char*)1};, the 1st element of talk is initialized with value 1, which is not a valid value for pointer. (And the other elements are still initialized as null pointer.) *(int*)talk[0] fails, because you're trying convert a char* to int*, which points to memory address 1, and then dereference on it.

You should initialize elements with valid value, such as:

int i;
float f;
void* talk[516] = {&i, &f, /*...*/};

#define testValue (*(int*)talk[0]) 
#define testValue2 (*(float*)talk[1]) 

testValue = 1; 
testValue2 = 1.5;

^{1. Note if use new to initialize elements of talk, you need to delete them at last.

2. You could use loop to initialize talk too.

3. Using char* instead of void* is confusing, IMO.

4. Consider about using std::map instead of macro magic.}

Interesting - however I was trying to access the first element of the array. — Shani de Leeuw, Sep 21 '16 at 06:53

score 0 · Answer 2 · answered Sep 21 '16 at 06:48

0

char* talk[516] = {0};

This means that the elements of the array talk are all NULL pointers. So when you try to deference it you get that exception.

answered Sep 21 '16 at 06:48

nishantsingh

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score 0 · Answer 3 · edited May 23 '17 at 12:08

0

See the following substitution sequence

testValue = 0;
#
(*(int*)talk[0]) = 0;
#
char* c = talk[0]; // == 1
(*(int*)c) = 0;
#
char* c = talk[0];
int* i = (int*)c; // == ? (*)
(*i) = 0;

* Alignment issues on pointer casting

Casting a char* to int* results in undefined behavior, since int* has stricter alignment requirements than char*.

edited May 23 '17 at 12:08

Community

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answered Sep 21 '16 at 06:53

grek40

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Hash define to the first element of an array causing a null point exception

3 Answers3