Basic types in C

Question

Such a quick, odd question: I passed the exam, and the question was:

char c = 'A' char *p =&c char **p2 = &p void *v = &p2 

Enter the type (i.e. int, void *, etc.) of the following expression:

• &v
• p2 + 1
• v[0]

I replied:

• void **
• char**
• ILLEGAL

and lost full credit. I just wanted to get a second opinion from someone before trying to get the points back.

• I was not sure about this, but I thought that since v was a void pointer to a memory address, v would be void** . If v were dereferenced, it would be char**** .

• p2 is char** , so adding 1 to it will still make it char** .

• v[0] does not exist.

If someone could help me with such a stupid problem, I would be very grateful. Thanks.

Answer 1

Your answers are correct for the reasons you gave (separately for the dereferencing bits of v , although it seems that you already know that you cannot play void * ).

Answer 2

p2+1 is pending segfault and v[0] == *(v+0) == *v , which is invalid and therefore invalid. It should be char ** if invented.