Pointer in C, don't understand how they got this result

Question

main(){
 char i[2];
 * i = 0;
 * (i + 1) = 1;
 printf("len = %d \n",sizeof(int *));  
 printf("i[0] = %d \n",*(int *)i); 
}

where the answer is not 16 the answer is 256 i use the turboc2.0 the hex is 100

Why the answer should be 16? and also, I never experienced **two** `printf()`s producing **one** answer. — Sourav Ghosh, Jan 16 '14 at 13:00
Predicting what this code does is pointless. It's all implementation specific and quite possibly UB too. Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. — David Heffernan, Jan 16 '14 at 13:02
i use turboc2 the the sizeof(int *) is 2 byte i is the array address — user3202574, Jan 16 '14 at 13:05
I can't help getting the feeling OP just bashed on his keyboard at random, without even remotely knowing/understanding what a pionter, an array and a type is — Elias Van Ootegem, Jan 16 '14 at 13:06

score 3 · Accepted Answer · edited May 23 '17 at 11:51

This code depends on your system, specifically on the size of an int.

After initializing, your i array looks like this:

 ------------
|0x00 | 0x01 |
 ------------

Assumed an int is 32 bits on your system:

When casting i to an *int and dereferencing it, there will be four bytes which are accessed (since an int is 32 bits or four bytes):

 --------------------------
|0x00 | 0x01 | 0x?? | 0x?? |
 --------------------------

So, the last two bytes are out of bounds of your array, will have any value, and you will observe undefined behavior (on my system, actually, it prints different values each time I execute the code, like 1762656512, -375848704, ...).

Assumed an int is 16 bits on your system, it gets a littlebit "better":

In this case, when casting i to an *int and dereferencing it, the two bytes will be accessed as a 16 bit value. But, it then still depends on the endianess which value you get:

Little endian: *(int*) i = 0x0100 = 256
Big endian: *(int*) i = 0x0001 = 1

So, if you expect 256, you need to make sure to be on a little endian 16 bit system ...

BTW: When using sizeof() with printf(), make sure to use the %zu format specifier.

i understand this i mistake the byte, one byte equal 8 bit,one hex equal 4 bit, it's big mistake,thanks @Andreas — user3202574, Jan 16 '14 at 13:34

Pointer in C, don't understand how they got this result

1 Answers1