2
int arr[10]={1,2,3,4,5,6,7,8,9,10};
printf("%p,%p\n", arr, &arr);
printf("%p,%p\n", arr+1, &arr+1);
return 0;

For this code, GCC compiler return 

0xbfe41348,0xbfe41348
0xbfe4134c,0xbfe41370

The first line is clear, no problem. But the second line makes me confused. The first address moves to the next int, so it is 4 bytes after arr, clear. However, for &arr+1 I thought it will point to the end of the whole array arr[10], so it should add 4*10 to the address. Do I misunderstand something?

Jonathan Leffler
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web rocker
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  • See http://stackoverflow.com/questions/16019009/find-size-of-array-without-using-sizeof-in-c, it should explain it. – nonsensickle Apr 10 '14 at 06:49

4 Answers4

6

The type of &arr is 'pointer to array of 10 int'. Therefore, when you add one to it (&arr+1), it moves to the start of the next array of 10 int, which is 40 bytes beyond the start of &arr.

Jonathan Leffler
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    I don't understand why an answer like this (that adds nothing, just a re-wording what's already inside the question body) would get so many upvotes. Does some people on Stack Overflow have a fan-base following them, praising them with upvotes for whatever they write? – Utkan Gezer Apr 10 '14 at 06:44
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    @ThoAppelsin Those re-wordings can be some times be oooohhhh!!! (i get it now) Factor....:) – Mysterious Jack Apr 10 '14 at 07:45
  • @MudassirHussain Definitely not in this case, *web rocker* already explicitly states that he/she would expect to have 40 added to the address. Probably just never for any case where the guy who asks uses his own genuine words; a re-wording could be useful only maybe when the guy who asks uses a quoted explanation that he's yet to correctly understand, which is not the case here, there aren't any quoted explanations here. – Utkan Gezer Apr 10 '14 at 08:15
6

What you think is right and it is done that way only.

Since &arr => 0xbfe41348 and

0xbfe41348 + 0x28(4*10 in decimal) = 0xbfe41370

I think you got confused due to addition of decimal to a hexadecimal number.

0xF1
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2

&arr+1 does, in fact, add 40 to the base address of the array arr but it's not obvious since the addresses are in hexadecimal or base 16, not decimal.

Now, for what it's worth, I'll add some explanation of each statement to make things clear. 

int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

The above statement defines arr to be of type int[10], i.e., an array of 10 integers. It then initializes arr with an array initializer list.

printf("%p,%p\n", arr, &arr);

In the above statement arr decays (evaluates or implicitly converted) to a pointer to its first element. Therefore its type is int *. &arr evaluates to a pointer to arr. Type of arr is int[10]. Therefore, type of &arr is int (*)[10], i.e., a pointer to an array of 10 integers. Parentheses are used because array subscript operator [] has higher precedence than * operator. So without parentheses int *[10] means an array of 10 pointers to integers. This is one of the cases where an array does not decay into a pointer to its first element.

Please note that both arr and &arr evaluate to the same value, i.e., the base address of the array in the above printf statement but their types are different and they have different pointer arithmetic. This shows up in the following statement - 

printf("%p,%p\n", arr+1, &arr+1);

arr+1 points to the next element. Here element type is int. Therefore arr+1 evaluates to

arr + (1 * sizeof(int))

&arr + 1 also points to the next element but here the element type is int[10] - an array of 10 integers. Therefore &arr + 1 evaluates to 

arr + (1 * sizeof(int[10]))
ajay
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  • Thanks a lot for your explanation, I'm much clearer now. Since I made a mistake about the data format, and the accepted answer point it out, I chose that. But still thank you very much. – web rocker Apr 10 '14 at 09:58
1
arr+1

Here arr is base pointer to integer array so results in incremented by sizeof(int)

&arr+1

Here &arr results in address of array so result is incremented by array size.

Jonathan Leffler
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LearningC
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