3d array using pointers

Question

I am new to C programming. Currently am trying to learn 3d array using pointers. Below is a program I am trying to debug. Can any one explain the difference between the two programs given below?

code1:

#include <stdio.h>
int main()
{
    int a;
    int d[2][2][2] = {1, -2, -3, 0, -9, -1, 3, -1};
    printf("%d\n",*(*(*(d +1)+1)+1));
    if(*(*(*(d +1)+1)+1) <(a= sizeof( int )))
        puts(" u got it ");
    else
        puts (" but wrong");
    return 0;
}

code2:

#include <stdio.h>
int main()
{
    int a;
    int d[2][2][2] = {1, -2, -3, 0, -9, -1, 3, -1};
    if(*(*(*(d +1)+1)+1) <(sizeof( int )))
        puts(" u got it ");
    else
        puts (" but wrong");
    return 0;
}

In the first code I am getting the […incomplete…]

Related: stackoverflow.com/questions/1683432 and certainly a duplicate (perhaps there is a better one, though). Look for integer promotion, (size_t)-1 is SIZE_MAX, which isn't smaller than the size of an int. The assignment in the first snippet converts the size_t to an int so the left operand of < isn't promoted. — mafso
– mafso, Commented Jul 30, 2014 at 16:04
Huh? I don't see size_t in this post, but I do see sizeof(int) which would probably be 4. — David Grayson
– David Grayson, Commented Jul 30, 2014 at 16:22
@david yes sizeof(int ) is 4.after assigning to a variable a ,what difference it make.What is the difference between the codes? — vipil
– vipil, Commented Jul 30, 2014 at 16:24
@DavidGrayson: The result of the sizeof operator is of type size_t, which is unsigned. — mafso
– mafso, Commented Jul 30, 2014 at 16:26
-1 < sizeof(int) : results sizeof Since unsigned, -1 is compared by converting into unsigned. -1 < (a=sizeof(int)) : results (a=sizeof(int)) Since signed(because type of a is signed int). — BLUEPIXY
– BLUEPIXY, Commented Jul 31, 2014 at 1:13

M.M · Accepted Answer · 2014-08-02 06:35:31Z

int d[2][2][2] = {1, -2, -3, 0, -9, -1, 3, -1};

The initializer is not fully braced but in this situation the initializers apply to the next element of the array in memory, so this assigns d[0][0][0] = 1, d[0][0][1] = -2, d[0][1][0] = -3, etc.

printf("%d\n",*(*(*(d +1)+1)+1));

The thing full of stars is an obfuscated way of writing d[1][1][1] . The definition of X[Y] is *(X+Y) .

(a= sizeof( int )))

The type of an assignment expression is the type of the left-hand operand. So the first program does (int)-1 < (int)4 . The second program does (int)-1 < (size_t)4 . (assuming your ints are 4 bytes big).

In the first case this is true. In the second case it is a type mismatch. The type mismatch has to be fixed before the comparison can occur. The rules of C say that in this case the signed type is converted to the unsigned type, giving (size_t)-1 < (size_t)4. Since (size_t)-1 is actually the largest possible size_t value, this comparison is false.

Community · Accepted Answer · 2017-05-23 12:06:43Z

1

It is basically an integer type comparison problem. First program compares a signed int(-1) with a signed int(a) and the second, a signed int(-1) with an unsigned int(sizeof()). Integer promotion happens in the second case, where signed int(-1) gets converted to unsigned int(-1) -> SIZE_MAX. For more details on type comparison check the thread, What are the general rules for comparing different data types in C?

edited May 23, 2017 at 12:06

CommunityBot

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answered Aug 2, 2014 at 4:21

Nandyy

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3 Comments

Jonathan Leffler Over a year ago

+1, but I have some concerns. The second program compares a (signed) int with a size_t (which is an unsigned integer type, but often not unsigned int — which is my main nitpick).

Nandyy Over a year ago

@JonathanLeffler rightly pointed, in the second case signed int(-1) is compared with size_t(-1). link explains this better.

Nandyy Over a year ago

bracing the initializer may help understand better int d[2][2][2] = { { {1, -2} , {-3, 0} }, { {-9, -1} , {3, -1} } };

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