Pointer of array output

Question

I have the below program in C:

main() {
    int arr[5]={100,200,300};
    int *ptr1=arr;
    char *ptr2=(char *)arr;
    printf("%d\n",sizeof(int));
    printf("%d   %d",*(ptr1+2),*(ptr2+4));
    return 0; 
}

Its output is as follows:

4
300 -56

Please explain the last output. As per my understanding the last output should be 44 rather than -56.

Ok I had run it in compiler where int is 2 bytes. In that case it gives output as 44. Well it depends also on how many bytes an integer takes. — Sarwan
– Sarwan, Commented Oct 2, 2014 at 11:48
Yea, with two byte ints, that code picks out the first octet of 300, which is stored as 44, 1. — David Heffernan
– David Heffernan, Commented Oct 2, 2014 at 11:54

David Heffernan · Accepted Answer · 2014-10-02 11:51:28Z

4

You are picking out the first octet of arr[1]. Your machine is little endian with 4 byte int and so the 4 octets that make up arr[1] are, written as decimal:

200, 0, 0, 0

So you are interpreting the octet 200 as a char. Your char is signed, and your machine uses two's complement.

The binary representation of 200 is

11001000

Since the most significant bit is set, this is a negative number when viewed as a signed value. So, to work out the magnitude of the negative value, we invert the bits, and add one.

00110111

is 55 in decimal. Hence the value output is -56.

edited Oct 2, 2014 at 11:51

answered Oct 2, 2014 at 11:44

David Heffernan

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1 Comment

Rudy Velthuis Over a year ago

One could also say that, since values wrap around, and char is obviously interpreted as being in the range -128 - 127 (by this particular compiler), 200 is interpreted as the negative value 200-256 = -56.

Serge Ballesta · Accepted Answer · 2014-10-02 11:58:02Z

2

In fact, you expected 44 what it would be if you had sizeof(int) = 2, because then *(ptr2 +4) would point to 300= 256 + 44 with a hex representation of 1002E and a last byte of 2E or 44 in decimal.

To get it independently of the size on an int, you should write :

printf("%d   %d",*(ptr1+2),*(ptr2+(2 + sizeof(int)));

But it will still depend on endianness, and give 0 on a big endian computer ...

answered Oct 2, 2014 at 11:58

Serge Ballesta

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