chars array in integer representation

Question

I'm reading post about pointers in C here>>>, and there is one example:

main() {
    ...
    char name[] = "Bill";
    int *q;
    ...
    q = name;
    printf("%d\n", *q);
    ...

Which gives me next result:

$ ./pointers_explained 
1819044162

So question is about next explanation from author:

(to see how, line up the binary representation of the ascii values for those 4 characters, and then run the 32 bits together, and convert that resultant binary number as an integer.)

Binary for the "Bill" will be:

B: 01000010
i: 01101001
l: 01101100
l: 01101100

(from an ASCII binary table here>>>)

What I can't get is:

and then run the 32 bits together, and convert that resultant binary number as an integer

So - how can I convert this "Bill"'s binary into the decimal (or any other ) integer "1819044162"?

note that this is undefined behaviour, the "explanation" is details of some particular compiler and system — M.M
– M.M, Commented May 22, 2016 at 8:28
@M.M Yeah, thanks for the clarification, just wondering about how it's working in total. — setevoy
– setevoy, Commented May 22, 2016 at 8:40

delta · Accepted Answer · 2016-05-22 08:29:55Z

3

In ascii code, "Bill" is

B: 0x42
i: 0x69
l: 0x6c
l: 0x6c

1819044162 in hex mode is 0x6c6c6942, this is because of the endianness

In memory, "Bill" is stored as:

     B   i   l   l
name 42  69  6c  6c

But if it is little endianness, and reads an 4-byte integer from address name, it reads backwards, results 0x6c6c6942, which is 1819044162.

If the system is big endian, it will results 0x42696c6c.

edited May 22, 2016 at 8:29

answered May 22, 2016 at 8:24

delta

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1 Comment

David Ranieri Over a year ago

Nice explanation, but it is good to say that this can have alignment problems.

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