How to input 8 byte hexadecimal number into char array?

Question

I want to generate sequence of hexadecimal numbers starting with 07060504003020100. Next number would be 0f0e0d0c0b0a0908 etc in that order.

When I use unsigned long long int and output the data the first 4 bits, meaning that 0 is truncated. and it prints 706050403020100.

So I was thinking of putting the number into a char array buffer(or some other buffer) and then print the output so that later if I want to compare I can do character/byte wise compare.

Can anybody help me out? char[8]="0x0706050403020100" doesn't look right.

Kricket · Accepted Answer · 2010-08-19 14:45:36Z

2

What are you using to print out your value? The syntax for printing 8 0-padded bytes to stdout would be something like

printf("%016X", value);

It breaks down like this:

%x    = lowercase hex
%X    = uppercase hex
%16X  = always 16 characters (adds spaces)
%016X = zero-pad so that the result is always 16 characters

For example, you could do the same thing in base 10:

printf("%05d", 3); // output: 00003

You can also play with decimals:

printf("%.05f", 0.5); // output: 0.50000

http://www.cplusplus.com/reference/clibrary/cstdio/printf/

edited Aug 19, 2010 at 14:45

answered Aug 19, 2010 at 11:16

Kricket

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7 Comments

Unicorn Over a year ago

No i wanted to know how can store my initial value 0x0706050403020100. can i store it using character array or dynamically using malloc or using unsigned long long int ??

Kricket Over a year ago

You can store your value anywhere where you've allocated at least 8 bytes of memory. This could be a long (depending on the hardware), a char[8], a long double (which might actually be 16 bytes), a struct that you make, etc. In your initial post you said your problem was that you don't see the first hex character when you output your number. This is normal - it's just like, when you output the value 3, you often don't want to see 0000000003. If you want leading zeroes, you have to tell printf to include them. But I can assure you that your machine has not "forgotten" about a nybble.

Kricket Over a year ago

I also might add that doing char[8]="0x0706050403020100"; will probably produce a segfault, because you're declaring 8 bytes and filling it with 18.

Unicorn Over a year ago

@Rider:Thanks.But hw to tell printf to include leading zeroes.

Unicorn Over a year ago

So how should i declare using char(static or dynamically)??

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