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Unsigned Long Int to Byte?

Posted on 2009-05-03
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Last Modified: 2013-11-05
Hi, I have an unsigned long int number (only 0 and 1) and I would like to take the first 8 digits and put them in a byte. Example:

unsigned long int code = 01100011001110;   //13 digits

char byte = ---8 first digits from code---

---remove the 8 first digits from code and place the rest of them at the beggining---

printf("%ld", code);  //this will printf "001110"  (the last 6 digits)

Thanks
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Question by:AL3X2
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by:evilrix
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You mean, like this?
#include <stdio.h>
 

int main(int argc, char **argv)

{

	unsigned long ul = 0x18CE; // 01100011001110

	unsigned char uc = static_cast<unsigned char>(ul >> 6);

	

	printf("0x%04X\n", uc); // 01100011 (0x0063)

	printf("0x%04X\n", (ul & 0x3F)); // 001110 (0x000E)

}

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evilrix earned 50 total points
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Sorry, same thing with a C and not C++ style cast :)
#include <stdio.h>
 

int main(int argc, char **argv)

{

	unsigned long ul = 0x18CE; // 01100011001110

	unsigned char uc = (unsigned char)(ul >> 6);

	

	printf("0x%04X\n", uc); // 01100011 (0x0063)

	printf("0x%04X\n", (ul & 0x3F)); // 001110 (0x000E)

}

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by:AL3X2
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Thanks for the fast reply!

But, this will work only when ul is 13 digits?
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Author Comment

by:AL3X2
Comment Utility
Nevermind, I think I found it ;)


Just one more thing.

If i have ul = 1001 (for example), how can I add digits ?

something like  ul = ul + 100, so printf("%ld", ul) will show "1001100" ?
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Expert Comment

by:evilrix
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>> But, this will work only when ul is 13 digits?
It is specific to the requires you stated. Line 6 cuts off the fist 6 bytes of the UL using the right shift operator and casts the return into a char and the result is printed at line 8.

01100011001110 >> 6 == 01100011

Line 9 uses the mask 111111 to ignore all but the first 6 bits of the UL

01100011001110 & 111111 == 001110

Once you understand how these 2 operations work you can tailor them to various needs.


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by:evilrix
Comment Utility
>> If i have ul = 1001 (for example), how can I add digits ?
>> something like  ul = ul + 100, so printf("%ld", ul) will show "1001100" ?

Read up on the << and >>  and the & and | bitwise C operators

http://www.cplusplus.com/doc/tutorial/operators/


#include <stdio.h>
 

int main(int argc, char **argv)

{

	unsigned long ul = 0x09; // 1001

	ul <<= 3; // 1001000

	ul |= 4; // 1001100
 

	printf("0x%04X\n", ul); // 1001100 (0x4C)

}

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