Array of boolean

10-23-2001, 08:28 AM

void main()
{
boolean number[8] ;

}

is this the proper way to declare an array of boolean characters?

Thanks

Procyon · 10-23-2001, 08:40 AM

Yes, except it's normally written as bool, not boolean.

10-23-2001, 12:25 PM

Arrays of bools are horibly inefficient. Use bitpacking instead. Each bool is a byte, while a byte can be used to store 8 boolean values as bits. That's a size savings of 8:1!

10-24-2001, 08:24 AM

hey, could you please briefly (or not) explain how to do the bitpacking thing?

Magos · 10-24-2001, 01:27 PM

unsigned char Number;

(Number & 1) Get 1:st bit out of Number
(Number & 2) Get 2:nd bit out of Number
(Number & 4) Get 3:rd bit out of Number
(Number & 8) Get 4:th bit out of Number
(Number & 16) Get 5:th bit out of Number
(Number & 32) Get 6:th bit out of Number
(Number & 64) Get 7:th bit out of Number
(Number & 128) Get 8:th bit out of Number

Ex:
if(Number & 1) cout << "First bit is true";
if(!(Number & 4)) cout << "Third bit is false";

gliptic · 10-24-2001, 02:24 PM

Setting a bit:

(Number | 1) Set 1:st bit in Number
(Number | 2) Set 2:nd bit in Number
(Number | 4) Set 3:rd bit in Number
(Number | 8) Set 4:th bit in Number
(Number | 16) Set 5:th bit in Number
(Number | 32) Set 6:th bit in Number
(Number | 64) Set 7:th bit in Number
(Number | 128) Set 8:th bit in Number

Toggle a bit:

(Number ^ 1) Toggle 1:st bit in Number
(Number ^ 2) Toggle 2:nd bit in Number
(Number ^ 4) Toggle 3:rd bit in Number
(Number ^ 8) Toggle 4:th bit in Number
(Number ^ 16) Toggle 5:th bit in Number
(Number ^ 32) Toggle 6:th bit in Number
(Number ^ 64) Toggle 7:th bit in Number
(Number ^ 128) Toggle 8:th bit in Number

In assembler, you can do much more efficient bitpacking but I guess most of you don't care about assembler (C++ board).

Aran · 10-24-2001, 04:04 PM

umm.. how would you declare that and use it? I'm looking to use a similar idea for a project for school...

would it be like

Code:

main()
{
 int a;
 
 // sets first to true
 a ^ 1
 // all the rest are false still
 if (a | 1)
 printf("a is true");
 else
 printf("a is false");
}

gliptic · 10-25-2001, 02:00 AM

No, the '^' is XOR. It toggles bits so you must use '|' to set it and you must write:

a ^= 1;

and not just 'a ^ 1' because that is removed by your compiler because it doesn't do anything.

The rest of the bits in 'a' can be anything. You have to initialize it to 0 to be sure it is 0.

This loop sets all the bits in a byte:

Code:

unsigned char a;
long b = 1; //The lowest bit is set
for(long i = 0; i < 8; i++)
{
 a |= b; //Set the bit
 b <<= 1; //Go to the next bit
}

Magos · 10-25-2001, 02:10 AM

Wow! I have never seen the 'b<<=1' operator before. Cool!

Multiplies with 2, right?

gliptic · 10-25-2001, 02:19 AM

Yes, it's the same as moving all bits to left one step.

(00001 << 1) = 00010

In that way I can set all the bits in the byte.

Nick · 10-25-2001, 11:50 AM

Well
0x00000001 XOR 0x000000001 is 0x00000000

I think it's about time
we stop using hex and start using 32 base.
With 64 bits comming up, hex is just as tedius to
write as binary was with 8 bits.

gliptic · 10-25-2001, 11:45 PM

Base 32 ? You mean like splitting up the numbers into small 5-bit parts? Why 5-bit? Isn't it better to use base 256?

10-23-2001, 08:28 AM	#1
DMaxJ Guest Posts: n/a	Array of boolean void main() { boolean number[8] ; } is this the proper way to declare an array of boolean characters? Thanks

10-23-2001, 08:40 AM	#2
Procyon Ethereal Raccoon Join Date: Aug 2001 Posts: 189	Yes, except it's normally written as bool, not boolean.
Procyon is offline

10-24-2001, 01:27 PM	#5
Magos Confused Join Date: Sep 2001 Location: Sweden Posts: 3,122	Reply unsigned char Number; (Number & 1) Get 1:st bit out of Number (Number & 2) Get 2:nd bit out of Number (Number & 4) Get 3:rd bit out of Number (Number & 8) Get 4:th bit out of Number (Number & 16) Get 5:th bit out of Number (Number & 32) Get 6:th bit out of Number (Number & 64) Get 7:th bit out of Number (Number & 128) Get 8:th bit out of Number Ex: if(Number & 1) cout << "First bit is true"; if(!(Number & 4)) cout << "Third bit is false"; __________________ MagosX.com Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime.
Magos is offline

10-24-2001, 02:24 PM	#6
gliptic Registered User Join Date: Sep 2001 Posts: 164	Setting a bit: (Number \| 1) Set 1:st bit in Number (Number \| 2) Set 2:nd bit in Number (Number \| 4) Set 3:rd bit in Number (Number \| 8) Set 4:th bit in Number (Number \| 16) Set 5:th bit in Number (Number \| 32) Set 6:th bit in Number (Number \| 64) Set 7:th bit in Number (Number \| 128) Set 8:th bit in Number Toggle a bit: (Number ^ 1) Toggle 1:st bit in Number (Number ^ 2) Toggle 2:nd bit in Number (Number ^ 4) Toggle 3:rd bit in Number (Number ^ 8) Toggle 4:th bit in Number (Number ^ 16) Toggle 5:th bit in Number (Number ^ 32) Toggle 6:th bit in Number (Number ^ 64) Toggle 7:th bit in Number (Number ^ 128) Toggle 8:th bit in Number In assembler, you can do much more efficient bitpacking but I guess most of you don't care about assembler (C++ board). __________________ // Gliptic
gliptic is offline

10-24-2001, 04:04 PM	#7
Aran Registered User Join Date: Aug 2001 Posts: 1,301	umm.. how would you declare that and use it? I'm looking to use a similar idea for a project for school... would it be like Code: main() { int a; // sets first to true a ^ 1 // all the rest are false still if (a \| 1) printf("a is true"); else printf("a is false"); }
Aran is offline

10-23-2001, 12:25 PM	#3
Unregistered Guest Posts: n/a	Arrays of bools are horibly inefficient. Use bitpacking instead. Each bool is a byte, while a byte can be used to store 8 boolean values as bits. That's a size savings of 8:1!

10-24-2001, 08:24 AM	#4
electricglidingman Guest Posts: n/a	hey, could you please briefly (or not) explain how to do the bitpacking thing?

10-25-2001, 02:00 AM	#8
gliptic Registered User Join Date: Sep 2001 Posts: 164	No, the '^' is XOR. It toggles bits so you must use '\|' to set it and you must write: a ^= 1; and not just 'a ^ 1' because that is removed by your compiler because it doesn't do anything. The rest of the bits in 'a' can be anything. You have to initialize it to 0 to be sure it is 0. This loop sets all the bits in a byte: Code: unsigned char a; long b = 1; //The lowest bit is set for(long i = 0; i < 8; i++) { a \|= b; //Set the bit b <<= 1; //Go to the next bit } __________________ // Gliptic
gliptic is offline

10-25-2001, 02:10 AM	#9
Magos Confused Join Date: Sep 2001 Location: Sweden Posts: 3,122	Wow Wow! I have never seen the 'b<<=1' operator before. Cool! Multiplies with 2, right? __________________ MagosX.com Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime.
Magos is offline

10-25-2001, 02:19 AM	#10
gliptic Registered User Join Date: Sep 2001 Posts: 164	Yes, it's the same as moving all bits to left one step. (00001 << 1) = 00010 In that way I can set all the bits in the byte. __________________ // Gliptic
gliptic is offline

10-25-2001, 11:50 AM	#11
Nick Blank Join Date: Aug 2001 Posts: 1,034	Well 0x00000001 XOR 0x000000001 is 0x00000000 I think it's about time we stop using hex and start using 32 base. With 64 bits comming up, hex is just as tedius to write as binary was with 8 bits.
Nick is offline

10-25-2001, 11:45 PM	#12
gliptic Registered User Join Date: Sep 2001 Posts: 164	Base 32 ? You mean like splitting up the numbers into small 5-bit parts? Why 5-bit? Isn't it better to use base 256? __________________ // Gliptic
gliptic is offline

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