Hex, Octal, and Negative Numbers

Base Conversion

/* Print the low 32 bits of this number */
void print_binary(int v) 
{
	for (int bit=31;bit>=0;bit--) {
		if ((v&(1<<bit))!=0) {
			std::cout<<"1";
		} else {
			std::cout<<"0";
		}
	}
	std::cout<<" binary\n";
}

int foo(void) {
	print_binary(6);
	return 0;
}

const int bits[]={1,1,0};
int nbits=sizeof(bits)/sizeof(bits[0]); /*<- funky trick to find size of static array!*/
int value=0;
for (int bit=0;bit<nbits;bit++) {
	if (bits[nbits-1-bit]) /*<- gotta index the bit array starting at the end */
		value = value | (1<<bit);
}
return value;

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Or you can do the same binary-to-integer conversion using digits from cin:

void print_binary(int v) {
	for (int i=0;i<32;i++)
	{
		int mask=1<<(32-1-i);
		if ((v&mask)==0) std::cout<<"0";
		else std::cout<<"1";
	}
	std::cout<<endl;
}
int read_binary(void) {
	int v=0;
	while (std::cin) {
		char c='?'; 
		std::cin>>c;
		//std::cout<<"I just read the character '"<<c<<"'."<<endl;
		if (c=='0') { 
			/* zero digit in that place */
			v=v<<1; 
		} else if (c=='1') {
			/* one digit in that place */
			v=(v<<1)+1;
		}
		else {
			break;
		}
	}
	return v;
}

int foo(void) {
	return read_binary();
}

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What's the deal with all this hex?

• Base 0: not counting at all.  (only representable number is zero)
  
• Base 1: counting on your fingers.  (inefficient use of fingers, though...)
  
• Base 2: binary.  Used by all known computers.
  
• Base 8: octal.  Popular back in 1970s, 1980s.  In C++, prefix the number with zero to make it octal, like "070" (==7*8 = 56).
  
• Base 10: decimal.  Matches number of human fingers.  In C++, numbers are decimal by default.
  
• Base 16: hexadecimal.  Fits nicely with bytes.  In C++, prefix the number with "0x" (zero-x) to make it hex, like "0xF3".
  
• Base 20, used by Mayans.  Matches number of human fingers and toes.
• Base 60, used by Sumerians six thousand years ago.
  

 804a516:       80 fa 3f                cmp    dl,0x3f
 804a519:       0f 84 7c 01 00 00       je     804a69b <exit@plt+0xc2b>

Hex & Bitwise Operations

int v=1024+15;
for (int digit=7;digit>=0;digit--) {
	char *digitTable="0123456789abcdef";
	int d=(v>>(digit*4))&0xF;
	std::cout<<digitTable[d];
}
std::cout<<std::endl;
return v;

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Arithmetic In Binary

Storage Sizes