Floating-Point Numbers

Overflow

For example:

int value=1; /* value to test, starts at first (lowest) bit */
for (int bit=0;bit<100;bit++) {
	std::cout<<"at bit "<<bit<<" the value is "<<value<<"\n";
	value=value+value; /* moves over by one bit */
	if (value==0) break;
}
return 0;

(Try this in NetRun now!)

Because "int" currently has 32 bits, if you start at one, and add a variable to itself 32 times, the one overflows and is lost completely. 

In assembly, there's a handy instruction "jo" (jump if overflow) to check for overflow from the previous instruction.  The C++ compiler doesn't bother to use jo, though!

mov edi,1 ; loop variable
mov eax,0 ; counter

start:
	add eax,1 ; increment bit counter

	add edi,edi ; add variable to itself
	jo noes ; check for overflow in the above add

	cmp edi,0
	jne start

ret

noes: ; called for overflow
	mov eax,999
	ret

(Try this in NetRun now!)

Notice the above program returns 999 on overflow, which somebody else will need to check for.  (Responding correctly to overflow is actually quite difficult--see, e.g., Ariane 5 explosion, caused by poor handling of a detected overflow.  Ironically, ignoring the overflow would have caused no problems!)

Signed versus Unsigned Numbers

If you watch closely right before overflow, you see something funny happen:

signed char value=1; /* value to test, starts at first (lowest) bit */
for (int bit=0;bit<100;bit++) {
	std::cout<<"at bit "<<bit<<" the value is "<<(long)value<<"\n";
	value=value+value; /* moves over by one bit (value=value<<1 would work too) */
	if (value==0) break;
}
return 0;

(Try this in NetRun now!)

This prints out:

at bit 0 the value is 1
at bit 1 the value is 2
at bit 2 the value is 4
at bit 3 the value is 8
at bit 4 the value is 16
at bit 5 the value is 32
at bit 6 the value is 64
at bit 7 the value is -128 
Program complete.  Return 0 (0x0)

Wait, the last bit's value is -128?  Yes, it really is!

This negative high bit is called the "sign bit", and it has a negative value in two's complement signed numbers.  This means to represent -1, for example, you set not only the high bit, but all the other bits as well: in unsigned, this is the largest possible value.  The reason binary 11111111 represents -1 is the same reason you might choose 9999 to represent -1 on a 4-digit odometer: if you add one, you wrap around and hit zero.

A very cool thing about two's complement is addition is the same operation whether the numbers are signed or unsigned--we just interpret the result differently.  Subtraction is also identical for signed and unsigned.  Register names are identical in assembly for signed and unsigned.  However, when you change register sizes using an instruction like "movsxd rax,eax", when you check for overflow, when you compare numbers, multiply or divide, or shift bits, you need to know if the number is signed (has a sign bit) or unsigned (no sign bit, no negative numbers).

Floats: Normalized Numbers

• You only need one bit to represent the sign--plus or minus.
• The exponent's just an integer, so you can store it as an integer.
• The 3.785746 part, called the "mantissa" or just "fraction" part, can be stored as the integer 3785746 (at least as long as you can figure out where the decimal point goes!).  Of course, we store the fraction part as a binary number.
  

1. There's exactly one digit to the left of the decimal point.
2. And that digit is not zero.

Bits in a Floating-Point Number

void print_bits(float f) {
	int i=*reinterpret_cast<int *>(&f); /* read bits with "pointer shuffle" */
	std::cout<<" float "<<std::setw(10)<<f<<" = ";
	for (int bit=31;bit>=0;bit--) {
		if (i&(1<<bit)) std::cout<<"1"; else std::cout<<"0";
		if (bit==31) std::cout<<" ";
		if (bit==23) std::cout<<" (implicit 1).";
	}
	std::cout<<std::endl;
}

int foo(void) {
	print_bits(0.0);
	print_bits(-1.0);
	print_bits(1.0);
	print_bits(2.0);
	print_bits(4.0);
	print_bits(8.0);
	print_bits(1.125);
	print_bits(1.25);
	print_bits(1.5);
	print_bits(1+1.0/10);
	return sizeof(float);
}

(Try this in NetRun now!)

/* IEEE floating-point number's bits:  sign  exponent   mantissa */
struct float_bits {
	unsigned int fraction:23; /**< Value is binary 1.fraction ("mantissa") */
	unsigned int exp:8; /**< Value is 2^(exp-127) */
	unsigned int sign:1; /**< 0 for positive, 1 for negative */
};

/* A union is a struct where all the fields *overlap* each other */
union float_dissector {
	float f;
	float_bits b;
};

float_dissector s;
s.f=8.0;
std::cout<<s.f<<"= sign "<<s.b.sign<<" exp "<<s.b.exp<<"  fract "<<s.b.fraction<<"\n";
return 0;

• Floats can store fractional numbers.
  
• Floats never overflow; they hit "infinity" instead.
  
• "double" has more bits than "int" (but less than "long").

Roundoff in Arithmetic

float f=0.73;
while (1) {
	volatile float g=f+1;
	if (g==f) {
		std::cout<<"f+1 == f  at f="<< f <<", or 2^"<< log(f)/log(2.0) <<std::endl;
		return 0;
	}
	else f=g;
}

(Try this in NetRun now!)

double lil=1.0;
double big=pow(2.0,53); //<- carefully chosen for IEEE 64-bit float (52 bits of fraction + implicit 1)
std::cout<<" big+(lil+lil) -big = "<<  big+(lil+lil) -big <<std::endl;
std::cout<<"(big+lil)+lil  -big = "<< (big+lil)+lil  -big <<std::endl;

(Try this in NetRun now!)

float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
for (int i=0;i<1000;i++)
	windshield += gnats;

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";

float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
volatile float gnatcup=0.0;
for (int i=0;i<1000;i++)
	gnatcup += gnats;
windshield+=gnatcup; /* add all gnats to the windshield at once */

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";

Roundoff in Representation

for (int i=1;i<1000000000;i*=10) {
	double mul01=i*0.1;
	double div10=i/10.0;
	double diff=mul01-div10;
	std::cout<<"i="<<i<<"  diff="<<diff<<"\n";
}

i=1  diff=5.54976e-18
i=10  diff=5.55112e-17
i=100  diff=5.55112e-16
i=1000  diff=5.55112e-15
i=10000  diff=5.55112e-14
i=100000  diff=5.55112e-13
i=1000000  diff=5.54934e-12
i=10000000  diff=5.5536e-11
i=100000000  diff=5.54792e-10
Program complete.  Return 0 (0x0)