I am new to C and trying to write a program that calculates the sum of the digits of 2^n, where n<10^8.
For example, for 2^10, we'd have 1+0+2+4, which is 7.
Here's what I came up with:
#include <stdio.h>
#include <math.h>
int main()
{
int n, t, sum = 0, remainder;
printf("Enter an integer\n");
scanf("%d", &n);
t = pow(2, n);
while (t != 0)
{
remainder = t % 10;
sum = sum + remainder;
t = t / 10;
}
printf("Sum of digits of 2 to the power of %d = %d\n", n, sum);
return 0;
}
The problem is: the program works fine with numbers smaller than 30. Once I set n to a number higher than 30, the result is always -47.
I really do not understand this error and what causes it.
An interesting problem to be sure, but I think the solution is way outside the scope of a simple answer if you wish to support large values of n, such as the 108 you mentioned. The number 2108 requires 108 + 1 (100,000,001) bits, or around 12 megabytes of memory, to store in binary. In decimal it has around 30 million digits.
Your int is 32 bits wide, which is why the signed int can't store 231 – the 32nd bit is the sign while 231 has a 1 followed by 31 zeros in binary, requiring 32 bits without the sign. So it overflows and is interpreted as a negative number. (Technically signed integer overflow is undefined behaviour in C.)
You can switch to an unsigned int to get rid of the sign and the undefined behaviour, in which case your new highest supported n will be 31. You almost certainly have 64-bit integers available, and perhaps even 128-bit, but 2127 is still way less than 2100000000.
So either you need to find an algorithm to compute the decimal digits of a power of 2 without actually storing them (and only store the sum), or forget about trying to use any scalar types in standard C and get (or implement) an arbitrary precision math library operating on arrays (of bits, decimal digits, or binary-coded decimal digits). Alternatively, you can limit your solution to, say, uint64_t, but then you have n < 64, which is not nearly as interesting… =)
For signed int t = pow(2,n), if n >= 31 then t > INT_MAX.
You can use unsigned long long t = pow(2,n) instead.
This will allow you to go as up as n == 63.
Also, since you're using base 2, you can use (unsigned long long)1 << n instead of pow(2,n).
http://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)FloatingPoint.html
I was looking into why there are sometimes rounding issues when storing a float. I read the above link, and see that floats are converted to scientific notation.
https://babbage.cs.qc.cuny.edu/IEEE-754/index.xhtml
Base is always 2. So, 8 is stored as 1 * 2^3. 9 is stored as 1.001 * 2^3.
What is the math algorithm to determine the mantissa/significand and exponent?
Here is C++ code to convert a decimal string to a binary floating-point value. Although the question is tagged C, I presume the question is more about the algorithm and calculations than the programming language.
The DecimalToFloat class is constructed with a string that contains solely decimal digits and a decimal point (a period, most one). In its constructor, it shows how to use elementary school multiplication and long division to convert the number from decimal to binary. This demonstrates the fundamental concepts using elementary arithmetic. Real implementations of decimal-to-floating-point conversion in commercial software using algorithms that are faster and more complicated. They involve prepared tables, analysis, and proofs and are the subjects of academic papers. A significant problem of quality implementations of decimal-to-binary-floating-point conversion is getting the rounding correct. The disparate nature of powers of ten to powers of two (both positive and negative powers) makes it tricky to correctly determine when some values are above or below a point where rounding changes. Normally, when we are parsing something like 123e300, we want to figure out the binary floating-point result without actually calculating 10300. That is a much more extensive subject.
The GetValue routine finishes the preparation fo the number, taking the information prepared by the constructor and rounding it to the final floating-point form.
Negative numbers and exponential (scientific) notation are not handled. Handling negative numbers is of course easy. Exponential notation could be accommodated by shifting the input—moving the decimal point right for positive exponents or left for negative exponents. Again, this is not the fastest way to perform the conversion, but it demonstrates fundamental ideas.
/* This code demonstrates conversion of decimal numerals to binary
floating-point values using the round-to-nearest-ties-to-even rule.
Infinities and subnormal values are supported and assumed.
The basic idea is to convert the decimal numeral to binary using methods
taught in elementary school. The integer digits are repeatedly divided by
two to extract a string of bits in low-to-high position-value order. Then
sub-integer digits are repeatedly multiplied by two to continue extracting
a string of bits in high-to-low position-value order. Once we have enough
bits to determine the rounding direction or the processing exhausts the
input, the final value is computed.
This code is not (and will not be) designed to be efficient. It
demonstrates the fundamental mathematics and rounding decisions.
*/
#include <algorithm>
#include <limits>
#include <cmath>
#include <cstring>
template<typename Float> class DecimalToFloat
{
private:
static_assert(std::numeric_limits<Float>::radix == 2,
"This code requires the floatng-point radix to be two.");
// Abbreviations for parameters describing the floating-point format.
static const int Digits = std::numeric_limits<Float>::digits;
static const int MaximumExponent = std::numeric_limits<Float>::max_exponent;
static const int MinimumExponent = std::numeric_limits<Float>::min_exponent;
/* For any rounding rule supported by IEEE 754 for binary floating-point,
the direction in which a floating-point result should be rounded is
completely determined by the bit in the position of the least
significant bit (LSB) of the significand and whether the value of the
trailing bits are zero, between zero and 1/2 the value of the LSB,
exactly 1/2 the LSB, or between 1/2 the LSB and 1.
In particular, for round-to-nearest, ties-to-even, the decision is:
LSB Trailing Bits Direction
0 0 Down
0 In (0, 1/2) Down
0 1/2 Down
0 In (1/2, 1) Up
1 0 Down
1 In (0, 1/2) Down
1 1/2 Up
1 In (1/2, 1) Up
To determine whether the value of the trailing bits is 0, in (0, 1/2),
1/2, or in (1/2, 1), it suffices to know the first of the trailing bits
and whether the remaining bits are zeros or not:
First Remaining Value of Trailing Bits
0 All zeros 0
0 Not all zeros In (0, 1/2)
1 All zeros 1/2
1 Not all zeros In (1/2, 1)
To capture that information, we maintain two bits in addition to the
bits in the significand. The first is called the Round bit. It is the
first bit after the position of the least significand bit in the
significand. The second is called the Sticky bit. It is set if any
trailing bit after the first is set.
The bits for the significand are kept in an array along with the Round
bit and the Sticky bit. The constants below provide array indices for
locating the LSB, the Round Bit, and the Sticky bit in that array.
*/
static const int LowBit = Digits-1; // Array index for LSB in significand.
static const int Round = Digits; // Array index for rounding bit.
static const int Sticky = Digits+1; // Array index for sticky bit.
char *Decimal; // Work space for the incoming decimal numeral.
int N; // Number of bits incorporated so far.
char Bits[Digits+2]; // Bits for significand plus two for rounding.
int Exponent; // Exponent adjustment needed.
/* PushBitHigh inserts a new bit into the high end of the bits we are
accumulating for the significand of a floating-point number.
First, the Round bit shifted down by incorporating it into the Sticky
bit, using an OR so that the Sticky bit is set iff any bit pushed below
the Round bit is set.
Then all bits from the significand are shifted down one position,
which moves the least significant bit into the Round position and
frees up the most significant bit.
Then the new bit is put into the most significant bit.
*/
void PushBitHigh(char Bit)
{
Bits[Sticky] |= Bits[Round];
std::memmove(Bits+1, Bits, Digits * sizeof *Bits);
Bits[0] = Bit;
++N; // Count the number of bits we have put in the significand.
++Exponent; // Track the absolute position of the leading bit.
}
/* PushBitLow inserts a new bit into the low end of the bits we are
accumulating for the significand of a floating-point number.
If we have no previous bits and the new bit is zero, we are just
processing leading zeros in a number less than 1. These zeros are not
significant. They tell us the magnitude of the number. We use them
only to track the exponent that records the position of the leading
significant bit. (However, exponent is only allowed to get as small as
MinimumExponent, after which we must put further bits into the
significand, forming a subnormal value.)
If the bit is significant, we record it. If we have not yet filled the
regular significand and the Round bit, the new bit is recorded in the
next space. Otherwise, the new bit is incorporated into the Sticky bit
using an OR so that the Sticky bit is set iff any bit below the Round
bit is set.
*/
void PushBitLow(char Bit)
{
if (N == 0 && Bit == 0 && MinimumExponent < Exponent)
--Exponent;
else
if (N < Sticky)
Bits[N++] = Bit;
else
Bits[Sticky] |= Bit;
}
/* Determined tells us whether the final value to be produced can be
determined without any more low bits. This is true if and only if:
we have all the bits to fill the significand, and
we have at least one more bit to help determine the rounding, and
either we know we will round down because the Round bit is 0 or we
know we will round up because the Round bit is 1 and at least one
further bit is 1 or the least significant bit is 1.
*/
bool Determined() const
{
if (Digits < N)
if (Bits[Round])
return Bits[LowBit] || Bits[Sticky];
else
return 1;
else
return 0;
}
// Get the floating-point value that was parsed from the source numeral.
Float GetValue() const
{
// Decide whether to round up or not.
bool RoundUp = Bits[Round] && (Bits[LowBit] || Bits[Sticky]);
/* Now we prepare a floating-point number that contains a significand
with the bits we received plus, if we are rounding up, one added to
the least significant bit.
*/
// Start with the adjustment to the LSB for rounding.
Float x = RoundUp;
// Add the significand bits we received.
for (int i = Digits-1; 0 <= i; --i)
x = (x + Bits[i]) / 2;
/* If we rounded up, the addition may have carried out of the
initial significand. In this case, adjust the scale.
*/
int e = Exponent;
if (1 <= x)
{
x /= 2;
++e;
}
// Apply the exponent and return the value.
return MaximumExponent < e ? INFINITY : std::scalbn(x, e);
}
public:
/* Constructor.
Note that this constructor allocates work space. It is bad form to
allocate in a constructor, but this code is just to demonstrate the
mathematics, not to provide a conversion for use in production
software.
*/
DecimalToFloat(const char *Source) : N(), Bits(), Exponent()
{
// Skip leading sources.
while (*Source == '0')
++Source;
size_t s = std::strlen(Source);
/* Count the number of integer digits (digits before the decimal
point if it is present or before the end of the string otherwise)
and calculate the number of digits after the decimal point, if any.
*/
size_t DigitsBefore = 0;
while (Source[DigitsBefore] != '.' && Source[DigitsBefore] != 0)
++DigitsBefore;
size_t DigitsAfter = Source[DigitsBefore] == '.' ? s-DigitsBefore-1 : 0;
/* Allocate space for the integer digits or the sub-integer digits,
whichever is more numerous.
*/
Decimal = new char[std::max(DigitsBefore, DigitsAfter)];
/* Copy the integer digits into our work space, converting them from
digit characters ('0' to '9') to numbers (0 to 9).
*/
for (size_t i = 0; i < DigitsBefore; ++i)
Decimal[i] = Source[i] - '0';
/* Convert the integer portion of the numeral to binary by repeatedly
dividing it by two. The remainders form a bit string representing
a binary numeral for the integer part of the number. They arrive
in order from low position value to high position value.
This conversion continues until the numeral is exhausted (High <
Low is false) or we see it is so large the result overflows
(Exponent <= MaximumExponent is false).
Note that Exponent may exceed MaximumExponent while we have only
produced 0 bits during the conversion. However, because we skipped
leading zeros above, we know there is a 1 bit coming. That,
combined with the excessive Exponent, guarantees the result will
overflow.
*/
for (char *High = Decimal, *Low = Decimal + DigitsBefore;
High < Low && Exponent <= MaximumExponent;)
{
// Divide by two.
char Remainder = 0;
for (char *p = High; p < Low; ++p)
{
/* This is elementary school division: We bring in the
remainder from the higher digit position and divide by the
divisor. The remainder is kept for the next position, and
the quotient becomes the new digit in this position.
*/
char n = *p + 10*Remainder;
Remainder = n % 2;
n /= 2;
/* As the number becomes smaller, we discard leading zeros:
If the new digit is zero and is in the highest position,
we discard it and shorten the number we are working with.
Otherwise, we record the new digit.
*/
if (n == 0 && p == High)
++High;
else
*p = n;
}
// Push remainder into high end of the bits we are accumulating.
PushBitHigh(Remainder);
}
/* Copy the sub-integer digits into our work space, converting them
from digit characters ('0' to '9') to numbers (0 to 9).
The convert the sub-integer portion of the numeral to binary by
repeatedly multiplying it by two. The carry-outs continue the bit
string. They arrive in order from high position value to low
position value.
*/
for (size_t i = 0; i < DigitsAfter; ++i)
Decimal[i] = Source[DigitsBefore + 1 + i] - '0';
for (char *High = Decimal, *Low = Decimal + DigitsAfter;
High < Low && !Determined();)
{
// Multiply by two.
char Carry = 0;
for (char *p = Low; High < p--;)
{
/* This is elementary school multiplication: We multiply
the digit by the multiplicand and add the carry. The
result is separated into a single digit (n % 10) and a
carry (n / 10).
*/
char n = *p * 2 + Carry;
Carry = n / 10;
n %= 10;
/* Here we discard trailing zeros: If the new digit is zero
and is in the lowest position, we discard it and shorten
the numeral we are working with. Otherwise, we record the
new digit.
*/
if (n == 0 && p == Low-1)
--Low;
else
*p = n;
}
// Push carry into low end of the bits we are accumulating.
PushBitLow(Carry);
}
delete [] Decimal;
}
// Conversion operator. Returns a Float converted from this object.
operator Float() const { return GetValue(); }
};
#include <iostream>
#include <cstdio>
#include <cstdlib>
static void Test(const char *Source)
{
std::cout << "Testing " << Source << ":\n";
DecimalToFloat<float> x(Source);
char *end;
float e = std::strtof(Source, &end);
float o = x;
/* Note: The C printf is used here for the %a conversion, which shows the
bits of floating-point values clearly. If your C++ implementation does
not support this, this may be replaced by any display of floating-point
values you desire, such as printing them with all the decimal digits
needed to distinguish the values.
*/
std::printf("\t%a, %a.\n", e, o);
if (e != o)
{
std::cout << "\tError, results do not match.\n";
std::exit(EXIT_FAILURE);
}
}
int main(void)
{
Test("0");
Test("1");
Test("2");
Test("3");
Test(".25");
Test(".0625");
Test(".1");
Test(".2");
Test(".3");
Test("3.14");
Test(".00000001");
Test("9841234012398123");
Test("340282346638528859811704183484516925440");
Test("340282356779733661637539395458142568447");
Test("340282356779733661637539395458142568448");
Test(".00000000000000000000000000000000000000000000140129846432481707092372958328991613128026194187651577175706828388979108268586060148663818836212158203125");
// This should round to the minimum positive (subnormal), as it is just above mid-way.
Test(".000000000000000000000000000000000000000000000700649232162408535461864791644958065640130970938257885878534141944895541342930300743319094181060791015626");
// This should round to zero, as it is mid-way, and the even rule applies.
Test(".000000000000000000000000000000000000000000000700649232162408535461864791644958065640130970938257885878534141944895541342930300743319094181060791015625");
// This should round to zero, as it is just below mid-way.
Test(".000000000000000000000000000000000000000000000700649232162408535461864791644958065640130970938257885878534141944895541342930300743319094181060791015624");
}
One of the surprising things about a real, practical computer -- surprising to beginning programmers who have been tasked with writing artificial little binary-to-decimal conversion programs, anyway -- is how thoroughly ingrained the binary number system is in an actual computer, and how few and how diffuse any actual binary/decimal conversion routines actually are. In the C world, for example (and if we confine our attention to integers for the moment), there is basically one binary-to-decimal conversion routine, and it's buried inside printf, where the %d directive is processed. There are perhaps three decimal-to-binary converters: atof(), strtol(), and the %d conversion inside scanf. (There might be another one inside the C compiler, where it converts your decimal constants into binary, although the compiler might just call strtol() directly for those, too.)
I bring this all up for background. The question of "what's the actual algorithm for constructing floating-point numbers internally?" is a fair one, and I'd like to think I know the answer, but as I mentioned in the comments, I'm chagrined to discover that I don't, really: I can't describe a clear, crisp "algorithm". I can and will show you some code that gets the job done, but you'll probably find it unsatisfying, as if I'm cheating somehow -- because a number of the interesting details happen more or less automatically, as we'll see.
Basically, I'm going to write a version of the standard library function atof(). Here are my ground rules:
I'm going to assume that the input is a string of characters. (This isn't really an assumption at all; it's a restatement of the original problem, which is to write a version of atof.)
I'm going to assume that we can construct the floating-point number "0.0". (In IEEE 754 and most other formats, it's all-bits-0, so that's not too hard.)
I'm going to assume that we can convert the integers 0-9 to their corresponding floating-point equivalents.
I'm going to assume that we can add and multiply any floating-point numbers we want to. (This is the biggie, although I'll describe those algorithms later.) But on any modern computer, there's almost certainly a floating-point unit, that has built-in instructions for the basic floating-point operations like addition and multiplication, so this isn't an unreasonable assumption, either. (But it does end up hiding some of the interesting aspects of the algorithm, passing the buck to the hardware designer to have implemented the instructions correctly.)
I'm going to initially assume that we have access to the standard library functions atoi and pow. This is a pretty big assumption, but again, I'll describe later how we could write those from scratch if we wanted to. I'm also going to assume the existence of the character classification functions in <ctype.h>, especially isdigit().
But that's about it. With those prerequisites, it turns out we can write a fully-functional version of atof() all by ourselves. It might not be fast, and it almost certainly won't have all the right rounding behaviors out at the edges, but it will work pretty well. (I'm even going to handle negative numbers, and exponents.) Here's how it works:
skip leading whitespace
look for '-'
scan digit characters, converting each one to the corresponding digit by subtracting '0' (aka ASCII 48)
accumulate a floating-point number (with no fractional part yet) representing the integer implied by the digits -- the significand -- and this is the real math, multiplying the running accumulation by 10 and adding the next digit
if we see a decimal point, count the number of digits after it
when we're done scanning digits, see if there's an e/E and some more digits indicating an exponent
if necessary, multiply or divide our accumulated number by a power of 10, to take care of digits past the decimal, and/or the explicit exponent.
Here's the code:
#include <ctype.h>
#include <stdlib.h> /* just for atoi() */
#include <math.h> /* just for pow() */
#define TRUE 1
#define FALSE 0
double my_atof(const char *str)
{
const char *p;
double ret;
int negflag = FALSE;
int exp;
int expflag;
p = str;
while(isspace(*p))
p++;
if(*p == '-')
{
negflag = TRUE;
p++;
}
ret = 0.0; /* assumption 2 */
exp = 0;
expflag = FALSE;
while(TRUE)
{
if(*p == '.')
expflag = TRUE;
else if(isdigit(*p))
{
int idig = *p - '0'; /* assumption 1 */
double fdig = idig; /* assumption 3 */
ret = 10. * ret + fdig; /* assumption 4 */
if(expflag)
exp--;
}
else break;
p++;
}
if(*p == 'e' || *p == 'E')
exp += atoi(p+1); /* assumption 5a */
if(exp != 0)
ret *= pow(10., exp); /* assumption 5b */
if(negflag)
ret = -ret;
return ret;
}
Before we go further, I encourage you to copy-and-paste this code into a nearby C compiler, and compile it, to convince yourself that I haven't cheated too badly. Here's a little main() to invoke it with:
#include <stdio.h>
int main(int argc, char *argv[])
{
double d = my_atof(argv[1]);
printf("%s -> %g\n", argv[1], d);
}
(If you or your IDE aren't comfortable with command-line invocations, you can use fgets or scanf to read the string to hand to my_atof, instead.)
But, I know, your question was "How does 9 get converted to 1.001 * 2^3 ?", and I still haven't really answered that, have I? So let's see if we can find where that happens.
First of all, that bit pattern 10012 for 9 came from... nowhere, or everywhere, or it was there all along, or something. The character 9 came in, probably with a bit pattern of 1110012 (in ASCII). We subtracted 48 = 1100002, and out popped 10012. (Even before doing the subtraction, you can see it hiding there at the end of 111001.)
But then what turned 1001 into 1.001E3? That was basically my "assumption 3", as embodied in the line
double fdig = idig;
It's easy to write that line in C, so we don't really have to know how it's done, and the compiler probably turns it into a 'convert integer to float' instruction, so the compiler writer doesn't have to know how to do it, either.
But, if we did have to implement that ourselves, at the lowest level, we could. We know we have a single-digit (decimal) number, occupying at most 4 bits. We could stuff those bits into the significand field of our floating-point format, with a fixed exponent (perhaps -3). We might have to deal with the peculiarities of an "implicit 1" bit, and if we didn't want to inadvertently create a denormalized number, we might have to some more tinkering, but it would be straightforward enough, and relatively easy to get right, because there are only 10 cases to test. (Heck, if we found writing code to do the bit manipulations troublesome, we could even use a 10-entry lookup table.)
Since 9 is a single-digit number, we're done. But for a multiple-digit number, our next concern is the arithmetic we have to do: multiplying the running sum by 10, and adding in the next digit. How does that work, exactly?
Again, if we're writing a C (or even an assembly language) program, we don't really need to know, because our machine's floating-point 'add' and 'multiply' instructions will do everything for us. But, also again, if we had to do it ourselves, we could. (This answer's getting way too long, so I'm not going to discuss floating-point addition and multiplication algorithms just yet. Maybe farther down.)
Finally, the code as presented so far "cheated" by calling the library functions atoi and pow. I won't have any trouble convincing you that we could have implemented atoi ourselves if we wanted/had to: it's basically just the same digit-accumulation code we already wrote. And pow isn't too hard, either, because in our case we don't need to implement it in full generality: we're always raising to integer powers, so it's straightforward repeated multiplication, and we've already assumed we know how to do multiplication.
(With that said, computing a large power of 10 as part of our decimal-to-binary algorithm is problematic. As #Eric Postpischil noted in his answer, "Normally we want to figure out the binary floating-point result without actually calculating 10N." Me, since I don't know any better, I'll compute it anyway, but if I wrote my own pow() I'd use the binary exponentiation algorithm, since it's super easy to implement and quite nicely efficient.)
I said I'd discuss floating-point addition and multiplication routines. Suppose you want to add two floating-point numbers. If they happen to have the same exponent, it's easy: add the two significands (and keep the exponent the same), and that's your answer. (How do you add the significands? Well, I assume you have a way to add integers.) If the exponents are different, but relatively close to each other, you can pick the smaller one and add N to it to make it the same as the larger one, while simultaneously shifting the significand to the right by N bits. (You've just created a denormalized number.) Once the exponents are the same, you can add the significands, as before. After the addition, it may be important to renormalize the numbers, that is, to detect if one or more leading bits ended up as 0 and, if so, shift the significand left and decrement the exponent. Finally, if the exponents are too different, such that shifting one significand to the right by N bits would shift it all away, this means that one number is so much smaller than the other that all of it gets lost in the roundoff when adding them.
Multiplication: Floating-point multiplication is actually somewhat easier than addition. You don't have to worry about matching up the exponents: the final product is basically a new number whose significand is the product of the two significands, and whose exponent is the sum of the two exponents. The only trick is that the product of the two M-bit significands is nominally 2M bits, and you may not have a multiplier that can do that. If the only multiplier you have available maxes out at an M-bit product, you can take your two M-bit significands and literally split them in half by bits:
signif1 = a * 2M/2 + b
signif2 = c * 2M/2 + d
So by ordinary algebra we have
signif1 × signif2 = ac × 2M + ad × 2M/2 + bc × 2M/2 + bd
Each of those partial products ac, ad, etc. is an M-bit product. Multiplying by 2M/2 or 2M is easy, because it's just a left shift. And adding the terms up is something we already know how to do. We actually only care about the upper M bits of the product, so since we're going to throw away the rest, I imagine we could cheat and skip the bd term, since it contributes nothing (although it might end up slightly influencing a properly-rounded result).
But anyway, the details of the addition and multiplication algorithms, and the knowledge they contain about the floating-point representation we're using, end up forming the other half of the answer to the question of the decimal-to-binary "algorithm" you're looking for. If you convert, say, the number 5.703125 using the code I've shown, out will pop the binary floating-point number 1.011011012 × 22, but nowhere did we explicitly compute that significand 1.01101101 or that exponent 2 -- they both just fell out of all the digitwise multiplications and additions we did.
Finally, if you're still with me, here's a quick and easy integer-power-only pow function using binary exponentiation:
double my_pow(double a, unsigned int b)
{
double ret = 1;
double fac = a;
while(1) {
if(b & 1) ret *= fac;
b >>= 1;
if(b == 0) break;
fac *= fac;
}
return ret;
}
This is a nifty little algorithm. If we ask it to compute, say, 1021, it does not multiply 10 by itself 21 times. Instead, it repeatedly squares 10, leading to the exponential sequence 101, 102, 104, 108, or rather, 10, 100, 10000, 100000000... Then it looks at the binary representation of 21, namely 10101, and selects only the intermediate results 101, 104, and 1016 to multiply into its final return value, yielding 101+4+16, or 1021, as desired. It therefore runs in time O(log2(N)), not O(N).
And, tune in tomorrow for our next exciting episode when we'll go in the opposite direction, writing a binary-to-decimal converter which will require us to do... (ominous chord)
floating point long division!
Here's a completely different answer, that tries to focus on the "algorithm" part of the question. I'll start with the example you asked about, converting the decimal integer 9 to the binary scientific notation number 1.0012×23. The algorithm is in two parts: (1) convert the decimal integer 9 to the binary integer 10012, and (2) convert that binary integer into binary scientific notation.
Step 1. Convert a decimal integer to a binary integer. (You can skip over this part if you already know it. Also, although this part of the algorithm is going to look perfectly fine, it turns out it's not the sort of thing that's actually used anywhere on a practical binary computer.)
The algorithm is built around a number we're working on, n, and a binary number we're building up, b.
Set n initially to the number we're converting, 9.
Set b to 0.
Compute the remainder when dividing n by 2. In our example, the remainder of 9 ÷ 2 is 1.
The remainder is one bit of our binary number. Tack it on to b. In our example, b is now 1. Also, here we're going to be tacking bits on to b on the left.
Divide n by 2 (discarding the remainder). In our example, n is now 4.
If n is now 0, we're done.
Go back to step 3.
At the end of the first trip through the algorithm, n is 4 and b is 1.
The next trip through the loop will extract the bit 0 (because 4 divided by 2 is 2, remainder 0). So b goes to 01, and n goes to 2.
The next trip through the loop will extract the bit 0 (because 2 divided by 2 is 1, remainder 0). So b goes to 001, and n goes to 1.
The next trip through the loop will extract the bit 1 (because 1 divided by 2 is 0, remainder 1). So b goes to 1001, and n goes to 0.
And since n is now 0, we're done. Meanwhile, we've built up the binary number 1001 in b, as desired.
Here's that example again, in tabular form. At each step, we compute n divided by two (or in C, n/2), and the remainder when dividing n by 2, which in C is n%2. At the next step, n gets replaced by n/2, and the next bit (which is n%2) gets tacked on at the left of b.
step n b n/2 n%2
0 9 0 4 1
1 4 1 2 0
2 2 01 1 0
3 1 001 0 1
4 0 1001
Let's run through that again, for the number 25:
step n b n/2 n%2
0 25 0 12 1
1 12 1 6 0
2 6 01 3 0
3 3 001 1 1
4 1 1001 0 1
5 0 11001
You can clearly see that the n column is driven by the n/2 column, because in step 5 of the algorithm as stated we divided n by 2. (In C this would be n = n / 2, or n /= 2.) You can clearly see the binary result appearing (in right-to-left order) in the n%2 column.
So that's one way to convert decimal integers to binary. (As I mentioned, though, it's likely not the way your computer does it. Among other things, the act of tacking a bit on to the left end of b turns out to be rather unorthodox.)
Step 2. Convert a binary integer to a binary number in scientific notation.
Before we begin with this half of the algorithm, it's important to realize that scientific (or "exponential") representations are typically not unique. Returning to decimal for a moment, let's think about the number "one thousand". Most often we'll represent that as 1 × 103. But we could also represent it as 10 × 102, or 100 × 101, or even crazier representations like 10000 × 10-1, or 0.01 × 105.
So, in practice, when we're working in scientific notation, we'll usually set up an additional rule or guideline, stating that we'll try to keep the mantissa (also called the "significand") within a certain range. For base 10, usually the goal is either to keep it in the range 0 ≤ mantissa < 10, or 0 ≤ mantissa < 1. That is, we like numbers like 1 × 103 or 0.1 × 104, but we don't like numbers like 100 × 101 or 0.01 × 105.
How do we keep our representations in the range we like? What if we've got a number (perhaps the intermediate result of a calculation) that's in a form we don't like? The answer is simple, and it depends on a pattern you've probably already noticed: If you multiply the mantissa by 10, and if you simultaneously subtract 1 from the exponent, you haven't changed the value of the number. Similarly, you can divide the mantissa by 10 and increment the exponent, again without changing anything.
When we convert a scientific-notation number into the form we like, we say we're normalizing the number.
One more thing: since 100 is 1, we can preliminarily convert any integer to scientific notation by simply multiplying it by 100. That is, 9 is 9×100, and 25 is 25×100. If we do it that way we'll usually get a number that's in a form we "don't like" (that is "nonnormalized"), but now we have an idea of how to fix that.
So let's return to base 2, and the rest of this second half of our algorithm. Everything we've said so far about decimal scientific notation is also true about binary scientific notation, as long as we make the obvious changes of "10" to "2".
To convert the binary integer 10012 to binary scientific notation, we first multiply it by 20, resulting in: 10012×20. So actually we're almost done, except that this number is nonnormalized.
What's our definition of a normalized base-two scientific notation number? We haven't said, but the requirement is usually that the mantissa is between 0 and 102 (that is, between 0 and 210), or stated another way, that the high-order bit of the mantissa is always 1 (unless the whole number is 0). That is, these mantissas are normalized: 1.0012, 1.12, 1.02, 0.02. These mantissas are nonnormalized: 10.012, 0.0012.
So to normalize a number, we may need to multiply or divide the mantissa by 2, while incrementing or decrementing the exponent.
Putting this all together in step-by-step form: to convert a binary integer to a binary scientific number:
Multiply the integer by 20: set the mantissa to the number we're converting, and the exponent to 0.
If the number is normalized (if the mantissa is 0, or if its leading bit is 1), we're done.
If the mantissa has more than one bit to the left of the decimal point (really the "radix point" or "binary point"), divide the mantissa by 2, and increment the exponent by 1. Return to step 2.
(This step will never be necessary if the number we started with was an integer.) If the mantissa is nonzero but the bit to the left of the radix point is 0, multiply the mantissa by 2, and decrement the exponent by 1. Return to step 2.
Running this algorithm in tabular form for our number 9, we have:
step mantissa exponent
0 1001. 0
1 100.1 1
2 10.01 2
3 1.001 3
So, if you're still with me, that's how we can convert the decimal integer 9 to the binary scientific notation (or floating-point) number 1.0012×23.
And, with all of that said, the algorithm as stated so far only works for decimal integers. What if we wanted to convert, say, the decimal number 1.25 to the binary number 1.012×20, or 34.125 to 1.000100012×25? That's a discussion that will have to wait for another day (or for this other answer), I guess.
Here's what I'm trying to do:
I need to print the fractional part of a floating number which has to be input as a float during user input.
The fractional part should be like: if float is 43.3423, the output should be 3423; and if number is 45.3400 output should be 3400.
This can be done easily with a string input but I need a way to make this work with float without losing the extra zeros or without appending zeros to user's original input.
Here's what I already tried :-
Take the fractional part by frac = num - (int)num and then multiplying frac until we get zero as the remainder. But this fails for cases like 34.3400 — the last two zeros won't get included with this method.
Convert the float number to a string by
char string[20];
sprintf(string, "%f", float_number);
The sprintf function puts the float number as a string but here also it doesn't automatically detect the user entered precision and fills the string with extra zeros at the end (6 total precision). So here also the information about the user's original entered precision is not obtained.
So, is there a way to get this done? The number must be taken as float number from user. Is there any way to get info about what's the user's entered precision? If it's not possible, an explanation would be very helpful.
I think I understand where you're coming from. E.g. in physics, it's a difference whether you write 42.5 or 42.500, the number of significant digits is implicitly given. 42.5 stands for any number x: 42.45 <= x < 42.55 and 42.500 for any x: 42.4995 <= x < 42.5005.
For larger numbers, you would use scientific notation: 1.0e6 would mean a number x with x: 950000 <= x < 1050000.
A floating point number uses this same format, but with binary digits (sometimes called bits ;)) instead of decimal digits. But there are two important differences:
The number of digits (bits) used depends only on the data type of the floating point number. If your data type has e.g. 20 bits for the mantissa, every number stored in it will have these 20 bits. The mantissa is always stored without a part after the "decimal" (binary?) point, so you won't know how many significant bits there are.
There's no direct mapping between bits and decimal digits. You will need roughly 3.5 bits to represent a decimal digit. So even if you knew a number of significant bits, you still wouldn't know how many significant decimal digits that would make.
To address your problem, you could store the number of significant digits yourself in something like this:
struct myNumber
{
double value;
int nsignificant;
};
Of course, you have to parse the input yourself to find out what to place in nsignificant. Also, use at least double here for the value, the very limited precision of float won't get you far. With this, you could use nsignificant to determine a proper format string for printing the number with the amount of digits you want.
This still has the problem mentioned above: you can't directly map decimal digits to bits, so there's never a guarantee your number can be stored with the precision you intend. In cases where an exact decimal representation is important, you'll want to use a different data type for that. C# provides one, but C doesn't. You'd have to implement it yourself. You could start with something like this:
struct myDecimal
{
long mantissa;
short exponent;
short nsignificant;
}
In this struct, you could e.g. place 1.0e6 like this:
struct myDecimal x = {
.mantissa = 1;
.exponent = 6;
.nsignificant = 2;
};
Of course, this would require you to write quite a lot of own code for parsing and formatting these numbers.
which has to be input as a float during user input.
So, is there a way to get this done.
Almost. The "trick" is to note the textual length of user input. The below will remember the offset of the first non-whitespace character and the offset after the numeric input.
scanf(" %n%f%n", &n1, &input, &n2);
n2 - n1 gives code the length of user input to represent the float. This method can get fooled if user input is in exponential notation, hexadecimal FP notation, infinity, Not-a-number, excessive leading zeros, etc. Yet works well with straight decimal input.
The idea is to print the number to a buffer with at least n2 - n1 precision and then determine how much of the fractional portion to print.
Recall that float typically has about 6-7 significant leading digits of significance, so attempting to input text like "123456789.0" will result in a float with the exact value of 123456792.0 and the output will be based on that value.
#include <float.h>
#include <math.h>
int scan_print_float(void) {
float input;
int n1, n2;
int cnt = scanf(" %n%f%n", &n1, &input, &n2);
if (cnt == 1) {
int len = n2 - n1;
char buf[len * 2 + 1];
snprintf(buf, sizeof buf, "%.*f", len, input);
char dp = '.';
char *p = strchr(buf, dp);
if (p) {
int front_to_dp = p + 1 - buf;
int prec = len - front_to_dp;
if (prec >= 0) {
return printf("<%.*s>\n", prec, p+1);
}
}
}
puts(".");
return 0;
}
int main(void) {
while (scan_print_float()) {
fflush(stdout);
}
return EXIT_SUCCESS;
}
Input/Output
43.3423
<3423>
45.3400
<3400>
-45.3400
<3400>
0.00
<00>
1234.500000
<500000>
.
.
To robustly handle this and the various edge cases, code should read user input as text and not as a float.
Note: float can typically represent about 232 numbers exactly.
43.3423 is usually not one of them. Instead it has an exactly value of 43.3423004150390625
43.3400 is usually not one of them. Instead it has an exactly value of 43.340000152587890625
The only way is to create a struct with the original string value and/ or required precision for rounding
I have a homework. The question is:
Write a function that takes as parameter a single integer, int input, and returns an unsigned character
such that:
a- if input is negative or larger than 11,111,111 or contains a digit that is not 0 or 1; then the function will
print an error message (such as “invalid input”) and return 0,
b- otherwise; the function will assume that the base-10 value of input represents a bit sequence and return the magnitude-only bit model correspondent of this sequence.
For Example: If input is 1011, return value is 11 and If input is 1110, return value is 14
This is my work for a, and I am stuck on b. How can I get bit sequence for given integer input?
int main()
{
int input = 0;
printf("Please type an integer number less than 11,111,111.\n");
scanf("%d",&input);
if(input < 0 || input > 11111111)
{
printf("Invalid Input\n");
system("PAUSE");
return 0;
}
for (int i = 0; i < 8; i++)
{
int writtenInput = input;
int single_digit = writtenInput%10;
if(single_digit == 0 || single_digit == 1)
{
writtenInput /= 10;
}
else
{
printf("Your digit contains a number that does not 0 or 1. it is invalid input\n");
system("PAUSE");
return 0;
}
}
printf("Written integer is %d\n",input);
system("PAUSE");
return 0;
}
The bit that you are missing is the base conversion. To interpret a number in base B, what you need to do is multiply digit N times B^N (assuming that you start counting digits from the least significative). For example, in base 16, A108 = (10)*16^3 + 1*16^2 + 0*16^1 + 8*16^0. Where in your base is 2 (binary).
Alternatively, you can avoid the exponentiation if you rewrite the expression as:
hex A008 = ((((10*16) + 1)*16 +0)*16 + 8
Which would be simpler if your input was stated in terms of an array of possibly unknown length as it is easily convertible into a loop of only additions and multiplications.
In the particular case of binary, you can use another direct solution, for each digit that is non-zero, set the corresponding bit in an integer type large enough (in your case, unsigned char suffices), and the value of the variable at the end of the loop will be the result of the conversion.
First off, the existing part of your code needs work. It doesn't report non-binary digits correctly. The question is "Are there any digits that are neither 0 nor 1". To answer this question in the negative, you need to check every single digit. Your code can terminate early. I also suggest renaming the function to something that clearly tells you what it does. For example haveNonBinaryDigit. The term check doesn't tell you what you should expect on the return value.
As for the second part, read up on the binary representation. It is fairly similar to decimal, except that instead of each digit being weighted by 1, 10, 100, .., 10^x, they are weighted by 1, 2, 4, ..., 2^n. Also the digits can only have values 0 and 1.
Your current code has a problem: it will terminate as soon as it finds one single binary bit (at the end) and call the entire integer correct.
I would take a different approach to this problem. What you want to do is get your hands on every single digit from the integer seperately. You made a good start in doing so by using modulo. Say you had this code:
for (int i = 0; i < 8; i++)
{
char single_digit = input%10;
input /= 10;
}
This makes it easy to start working with the digits. You would be checking 8 bits because that seems to be the maximum allowed (11,111,111). Check if each digit is either 0 or 1. Then you can start pushing it in to an unsigned character using bitwise operations. Shift every digit to the left by i, then use a bitwise OR.