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This program works in handling positive integers, but not on negative integers. How can I fix this one? Thanks!
By the way, is my code good or not? Is there a better way on getting the quotient and remainder without using the '/', '%', and '*' operators?
#include <stdio.h>
int divide(int x, int y, int quotient);
int getRem(int x, int y, int quotient, int product, int count,
int remainder);
int main()
{
int dividend, divisor, quotient = 0, product = 0;
int remainder, count = 0;
scanf("%d %d", ÷nd, &divisor);
printf("\nQuotient: %d", divide(dividend, divisor, quotient));
quotient = divide(dividend, divisor, quotient);
printf("\nRemainder: %d", getRem(dividend, divisor, quotient, product, count, remainder));
}
int divide(int x, int y, int quotient)
{
while (x > 0)
{
x -= y;
quotient++;
}
if (x != 0)
return quotient - 1;
else
return quotient;
}
int getRem(int x, int y, int quotient, int product, int count, int remainder)
{
while (count != y)
{
product += quotient;
count++;
remainder = x - product;
}
return remainder;
}
By the way, is my code good or not?
Well, there's room for improvements...
First of all - don't pass unnecessary variables to your function!
A function that shall divide x by y shall only take x and y as arguments. Whatever variables you need inside the function shall be defined inside the function.
So the first step is to change your divide function to be:
int divide(int x, int y)
{
int quotient = 0; // use a local variable
while (x > 0)
{
x -= y;
quotient++;
}
if (x != 0)
return quotient - 1;
else
return quotient;
}
Another (minor) issue is the two return statements. With a simple change of the while statement that can be avoided.
int divide(int x, int y)
{
int quotient = 0; // use a local variable
while (x >= y) // notice this change
{
x -= y;
quotient++;
}
return quotient;
}
Also notice that a call like divide(42, 0); will cause an infinite loop. So perhaps you should check for y being zero.
The algorithm can be improved - especially for large numbers - but I guess you want a simple approach so I stick to your basic algorithm.
... but not on negative integers. How can I fix this one?
A simple approach is to convert any negative input before entering the loop and maintain a counter to remember the number of negative numbers. Something like:
int divide(int x, int y)
{
int quotient = 0;
int negative = 0;
if (x < 0)
{
x = -x; // Make x positive
++negative;
}
if (y < 0)
{
y = -y; // Make y positive
++negative;
}
while (x >= y) // Both x and y are positive here
{
x -= y;
quotient++;
}
return (negative == 1) ? -quotient : quotient;
}
int main(void)
{
printf("%d\n", divide( 5, 2));
printf("%d\n", divide( 5,-2));
printf("%d\n", divide(-5, 2));
printf("%d\n", divide(-5,-2));
printf("%d\n", divide( 6, 2));
printf("%d\n", divide( 6,-2));
printf("%d\n", divide(-6, 2));
printf("%d\n", divide(-6,-2));
return 0;
}
Output:
2
-2
-2
2
3
-3
-3
3
You can apply the same kind of changes to the function getRem and I'll leave that part for you as an exercise...
However, notice that your current function uses quotient without any benefit. The function (only handling positive numbers) could simply be:
int getRem(int x, int y) // requires x >= 0 and y > 0
{
while (x >= y)
{
x -= y;
}
return x;
}
Is there a better way of getting the quotient and remainder without using the '/', '%', and '*' operators?
The majority of the time, I think, by far the best way of computing quotient and remainder is with the / and % operators. (It's their job, after all!)
If you need both quotient and remainder at the same time, there's also the div function, which does exactly that. (Its return value is a struct containing quot and rem members.) It's a bit cumbersome to use, but it might be more efficient if it means executing a single divide instruction (which tends to give you both quotient and remainder at the machine level anyway) instead of two. (Or, these days, with a modern optimizing compiler, I suspect it wouldn't make any difference anyway.)
But no matter how you do it, there's always a question when it comes to negative numbers. If the dividend or the divisor is negative, what sign(s) should the quotient and remainder be? There are basically two answers. (Actually, there are more than two, but these are the two I think about.)
In Euclidean division, the remainder is always positive, and is therefore always in the range [0,divisor) (that is, between 0 and divisor-1).
In many programming languages (including C), the remainder always has the sign of the dividend.
See the Wikipedia article on modulo operation for much, much more information on these and other alternatives.
If your programming language gives you the second definition (as C does), but you want a remainder that's never negative, one way to fix it is just to do a regular division, test whether the remainder is negative, and if it is, increment the remainder by the divisor and decrement the quotient by 1:
quotient = x / y;
remainder = x % y;
if(remainder < 0) {
reminder += y;
quotient--;
}
This works because of the formal definition of integer division with remainder:
div(x, y) → q, r such that y × q + r = x
If you subtract 1 from q and add y to r, you get the same result, and it's still x.
Here's a sample program demonstrating three different alternatives. I've encapsulated the little "adjust quotient for nonnegative remainder" algorithm as the function euclid(), which returns the same div_t type as div() does:
#include <stdio.h>
#include <stdlib.h>
div_t euclid(int, int);
int main()
{
int x = -7, y = 3;
printf("operators: q = %d, r = %d\n", x/y, x%y);
div_t qr = div(x, y);
printf("div: q = %d, r = %d\n", qr.quot, qr.rem);
qr = euclid(x, y);
printf("euclid: q = %d, r = %d\n", qr.quot, qr.rem);
}
div_t euclid(int x, int y)
{
div_t qr = div(x, y);
if(qr.rem < 0) {
qr.quot--;
qr.rem += y;
}
return qr;
}
To really explore this, you'll want to try things out for all four cases:
positive dividend, positive divisor (the normal case, no ambiguity)
negative dividend, positive divisor (the case I showed)
positive dividend, negative divisor
negative dividend, negative divisor
This program works in handling positive integers, but not on negative integers. How can I fix this one?
When you call :
divide(-10, 3) or divide(-10, -3) the while loop condition while(x > 0) will be false
divide(10, -3) the while loop will be infinite loop because of the statement x -= y. (x) will be increase by the value of (y)
is my code good or not?
Your code is need to be organized:
in the main function you need to pass only the necessary parameters. in the divide function why you pass the quotient. the quotient will be calculated inside the function. so that you should edit the prototype to int divide(int x, int y);.
the same thing with getRem function you should edit the prototype to int getRem(int x, int y);.
in the main function you are calling the divide function twice to print the quotient and to save the quotient in the variable quotient. instead you should call the function only one time and reuse the returned value.
After the previous points your main and functions prototypes should be as follow:
#include <stdio.h>
int divide(int x, int y);
int getRem(int x, int y);
int main()
{
int dividend, divisor, quotient = 0;
scanf("%d %d", ÷nd, &divisor);
quotient = divide(dividend, divisor);
printf("\nQuotient: %d", quotient);
printf("\nRemainder: %d", getRem(dividend, divisor));
}
Now lets analyze the function divide. The first point, In math when multiplying or dividing, you actually do the operation on the sign as doing on the number. the following is the math rules for multiply or divide the sign.
negative * positive -> negative
positive * negative -> negative
negative * negative -> positive
negative / positive -> negative
positive / negative -> negative
negative / negative -> positive
The second point is the while loop. You should divide until (x) is less than (y).
As an example: suppose x = 7, y = 3 :
after first loop x -> 4 and y -> 3
after second loop x -> 1 and y -> 3 so that the condition should be while(x >= y)
so now you should first manipulate the sign division after that the numbers division. as follows.
int divide(int x, int y)
{
int quotient = 0;
//save the resulting sign in the sign variable
int sign = 1;
int temp = 0;
/* sign division */
// negative / positive -> negative
if((x < 0) && (y >= 0)){
sign = -1;
temp = x;
x -= temp;
x -= temp;
}// positive / negative -> negative
else if((x >= 0) && (y < 0)){
sign = -1;
temp = y;
y -= temp;
y -= temp;
}// negative / negative -> positive
else if((x < 0) && (y < 0)){
temp = x;
x -= temp;
x -= temp;
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
quotient++;
}
if(sign < 0){
temp = quotient;
quotient -= temp;
quotient -= temp;
}
return quotient;
}
Lets go into the getRem function. The remainder(%) operation rules is as follows:
negative % (negative or positive) -> negative
positive % (negative or positive) -> positive
note: the result follows the sign of the first operand
As an Example:
suppose x = -10 and y = 3, x / y = -3, to retrieve (x) multiply y and -3, so x = y * -3 = -9 the remainder is equal to -1
suppose x = 10 and y = -3, x / y = -3',x = y * -3 = 9` the remainder is equal to 1
now apply the previous points on the getRem function to get the following result:
int getRem(int x, int y)
{
int temp, remSign = 1;
if(x < 0){
remSign = -1;
temp = x;
x -= temp;
x -= temp;
}
if(y < 0){
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
}
if(remSign < 0){
x = -x;
}
return x;
}
You can to combine the two operation in one function using pointers for remainder as follows:
#include <stdio.h>
int divide(int x, int y,int * rem);
int getRem(int x, int y);
int main()
{
int dividend, divisor, quotient = 0;
int rem;
scanf("%d %d", ÷nd, &divisor);
quotient = divide(dividend, divisor, &rem);
printf("\nQuotient: %d", quotient);
printf("\nRemainder: %d", rem);
}
int divide(int x, int y, int * rem)
{
int quotient = 0;
int sign = 1;
int remSign = 1;
int temp = 0;
if((x < 0) && (y >= 0)){
sign = -1;
remSign = -1;
temp = x;
x -= temp;
x -= temp;
}
else if((x >= 0) && (y < 0)){
sign = -1;
temp = y;
y -= temp;
y -= temp;
}
else if((x < 0) && (y < 0)){
temp = x;
x -= temp;
x -= temp;
temp = y;
y -= temp;
y -= temp;
}
while (x >= y)
{
x -= y;
quotient++;
}
if(remSign < 0){
*rem = -x;
}else{
*rem = x;
}
if(sign < 0){
temp = quotient;
quotient -= temp;
quotient -= temp;
}
return quotient;
}
I want to make a bitwise AND computation over integers, but without converting them to binary numbers. For example, I have a integer "10111" (it is integer, not binary) and another integer "01001". I want bitwise AND of these numbers without converting them to binary and then making bitwise AND. I know it is not bitwise what I ask, but I want something similar to this. I know it can be interpreted initially as binary, converted to decimal and then do bitwise AND, but I do not want that. I want something like this:
int a;
int b;
int temp;
double result;
temp = a & b;
while (result != 0) {
if (result % 10 == 1)
count++;
result /= 10;
}
int length = floor(log10(abs(a))) + 1;
result = count / length;
return result;
I want this to check similarity of the Bag of Words(from natural language processing, string of 0s and 1s). I am importing Bag of Words in Monetdb, Column type should be Integer (Not string). If I have for example "10111" and "01001" in the Integer type cells, I want to get "00001" and fraction 1/5, because only 1 positions matches.
Thanks in advance
Might be a bit bulky, but it kind of works) You can optimize it yourself. I hope that I get you correctly.
IDEOne demo
#include <stdio.h>
unsigned int weirdAnd(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a && b) {
result += ((a % 10) && (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
unsigned int weirdOr(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a || b) {
result += ((a % 10) || (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
int main(void) {
// your code goes here
unsigned int a = 10110;
unsigned int b = 10011;
printf("%u and \n%u = \n%u\n\n", a, b, weirdAnd(a, b));
printf("%u or \n%u = \n%u\n\n", a, b, weirdOr(a, b));
return 0;
}
Output:
10110 and 10011 = 10010
10110 or 10011 = 10111
The problem is that and works on bits only, and it does not care if the input numbers are given in decimals, octal, hexadecimal, or any other way. To force and to work correctly, you must give it an input in 'binary', that is, only ones and zeroes. To do so you need to grab each digit of the input numbers as if they are binary digits.
The following works as expected:
#include <stdio.h>
int cheap_pow (int base, int exponent)
{
int result = base;
while (exponent-- > 0)
result *= base;
return result;
}
int main (void)
{
int a = 10111, b = 1001 ;
int result, factor;
printf ("BAD: %05d AND %05d = %05d\n", a, b, a & b);
printf ("GOOD: %05d AND %05d = ");
result = 0;
factor = 1;
while (a | b)
{
result += factor * ((a & 1) and (b & 1));
factor *= 10;
a /= 10;
b /= 10;
}
printf ("%05d\n", result);
}
but you must be careful when defining your inputs. When written directly into your source code, a value b = 01001 will be interpreted by the C compiler as octal (base 8), rather than decimal; and it will have the (decimal) value 513. It's just a Rule of C.
If your input comes from elsewhere, you don't need to take this into account – except when you use a standard function such as strtol, where you should carefully read the documentation because it does the same thing:
If base is zero or 16, the string may then include a "0x"
prefix, and the number will be read in base 16; otherwise, a zero
base is taken as 10 (decimal) unless the next character is '0', in
which case it is taken as 8 (octal).
An additional note is this program only will work in the range for signed int, that is (currently) up to 10 "binary" digits. If you need some more, you could switch to a larger type; but beyond that, you are better off with a not-numerical solution and use a character strings throughout.
int a, b;
int tmp = a & b;
int res = 0;
while ((tmp != 0 &&) (tmp / 10 != 0)){
int dig = tmp % 10;
res = (dig == 1)? ++res: res;
tmp /= 10;
}
Try this.
I have a logical problem in my code, maybe it is caused by overflowing but I can't solve this on my own, so I would be thankful if anyone can help me.
In the following piece of code, I have implemented the function taylor_log(), which can count "n" iterations of taylor polynomial. In the void function I am looking for number of iterations (*limit) which is enough to count a logarithm with desired accuracy compared to log function from .
The thing is that sometimes UINT_MAX is not enough iterations to get the desired accuracy and at this point I want to let the user know that the number of needed iterations is higher than UINT_MAX. But my code don't work, for example for x = 1e+280, eps = 623. It just counts, counts and never give result.
TaylorPolynomial
double taylor_log(double x, unsigned int n){
double f_sum = 1.0;
double sum = 0.0;
for (unsigned int i = 1; i <= n; i++)
{
f_sum *= (x - 1) / x;
sum += f_sum / i;
}
return sum;
}
void guessIt(double x, double eps, unsigned int *limit){
*limit = 10;
double real_log = log(x);
double t_log = taylor_log(x, *limit);
while(myabs(real_log - t_log) > eps)
{
if (*limit == UINT_MAX)
{
*limit = 0;
break;
}
if (*limit >= UINT_MAX/2)
{
*limit = UINT_MAX;
t_log = taylor_log(x, *limit);
}
else
{
*limit = (*limit) *2;
t_log = taylor_log(x, *limit);
}
}
}
EDIT: Ok guys, thanks for your reactions so far. I have changed my code to this:
if (*limit == UINT_MAX-1)
{
*limit = 0;
break;
}
if (*limit >= UINT_MAX/2)
{
*limit = UINT_MAX-1;
t_log = taylor_log(x, *limit);
}
but it still doesn't work correctly, I have set printf to the beggining of taylor_log() function to see the value of "n" and its (..., 671088640, 1342177280, 2684354560, 5, 4, 3, 2, 2, 1, 2013265920, ...). Don't understand it..
This code below assigns the limit to UINT_MAX
if (*limit >= UINT_MAX/2)
{
*limit = UINT_MAX;
t_log = taylor_log(x, *limit);
}
And your for loop is defined like this:
for (unsigned int i = 1; i <= n; i++)
i will ALWAYS be less than or equal to UINT_MAX because there is never going to be a value of i that is greater than UINT_MAX. Because that's the largest value i could ever be. So there is certainly overflow and your loop exit condition is never met. i rolls over to zero and the process repeats indefinitely.
You should change your loop condition to i < n or change your limit to UINT_MAX - 1.
[Edit]
OP coded correctly but must insure a limited range (0.5 < x < 2.0 ?)
Below is a code version that self determines when to stop. Iteration count goes high near x near 0.5 and 2.0. The iteration count needed goes into the millions. Such the alternative coded far below.
double taylor_logA(double x) {
double f_sum = 1.0;
double sum = 0.0;
for (unsigned int i = 1; ; i++) {
f_sum *= (x - 1) / x;
double sum_before = sum;
sum += f_sum / i;
if (sum_before == sum) {
printf("%d\n", i);
break;
}
}
return sum;
}
Wrongalternative implementation of the series: Ref
Sample alternative - it converges faster.
double taylor_log2(double x, unsigned int n) {
double f_sum = 1.0;
double sum = 0.0;
for (unsigned int i = 1; i <= n; i++) {
f_sum *= (x - 1) / 1; // / 1 (or remove)
if (i & 1) sum += f_sum / i;
else sum -= f_sum / i; // subtract even terms
}
return sum;
}
A reasonable number of terms will converge as needed.
Alternatively, continue until terms are too small (maybe 50 or so)
double taylor_log3(double x) {
double f_sum = 1.0;
double sum = 0.0;
for (unsigned int i = 1; ; i++) {
double sum_before = sum;
f_sum *= x - 1;
if (i & 1) sum += f_sum / i;
else sum -= f_sum / i;
if (sum_before == sum) {
printf("%d\n", i);
break;
}
}
return sum;
}
Other improvements possible. example see More efficient series
First, using std::numeric_limits<unsigned int>::max() will make your code more c++-ish than c-ish. Second, you can use the integral type unsigned long long and std::numeric_limits<unsigned long long>::max() for the limit, which is pretty mush the limit for an integral type. If you want a higher limit, you may use long double. floating points also allows you to use infinity with std::numeric_limits<double>::infinity() note that infinity work with double, float and long double.
If neither of these types provide you the precision you need, look at boost::multiprecision
First of all, the Taylor series for the logarithm function only converges for values of 0 < x < 2, so it's quite possible that the eps precision is never hit.
Secondly, are you sure that it loops forever, instead of hitting the *limit >= UINT_MAX/2 after a very long time?
OP is using the series well outside its usable range of 0.5 x < 2.0 with calls like taylor_log(1e280, n)
Even within the range, x values near the limits of 0.5 and 2.0 converge very slowly needing millions+ of iterations. A precise log() will not result. Best to use the 2x range about 1.0.
Create a wrapper function to call the original function in its sweet range of sqrt(2)/2 < x < sqrt(2). Converges, worst case, with about 40 iterations.
#define SQRT_0_5 0.70710678118654752440084436210485
#define LN2 0.69314718055994530941723212145818
// Valid over the range (0...DBL_MAX]
double taylor_logB(double x, unsigned int n) {
int expo;
double signif = frexp(x, &expo);
if (signif < SQRT_0_5) {
signif *= 2;
expo--;
}
double y = taylor_log(signif,n);
y += expo*LN2;
return y;
}
I'm trying to input two numbers and remove 8s and 9s (not convert them, only remove unneeded numbers) so they can fit in the octal base.
When I run the program and input the numbers, they are almost correctly returned, but they are deducted by a bit.
I still don't know why and how to fix it. It might be something with zeroes, but I don't know.
int octal(int a)
{
int b = 0;
int i = 0;
while(a > 0){
if((a % 10) <= 7){
b= pow(10,i) * (a%10)+b;
i++;
}
a=a/10;
}
return b;
}
int main()
{
int j, o, a, b;
scanf(" %d %d", &j, &o);
a=octal(j);
b=octal(o);
printf("%d\n%d\n",a,b);
return 0;
}
edit: Example
Input: 72349 and 91238
Output: 7233 and 122
Code is encountering a weak double pow() that is returning values near the expected mathematical result. When the result is just under a whole number, the conversion back to int results in fraction truncation - results appear 1 less then expected.
Could use round(pow(10,i)) or better yet, stay with int math as follows.
Other simplifications possible, but minimal code changes to outline OP's issue.
int octal(int a) {
int b = 0;
int i = 0;
int pow10 = 1;
while (a > 0) {
if ((a % 10) <= 7) {
//b = pow(10,i) * (a%10)+b;
b = pow10 * (a % 10) + b;
pow10 *= 10;
i++;
}
a = a / 10;
}
return b;
}
Care for a walk on the recursive side of life?
int octal(int a) {
assert(a >= 0);
while (a >= 8) {
int digit = a%10;
a /= 10;
if (digit < 8) {
return octal(a)*10 + digit;
}
}
return a;
}
Let's assume that I have n=1234 and I want to get the first x digits of n. Assume x=2, in C math I just compute 1234/100 and I will get 12. But how can I do it programatically? I.e., using math.
I have implemented it by the horrible way, converting to string and putting a 0 at x position.
If possible, I want to avoid built-in C functions because my goal is to convert the algorithm to assembly language later.
Without using any library functions, the best way to do it is with brute force. The maximum value an integer can take is 2147483648 so we won't deal with anything over that.
int first_two(int value)
{
assert(value >= 0); // unspecified for negative numbers
if (value >= 1000000000)
return value / 100000000;
if (value >= 100000000)
return value / 10000000;
if (value >= 10000000)
return value / 1000000;
if (value >= 1000000)
return value / 100000;
if (value >= 100000)
return value / 10000;
if (value >= 10000)
return value / 1000;
if (value >= 1000)
return value / 100;
if (value >= 100)
return value / 10;
return value;
}
You can do it programmatically by taking the floor of the decimal logarithm of your number (in case of 1234, it's floor(3.091315), which is 3), adding one, and subtracting n - the desired number of decimal digits. This would give you x such that integer-dividing the original value by 10^x gives you the desired result:
#include <math.h>
...
int num = 12345;
int n = 3;
int log10 = (log(num)/log(10))+1;
int divisor = pow(10, log10-n);
int res = num / divisor;
printf("%d\n", res);
Here is a demo on ideone.
Converting the above to assembly would be tricky because of the math functions involved. You can simplify it by defining a table of powers of ten, searching it for the first item that's greater than or equal to the desired number (giving you log10 above) and then grabbing the log10-n-th entry, giving you pow(10, log10-n):
int pow10[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};
int main(void) {
int num = 12345;
int n = 3;
int log10 = 0;
while (pow10[log10] < num) {
log10++;
}
int divisor = pow10[log10-n];
int res = num / divisor;
printf("log10(num)+1=%d, divisor=%d, result=%d\n", log10, divisor, res);
return 0;
}
Here is the modified demo.
You can use the following algorithm: keep divide the n per 10 till you get n < 10^x
here after the code
int power10(int x) {
int p = 1;
while (x) {
p *= 10;
x--;
}
return p;
}
int main (void) {
int x = 2;
int n = 1234;
int max = power10(x);
int res = n;
while(res>=max)
res = res/10;
printf("%d\n",res);
}
int getLeftDigits(double num, double numOfDigits)
{
double divider = pow(10, numOfDigits);
if (num < divider)
return num;
getLeftDigits(num/10, numOfDigits);
}
If you want to avoid the use of the pow function, you can just implement it by yourself as shown in one of the other comments here.