generate random number 0-1, when I type the input number less than 4, the code works fine. However when the input number above 4, the eclipse stop working. what's wrong with my code
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
double ran(double x0, double x1){
return x0+(x1-x0)*rand()/((double)RAND_MAX);
}
int main(void) {
int a,i;
double *b;
printf("input the size\n");
scanf("%d", &a);
b=(double*)malloc(sizeof(int)*a);
srand((unsigned)time(NULL));
for(i=0;i<a;i++)
{
b[i]=ran(0,1);
printf("\n %f", b[i]);
}
free (b);
return 1;
}
A double is bigger than an integer. A double is eight bytes, while an integer is 4 bytes.
You should replace
b = (double*)malloc(sizeof(int)*a);
with
b = (double*)malloc(sizeof(double)*a);
or even better (thaks to Lưu Vĩnh Phúc)
b = malloc(a * sizeof b[0]);
Related
Examples are
A B C
1 2 3
369 428 797
We can consider A=01 and B=10 in binary. Then, there are two possible ways: swapping the two bits of A and the two bits of B (As=10, Bs=01 in binary, 2 and 1 in decimal representation) or not shuffling any bits.
number 2 is done in same way but dont know how to do it.... in c++ code
is there any basic idea I am missing
I can use the code to find all combinations of string to solve this...
but this will take too much time for 10^9
time limit is 1s
this code shows all combinations of string
#include <stdio.h>
#include <string.h>
/* Function to swap values at two pointers */
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
}
/* Function to print permutations of string
This function takes three parameters:
1. String
2. Starting index of the string
3. Ending index of the string. */
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf("%s\n", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r);
swap((a+l), (a+i)); //backtrack
}
}
}
/* Driver program to test above functions */
int main()
{
char str[] = "10000000000111111110101010101010101";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
I am trying to print a number up to say 5 decimal places.
I have:
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
float a = 987.65;
float b = 1.23456789;
scanf("%f %f", &a, &b);
printf("%.5f %.5f", a, b);
return 0;
}
I get the result as 987.65000 and 1.23456
I want the result to be 987.65 and 1.23456, so basically I want up to 5 i.e <=5 decimal digits in my answer.
A slightly less technical way to do it would be using setprecision, as displayed below:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float a = 987.65;
float b = 1.23456789;
cout << setprecision(5);
cout << a << " " << b << endl;
return 0;
}
Output:
987.65 1.2346
The fundamental problem is that a computer can't exactly represent most floating point numbers.
Also, you want a complex formatting rule: if ending digits are zero, print spaces.
The problem is that your number 987.65000 could be represented as 98.6500001 and so wouldn't work with your formatting rule.
I believe you will have to write your own formatting function to achieve the functionality you are looking for.
Here's a solution that does what you want:
#include <iostream>
#include <cstdio>
using namespace std;
unsigned int numPlaces = 5; //number of decimal places
unsigned int determinePlaces (float number) {
unsigned int myPlaces = 0;
while (number != (int)(number) && myPlaces<=numPlaces-1) {
number = number*10;
myPlaces++;
}
return myPlaces;
}
int main() {
float a = 987.65;
float b = 1.23456789;
printf("%.*f %.*f\n", determinePlaces(a), a, determinePlaces(b), b);
return 0;
}
Output:
987.65 1.23457
Basically, the code keeps on multiplying the goat by 10 until the cast to an integer (essentially taking the floor value) matches the float. If it doesn't when it reaches the fifth multiplication by 10, we satisfy ourselves with a printf of 5 decimal places. Otherwise, we print the amount that was necessary to make the match to the floor value.
This program compiles fine, but it returns a message "Floating Point Exception" when I run it. I've looked at other threads and the problem appears to be dividing by 0, but I have looked over the program and there's no division by zero in my entire program. I even used the absolute value function in case.
By the way, the program is meant to reduce fractions.
Example input: 6 12, representing the fraction 6/12
Expected output: 1/2
#include <stdio.h>
/*declaring variables*/
int num1, num2, num1b, num2b, gcd, x;
int higher, lower, higher_2, lower_2;
/*declaring functions*/
int find_gcd(int num1, int num2);
void reduce(int numerator, int denominator, int *reduced_numerator, int *reduced_denominator);
int main(void)
{
do
{
printf("enter 2 numbers: ");
scanf("%d %d", &num1, &num2);
reduce(higher, lower, &higher_2, &lower_2);
printf("enter 0 to end program and any number continue: \n");
scanf("%d", &x);
} while(x != 0);
return 0;
}
void reduce(int numerator, int denominator, int *reduced_numerator, int *reduced_denominator)
{
num1=numerator;
num2=denominator;
gcd =find_gcd(numerator, denominator);
*reduced_numerator = (numerator/abs(gcd));
*reduced_denominator = (denominator/abs(gcd));
printf("The GCD is %d/%d\n", *reduced_numerator, *reduced_denominator);
}
int find_gcd(int m, int n)
{
while (n != 0) {
int remainder = m % n;
m = n;
n = remainder;
}
return m;
}
Your main problem is that you are not passing your input values num1 and num2 into your reduce() function. Instead you are passing in the global variables higher and lower. You didn't assign any values to them, but global variables are always initialized to 0 by default. Therfore, you run into the exception, because in reduce() you divide 0 by 0. You can verify that with a debugger.
If I change your main() as follows, then your code is at least working for your test case with 6 and 12 as input:
int main(void)
{
do
{
printf("enter 2 numbers: ");
scanf("%d %d", &num1, &num2);
reduce(num1, num2, &higher_2, &lower_2);
printf("enter 0 to end program and any number continue: \n");
scanf("%d", &x);
} while(x != 0);
return 0;
}
Output:
enter 2 numbers: 6
12
The GCD is 1/2
enter 0 to end program and any number continue:
As indicated in the comments you should also get rid of global and spurious variables. Therefore, you should first delete the following lines in your code:
/*declaring variables*/
int num1, num2, num1b, num2b, gcd, x;
int higher, lower, higher_2, lower_2;
Then let your main() function start the following way:
int main(void)
{
int num1, num2, higher_2, lower_2, x;
...
}
And your reduce() function should read like this:
void reduce(int numerator, int denominator, int *reduced_numerator, int *reduced_denominator)
{
int gcd = find_gcd(numerator, denominator);
*reduced_numerator = (numerator/abs(gcd));
*reduced_denominator = (denominator/abs(gcd));
printf("The GCD is %d/%d\n", *reduced_numerator, *reduced_denominator);
}
So far, you don't use your variables higher_2 and lower_2 in the main() function, but I guess you plan to do so. If not, you can also get rid of them together with parameters 3 and 4 of your reduce() function.
There is another issue with the code you provided (thanks to #user3629249 for pointing it out): You are missing an include for the abs() function. So you need to add the line #include <stdlib.h> at the beginning of your code (include <math.h> will also so the trick, as well as include <Windows.h> on Windows).
#include <iostream>
using namespace std;
double calc(int a, int b);
int main()
{
int n1, n2;
cout << "Enter a number for a: ";
cin >> n1;
cout << "Enter a number for b: ";
cin >> n2;
cout << calc(n1, n2) << endl;
system("PAUSE");
return 0;
}
double calc(int a, int b)
{
double s;
s = (a) / ((sqrt(a / b)));
return s;
}
This program is meant to check whether the two integers are greater than zero. If it is it will calcualte the formula. Otherwise if one of the integers is zero or less than zero it will not return anything and exit the program.
My question here is that no matter what I input for a and b, i keep getting 1.#INF as the output and I have no idea why. I've checked the formula in a seperate program and it worked fine.
Any ideas?
Here, you are operating with int numbers:
s = (a) / ((sqrt(a / b)));
If a is less then b, then a/b (both are integers, remember, so the fractional part of the result will simply be lost) will be equal to 0, which leads to division by 0. You need to cast one of the numbers to double:
s = (a) / ((sqrt(static_cast<double>(a) / b)));
sqrt takes and returns a double. When you call it with integer arguments it will be converted in a double, and will thus get the value of infinity.
change your function signature to:
double calc(double a, double b);
and declare n1 and n2 as double.
You say that the function will exit the program when one of the integers are 0 or less, but where?
Try to check them like this:
Additionally, you should have a check whether a is greater than b
double calc(int a, int b)
{
double s;
if(a <= 0) exit(-1);
if(b <= 0) exit(-1);
if(a < b) exit(-1);
s = (a) / ((sqrt(a / b)));
return s;
}
You are having problems with infinity. For it use isinf. Here is some sample usage:
#include <stdio.h> /* printf */
#include <math.h> /* isinf, sqrt */
int main()
{
printf ("isinf(-1.0/0.0) : %d\n",isinf(-1.0/0.0));
printf ("isinf(sqrt(-1.0)): %d\n",isinf(sqrt(-1.0)));
return 0;
}
output:
isinf(-1.0/0.0) : 1 isinf(sqrt(-1.0): 0
Suppose I have a float. I would like to round it to a certain number of significant digits.
In my case n=6.
So say float was f=1.23456999;
round(f,6) would give 1.23457
f=123456.0001 would give 123456
Anybody know such a routine ?
Here it works on website: http://ostermiller.org/calc/significant_figures.html
Multiply the number by a suitable scaling factor to move all significant digits to the left of the decimal point. Then round and finally reverse the operation:
#include <math.h>
double round_to_digits(double value, int digits)
{
if (value == 0.0) // otherwise it will return 'nan' due to the log10() of zero
return 0.0;
double factor = pow(10.0, digits - ceil(log10(fabs(value))));
return round(value * factor) / factor;
}
Tested: http://ideone.com/fH5ebt
Buts as #PascalCuoq pointed out: the rounded value may not exactly representable as a floating point value.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char *Round(float f, int d)
{
char buf[16];
sprintf(buf, "%.*g", d, f);
return strdup(buf);
}
int main(void)
{
char *r = Round(1.23456999, 6);
printf("%s\n", r);
free(r);
}
Output is:
1.23457
Something like this should work:
double round_to_n_digits(double x, int n)
{
double scale = pow(10.0, ceil(log10(fabs(x))) + n);
return round(x * scale) / scale;
}
Alternatively you could just use sprintf/atof to convert to a string and back again:
double round_to_n_digits(double x, int n)
{
char buff[32];
sprintf(buff, "%.*g", n, x);
return atof(buff);
}
Test code for both of the above functions: http://ideone.com/oMzQZZ
Note that in some cases incorrect rounding may be observed, e.g. as pointed out by #clearScreen in the comments below, 13127.15 is rounded to 13127.1 instead of
13127.2.
This should work (except the noise given by floating point precision):
#include <stdio.h>
#include <math.h>
double dround(double a, int ndigits);
double dround(double a, int ndigits) {
int exp_base10 = round(log10(a));
double man_base10 = a*pow(10.0,-exp_base10);
double factor = pow(10.0,-ndigits+1);
double truncated_man_base10 = man_base10 - fmod(man_base10,factor);
double rounded_remainder = fmod(man_base10,factor)/factor;
rounded_remainder = rounded_remainder > 0.5 ? 1.0*factor : 0.0;
return (truncated_man_base10 + rounded_remainder)*pow(10.0,exp_base10) ;
}
int main() {
double a = 1.23456999;
double b = 123456.0001;
printf("%12.12f\n",dround(a,6));
printf("%12.12f\n",dround(b,6));
return 0;
}
If you want to print a float to a string use simple sprintf(). For outputting it just to the console you can use printf():
printf("My float is %.6f", myfloat);
This will output your float with 6 decimal places.
Print to 16 significant digit.
double x = -1932970.8299999994;
char buff[100];
snprintf(buff, sizeof(buff), "%.16g", x);
std::string buffAsStdStr = buff;
std::cout << std::endl << buffAsStdStr ;