I have this problem with this homework I'm supposed to do.
[ It says Create a program that's able to calculate and show the sum of S]
Like S=1+1/4+1/8+1/16 ... till 1/ [2 pow n]
So I worked on it and came up with this code
#include <stdio.h>
void main()
{
int n,i;
float p,s;
printf("Enter the maximum power n :");
scanf("%d",&n);
s=0;
p=0;
for (i=0;i<n;i++)
{
p+=1/pow(2, i);
s+=p;
printf("s = %f\n",s);
}
printf("The sum of this equation is :%f",&s);
}
But when I execute it is always like S=0.
What am I doing wrong?
You are printing an address ('&s) with%f` variable. Using a wrong specifier invokes undefined behavior. You may get anything.
Also, No need of variable s. Remove the line
s+=p;
It should be like:
#include <stdio.h>
int main(void)
{
int n,i;
float p;
printf("Enter the maximum power n :");
scanf("%d",&n);
p=0;
for (i=0;i<n;i++)
{
p+=1/pow(2, i);
printf("p = %f\n",p);
}
printf("The sum of this equation is :%f",p);
}
You need to include <math.h> to get the proper prototype of pow().
You might need to link to the math library too gcc main.c -Wall -lm
#include <math.h>
....
for (i=0;i<n;i++)
{
p=1/pow(2, i);
s+=p;
printf("s = %f\n",s);
}
printf("The sum of this equation is :%f",s);
Your program has multiple problems. Enabling compiler warnings should tell you about some of them.
You should include the C header which contains the declaration of the pow function.
You add each addend twice.
In your second printf statement, you pass a float. But the %f format specifier expects a double argument. In your third printf statement, you pass a pointer to a float.
Another cosmetic problem is that your main function should return an int.
just a guess, may you replace the line
p+=1/pow(2, i);
with
p+=1.0f/(float)pow(2, i);
and
printf("The sum of this equation is :%f",&s);
with
printf("The sum of this equation is :%f",s);
Typo may be.. but you will have say %f and s (not &s)
printf("The sum of this equation is :%f",s);
On side note:
Once you include <math.h> you will get compiler warning for using correct pow(..) prototype. Below code would be relevant.
p+=1.0f/(float)pow(2.0f, i);
Related
I'm currently working in a c program that calculates the pi value using Leibniz summation, but I'm getting the error "segment violation (`core 'generated)" in the console once the program is executed. Also I've checked similar questions but I'm still unable to fit that info into my problem. This is the code I have:
#include <stdio.h>
double calPi(int x);
double calPi(int x)
{
double sum;
int i;
sum=0.0;
for (i=0; i<=x; i++)
{
if ((i=2)||((i%2)==0))
{
sum=sum+(1/(2*i+1));
}
else sum=sum-(1/(2*i+1));
}
return 4*sum;
}
int main(void)
{
int x;
double PI;
printf("Enter the number of terms you want to calculate");
scanf("%i",x);
PI=calPi(x);
printf("Pi value is: %f", PI);
system("pause");
return 0;
}
Thanks in advance for any hint, help or correction.
This is your problem:
scanf("%i", x);
Should be
scanf("%i", &x);
You need to take the address of x, otherwise just using x , will be treated as a pointer pointing to who knows what. Make sure you compile with warnings on, as your compiler should warn you about stuff like that. My compiler complains with
"format string '%i' requires an argument of type 'int *', but variadic argument 1 has type 'int'"
I am currently attempting to learn C, and have made this program to calculate the area of a regular hexagon:
#include <stdio.h>
#include <math.h>
void main(){
int a;
float ans;
scanf("%d", &a); // get length of side
ans = ((pow(a, (1/3)))/2)*(a*a);
printf("%f", ans);
}
However, it outputs seemingly random numbers.
Firstly your code doesn't compile (Missing semicolon) and also you should use int main() instead of void main().
Secondly your formula also wrong, the area of a regular hexagon of side length a is calculated as ((3√3)/2)*a².
Thirdly Expression like 1/3 always yield zero as both are integer, to get expected behavior make one of them float/double. like 1.0/3 or (float)1/3 etc.
#include <stdio.h>
#include <math.h>
int main()
{
int a;
float ans;
scanf("%d", &a); // get length of side
ans = (3*sqrt(3)/2.0)*a*a;
printf("%f", ans);
}
Here is my code:
#include <stdio.h>
#include <math.h>
int main(void)
{
double x, y, z;
double numerator;
double denominator;
printf("This program will solve (x^2+y^2)/(x/y)^3\n");
printf("Enter the value for x:\n");
scanf("%lf", x);
printf("Enter the value for y:\n");
scanf("%lf", y);
numerator = sqrt(x) + sqrt(y);
denominator = pow((x/y),3);
z = (numerator/denominator);
printf("The solution is: %f\n", z);
return(0);
}
Can anyone give me a (hopefully) quick pointer to fix my infinite loop?
There's no loop in your function, so I think it's your calls to scanf() that are causing the error:
You need to pass reference to scanf(), i.e. use scanf("%lf",&x) instead of scanf("%lf",x).
BTW, according to your fonction definition, you should use pow(x,2) instead of sqrt(x) which returns the square root.
Since this is your first question
**Welcome to stack overflow**
Your code doesnt go into an infinite loop,there is a runtime error.
Your scanf Code is flawed use this:
scanf("%lf",&x);
scanf("%lf",&y);
you want scanf to modify the value contained at the address field of your value.Please read tutorials.
Also use
numerator=pow(x,2) + pow(y,2);//numerator=x^2+y^2
It's not infinite loop, your code just returns infinity. And that is because scanf() needs a pointer to variable where it should put the read number. To get an address of variable you can use & operator like this:
scanf("%lf", &x);
scanf("%lf", &y);
As I am new to programming, I was trying to write a simple code using functions which will give me the addition of three numbers. Here's the code!
/* Your includes go here */
#include <stdio.h>
int addThreeNumbers(int a, int b, int c)
{
int d;
d = a + b + c;
return(d);
/* Complete this function only
DO NOT write main function.
*/
}
int main()
{
int x, y, z, sum;
printf("Enter the three numbers: ");
scanf(" %d %d %d", &x, &y, &z);
sum = addThreeNumbers(x, y, z);
printf("The sum is %d", sum);
return 0;
}
And the error was as follows:
solution.c:30:5: error: redefinition of ‘main’
solution.c:15:9: note: previous definition of ‘main’ was here
You have another main function in the code somewhere. Post the complete code and I will take a closer look. But that is the only way you can receive this error
In modern C, empty argument parentheses mean that the type and number of arguments is unknown.
Although this segment runs fine with most compilers, yours might be picky. Try declaring main with zero arguments explicitly, like this:
int main(void) {
//code
}
Pretty sure this is one of the online coding sites' question. They put in the main function themselves by appending it to the code, you don't have to explicitly write it. Delete the main function you have written and check if that works out.
i have made a program to compute roots of quauation but it does not simplify the roots.can anyone help me to simplify them
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main(void)
{
int a,b,c;
float d,d2;
printf(" Enter a,b and c:");
scanf("%d %d %d",&a,&b,&c);
d=b*b-4*a*c;
if(d<0)
{
printf("(%d+i%d)/%d\n",-b,sqrt(-d),2*a) ;
printf("(%d-i%d)/%d\n",-b,sqrt(-d),2*a);
}
else
{
printf("(%d+%d)/%d\n",-b,sqrt(d),2*a);
printf("(%d-%d)/%d\n",-b,sqrt(d),2*a);
}
getch();
}
You can't compute the square root of a negative number. d is negative and you're trying to find its square root. The whole point of complex solutions and the imaginary unit i is to write -1 as i^2, and then when d < 0 you have:
sqrt(d) = sqrt(i^2 * (-d)) = i*sqrt(-d)
So change to this:
if(d<0)
{
printf("(%d+i%lf)/%d",-b,sqrt(-d),2*a);
printf("(%d-i%lf)/%d",-b,sqrt(-d),2*a);
}
I don't know why you had parantheses around your printf arguments, I removed those.
The second %d should also be changed to %lf since sqrt returns a double.
If you want to compute square roots tof negative numbers, find a C99 compiler (basically, anything besides MSVC will do), include <complex.h> header, use complex data type and csqrt function.
http://en.wikipedia.org/wiki/Complex.h