I would like to find all the primes within 100. Here is my codes.
// Find all the prime number within 100.
#include<stdio.h>
#include<math.h>
#include<stdbool.h>
int main() {
int i,n;
int j = 0;
for ( n = 2; n <= 100; ++n) {
bool isPrime = true;
for (i = 2; i <= sqrt(float(n)); ++i) {
if(n % i == 0) {
isPrime = false;
break;
}
}
if(isPrime) {
++j;
printf("%d is a prime number\n",n);
}
}
printf("The total number of prime number within 100 is %d\n",j);
return 0;
}
When compile it, there is one error.
prime.c:14:8: error: expected expression before ‘float’
m = float(n);
^
Could anyone help solve this problem？ Thanks.
You're using the wrong syntax when casting (you're using one of C++'s many styles of casting, but for C there is only one way). Change:
sqrt(float(n))
to
sqrt((float)n)
Note however that sqrt takes a double, so strictly speaking this should be:
sqrt((double)n)
Note also that the cast is not necessary, and you can just write:
sqrt(n)
Change this
sqrt(float(n))
to this
sqrt((float)n)
You want to cast n to float.
You should use this function:
float sqrtf (float x);
which in C99 receives a float as an argument. Otherwise, it would be better to cast into double (if you use sqrt()).
sqrt-ref
What you have written:
float(n)
is like saying that float is a name of a function and you pass to it the parameter n.
Notice, that in your case, you don't need casting, since it's going to be performed automatically (to float if you use sqrtf() or to double if you use sqrt()).
Other notes, irrelevant with your syntax error.
Why not start the loop from 3 and increase the counter by two? If you think about it, this will faster and will produce the same results. If you want to test yourself, check my example here.
Also, what I had found pretty exciting when I was searching for primes, is the sieve of Eratosthene's (Κόσκινο του Ερατοσθένη) . Here is an example of it.
If you want to cast n to a float, use (float)n.
Just do:
sqrt(n);
You'll be having the exam same result as the casting for your case.
Related
This is my homework:
I haven't tried to write the part of Natural Logarithm because I can't solve the part of Exponential.
This is the the approximations of Exponential in C using Taylor Series expansion I wrote.
However, it returns inf. What did I do wrong?
#include <stdio.h>
// Returns approximate value of e^x
// using sum of first n terms of Taylor Series
float exponential(int n, float x)
{
float sum = 1.0f; // initialize sum of series
for (int a = n; a >= 0; ++a ) {
while (x * sum / a < 0.00001) {
break;
}
sum = 1 + x * sum / a;
return sum;
}
}
int main()
{
int n = 0;
float x = 1.0f;
printf("e^x = %.5f", exponential(n, x));
return 0;
}
With How do I ask and answer homework questions? in mind, I will give you a few things to have a careful look at.
From comment by Spektre:
from a quick look you are dividing by zero in while (x * sum / a < 0.00001) during first iteration of for loop as a=n and you called the function with n=0 ... also your code does not match the expansion for e^x at all
Have a look at the for loop:
for (int a = n; a >= 0; ++a )
What is the first value of a? The second? The third?
Keep in mind that the values are determined by ++a.
When will that loop end? It is determined by a >= 0. When is that false?
What is this loop doing?
while (x * sum / a < 0.00001) {
break;
}
I suspect that you programmed "English to C", as "do the outer loop while ...", which is practically from the assignment.
But the loop does something else. Apart from risking the division by 0 mentioned above, if the condition is true it will stay true and cause an endless loop, which then however is immediatly canceled in the first iteration.
The head of your function float exponential(int n, float x) expects n as a parameter. In main you init it with 0. I suspect you are unclear about where that value n is supposed to come from. In fact it is unknown. It is more a result of the calculation than an input.
You are supposed to add up until something happens.
You do not actually ever need the value of n. This means that your for loop is meaningless. The inner loop (though currently pointless) is much closer to your goal.
I will leave it at this for now. Try to use this input.
Feel free to edit the code in your question with improvements.
(Normally that is not appreciated, but in case of homework dialog questions I am fine with it.)
Your current implementation attempt is quite a bit off. Therefore I will describe how you should approach calculating such a series as given in your quesiton.
Let's look at your first formula:
You need to sum up terms e(n) = x^n / n!
To check with your series: 1 == x^0 / 0! - x == x^1 / 1! - ...
To calculate these terms, you need a simple rule how to get from e(n) to e(n+1). Looking at the formula above we see that you can use this rule:
e(n+1) = e(n) * x / (n+1)
Then you need to create a loop around that and sum up all the bits & pieces.
You are clearly not supposed to calculate x^n/n! from scratch in each iteration.
Your condition to stop the loop is when you reach the limit of 1e-5. The limit is for the new e(n+1), not for the sum.
For the other formulas you can use the same approach to find a rule how to calculate the single terms.
You might need to multiply the value by -1 in each step or do something like *x*n/(n+1) instead of *x/(n+1) etc.
Maybe you need to add some check if the formula is supposed to converge. Then maybe print some error message. This part is not clear in your question.
As this is homework, I only point into the direction and leave the implementation work to you.
If you have problems with implementation, I suggest to create a new question.
#include <stdio.h>
int main() {
float power;
printf("Enter the power of e\n");
scanf("%f", &power);
float ans = 1;
float temp = 1;
int i = 1;
while ((temp * power) / i >= 0.00001) {
temp = (temp * power) / i;
ans = ans + temp;
i++;
}
printf("%.5f", ans);
return 0;
}
I think I solved the problem
But the part about Natural Log is not solved, I will try.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
int i,a,n,r;
n=12345;
r=0;
for(i=4;i>=0;i--)
{
a=n%10;
n=n/10;
r=r+a*pow(10,i);
}
printf("%d",r);
return 0;
}
Current output - 54320
Expected output - 54321
Please advise on what I may change in my code to reflect the correct output.
The pow function returns a value of type double. Because this is a floating point type, the result it returns will not always be exact.
What's happening in this case is that on the last iteration of the loop pow(10, 0) returns a value slightly less than 1. This results in the right hand side of r=r+a*pow(10,i); to similarly be slightly less than 54321. When this value is then assigned to r, which is of type int, it gets truncated.
Rather than using the pow function here, use the following:
r=r*10+a;
This shifts the current digits in r over by 1, then adds the newest digit to the end. Also, rather than using a for loop, use while (n>0) instead. Then it doesn't matter how many digits you have.
while (n>0)
{
a=n%10;
n=n/10;
r=r*10+a;
}
Here is a simplified version of your algorithm:
void reverse_digits(int a) {
int b = 0;
while (a > 0) {
b = b * 10 + a % 10;
a /= 10;
}
printf("%d\n", b);
}
As for converting to character arrays as mentioned in the comments it's worth to notice that the convertion function will do similar arithmetic operations in order to convert the integer to character array, so doing the reversing using integers seems more convenient.
I have some problem with that. I am trying to learn C programming. Please help me
#include<stdio.h>
int main()
{
int a, factorial;
printf("Please enter a value :" );
scanf("%d", &a);
for (int i = 1; i<=a; i++)
{
a = (a - 1)*a;
}
printf("%d", factorial);
return 0;
}
Well in your code line a = (a - 1)*a; you actually changed your input for getting the factorial. It also will blow your loop. See your for loop will continue as long as your i is less than a, lets say you choose a=3 after first iteration the a itself will become 6, so the for loop will continue until it reach the integer limit and you will get overflow error.
What you should do?
First of all you should use a second variable to store the factorial result, you introduced it as factorial, the way that #danielku97 said is a good way to write a factorial since if you present 0 as input it will also give the correct result of 1. so a good code is:
factorial = 1;
for (int i = 1; i<=a; i++)
{
factorial *= i;
}
But lets say you insist of subtraction, the way you just tried to use, then you need to change the code like:
scanf("%d", &a);
if (a==1 || a==0){
printf("1");
return 0;
}
factorial = a;
for (int i = 1; i<a; i++)
{
factorial *= (a - i)*factorial;
}
You can see that the code just got unnecessarily longer. An if included to correct the results for 1 and 0. Also you need to make sure that i never become like i =a since in that case a-i will be equal to zero and will make the factorial result equal to zero.
I hope the explanations can help you on learning C and Algorithm faster.
Your for loop is using your variable 'a' instead of the factorial variable and i, try something like this
factorial = 1;
for (int i = 1; i<=a; i++)
{
factorial *= i;
}
You must initialize your factorial to 1, and then the for loop will keep multiplying it by 'i' until 'i' is greater than 'a'.
You are modifying the input a rather than factorial and also wrong (undefined behaviour) because you are using factorial uninitialized. You simply need to use the factorial variable you declared.
int factorial = 1;
...
for (int i = 1; i<=a; i++) {
factorial = i*factorial;
}
EDIT:
Also, be aware that C's int can only hold limited values. So, beyond a certain number (roughly after 13! if sizeof(int) is 4 bytes), you'll cause integer overflow.
You may want to look at GNU bugnum library for handling large factorial values.
I'm trying to calculate a simple approximation to an integral by dividing it up into a step function, easy stuff. The problem starts when I simply try to do a division. Here is my code:
double integrand(int a, double div, int n) {
int i;
double sum, val;
val = 1.0/div;
for(i = 0; i < div; i++) {
sum = sum + (pow(i*val, n)/(i*val + a)) * val;
}
return sum;
}
Here div is actually an integer, I tried bringing it into the integrand function originally as an integer and typecasting it to a double inside the function, with the same result. When I debug the code, div can be say, 100, yet val will return something ludicrous like -7.2008557565654656e+304. As far as I'm aware the rest of the code is correct, but I just cannot figure this out, what's going on?!
You never initialized sum:
double sum = 0, val;
Right now, you're using it in your calculation with an uninitialized value, and hence getting some garbage results.
First initialize sum and then use it. Otherwise it will invoke undefined behavior.
so I wrote this simple recursion program and am getting an error when I compile it with GCC
error: lvalue required as left
operand of assignment
Hopefully this isnt anything to serious, any insight is appreciated
THanks!
#include <stdio.h>
int factorial (int);
int main (void)
{
int i = 0;
int a = 0;
printf("Please enter an integer: ");
scanf("%d", &i);
a = factorial (i);
printf("\n\n%d factorial equals: %d \n", i, a);
return 0;
}
int factorial ( int n )
{
if ( n <= 0 )
return 0 ;
else
f(n) = f( n-1) + 2;
}
The following statement is not valid C:
f(n) = f( n-1) + 2;
(I assume this is the line on which you got the error; you didn't say.)
You might want to try the following:
return factorial(n-1) + 2;
but then the name factorial is misleading because that is not the correct formula for the factorial function.
Why are you writing this
f(n) = f( n-1) + 2;
I can't see any function named f().
This is not the correct formula for calculation factorial of any number. Look at Greg's provided link.
Change it to
int factorial (int n)
{
if (n==1||n==0)
return 1;
else
return n*factorial(n-1);
}
The error is with f(n) = f(n+1) in your factorial function. Anything with parenthesis is a function in c and a function cannot be assigned a value. You probably want n = factorial(n+1);
The assignment operator = needs a variable on the left hand side, to which the value on the right hand side is assigned to. You can't assign something to a function, which is what f(n) is according to C syntax. This is assigning a value to lines of code, which makes no sense. The only thing that makes sense on the left hand side of a function is something that can store a value.
Functions can go on the right hand side of the assignment though, as long as they return something (they are not type void).
To get the factorial right you need to think through it a little more... first of all remember that you want the last value to be 1, not zero. And all of the numbers in the factorial are multiplied.
replace f(n) = f( n-1) + 2; with
return n*factorial(n-1)
and yes, 0! is one so add
if(n==0) return 1;