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I'm having trouble with arrays. I copied this code from a book:
#include <stdio.h>
#include <stdbool.h>
int main (void)
{
int p, i, primes[50], primeIndex = 2;
bool isPrime;
primes[0] = 2;
primes[1] = 3;
for (p = 5; p <= 50; p = p + 2) {
isPrime = true;
for (i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
if (p % primes[i] == 0)
isPrime = false;
if (isPrime == true) {
primes[primeIndex] = p;
++primeIndex;
}
}
for (i = 0; i < primeIndex; ++i)
printf ("%i ", primes[i]);
printf ("\n");
return 0;
}
In particular, I'm having trouble understanding the difference between the primeIndex and the i variables. The primeIndex refers to the array number and i refers to the number placed into the array. Right?
primeIndex is the place where the next found prime is written in the prime array, and also the number of primes known so far. i is the index of the prime used for trial division. For each candidate, i loops from 1 (we don't need to try out primes[0] = 2 because only odd numbers are checked) to the index of the first prime larger than the square root of the candidate.
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Phenomenon: Since Fabi is passed through the pointer parameter, the address of Fabi enters the function normally for the first time. The second time Fabi becomes an inaccessible address 0x384(Watcher), but it has not been changed by itself and passed to the function as const.
The fibonacci_search function uses the global variable Fabi instead of the Fabi parameter, so there is no problem.
It is precisely because the inaccessible address is accessed, a Segmentation fault is reported, and the program terminates.
#include <stdio.h>
#include <stdlib.h>
int fibonacci_search( const int* const F,int* a,int n,int val){
int low = 0, high=n-1,maxn = n-1;
int mid;
int k = 0;
while(maxn>F[k]-1)
k++;
for(int i=maxn;i<F[k]-1;i++)
a[i]=a[maxn];
int i=1;
while(low <= high){
mid = low + F[k-1] - 1; //mid = low + F[k] - 1;
printf("The %dth search value is:%d\n",i++,a[mid]);
if( val < a[mid]){
high = mid-1;
k = k-1;
}else if( val > a[mid]){
low = mid + 1;
k = k-2;
}else{
if(mid<=n)
return mid;
else
return maxn;
}
}
return -1;
}
int main(void){
int biarr[] = {0,1,16,24,35,47,59,62,73,88,99,102,304,758,777,801,900};
int amount = 15;
int* Fabi= (int*) malloc(sizeof(int)*amount);
Fabi[0]=0;Fabi[1]=1;
for(int i=2;i<amount;i++){
Fabi[i]=Fabi[i-1]+Fabi[i-2];
}
for(int i =5;i<17;i++){
int result = fibonacci_search(Fabi,biarr,17,biarr[i]);
if(result == -1)
printf("Can not find the result.\n");
else
printf("Find the Value,the Index is:%d\n",result);
}
return 0;
}
The picture
You have a (possible) overflow here:
for(int i=maxn;i<F[k]-1;i++)
a[i]=a[maxn];
How could you be sure that F[k]-1 is a valid bound for your array a?
OMG.
when change
int biarr[] = {0,1,16,24,35,47,59,62,73,88,99,102,304,758,777,801,900};
to
int biarr[50] = {0,1,16,24,35,47,59,62,73,88,99,102,304,758,777,801,900};
Its right.
Thanks to: When you debug, have you made sure that there's no out-of-bounds writes to any of the arrays? Some programmer dude's remind.
Its really a peculiar phenomenon！！！
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This might be simple but I'm new to recursion in c. I want to find the sum of odd integers based on user's input. For example if user inputs 3, function returns 9 (1 + 3 + 5 = 9)
int recursiveSumNOdd(int n)
{
int start = -2; //later start = start+2, so it starts at 0
int n1 = n*2; //total number of digits with rec
int num = 0;
int sum=0;
int count=0;
if(start>=n1)
{
return 0;
}
else
{
start = start+2;
count++;
sum = sum +recursiveSumNOdd(start);
}
return sum;
}
Explanations in comment:
int recursiveSumNOdd(int n) {
if (n == 1 || n == 0)// first "if" in a recursive is its stop condition
return n;
return 2 * n - 1 + recursiveSumNOdd(n-1); // formula for 2->3, 3->5 etc
}
int main(void) {
printf("%d\n", recursiveSumNOdd(3));
return 0;
}
NB: You may want to handle integer overflow
NB2: You can have a mathematics formula to return instantly the result, it is way better, but I guess it was to understand better recursion?
return n * n; // the sum of odd numbers is the square of user's input
You are over-complicating things.
You cannot have the sum of negative elements.
int sum_n_odd(unsigned n)
{
What is the sum of 0 (zero) elements?
if (n == 0) return 0;
If you knew the sum of n - 1 numbers, what is the sum of n numbers?
return sum_n_odd(n - 1) + something; // something is easy to figure out
}
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I was given code and I don't understand why the function returns -1. I feel that it is a typo. The code is given below:
int equilibrium(int array[], int size)
{
int sum = 0;
int left_sum = 0;
int i;
for (i = 0; i < size; i++) {
sum += array[i];
}
for (i = 0; i < size; i++) {
if (array[i] == sum - 2 * left_sum) {
return i;
}
left_sum += array[i];
}
return -1;
}
It looks like it uses -1 to mean "not found", it looks like a search function trying to find an index i where the condition in the innermost if is true.
equilibrium returns the equilibrium point of an array as an index in the array. When the array doesn't have an equilibrium point, the function returns -1 instead. It is used as followed:
int equilibrium_idx = equilibrium(somearrayvalue, somesize);
if( equilibrium_idx == -1 )
printf("It isn't balanced\n");
else
printf("The equilibrium index is %d\n", equilibrium_idx);
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I try to do a function in order to sort an array and display after that only the prime number. But all the elements of my array are random numbers, and the problem is that the function display only negative prime numbers and not positive like 7 and 3, what can I do in order to solve the problem
int prime_arr(int size, int *arr, int *sort_arr)
{
int i, j, k = 0, flag;
for (i = 0; i < size; i++)
{
flag = 0;
for (j = 2; j < arr[i]/2; j++)
{
if (arr[i] % j == 0){
flag = 1;
break;
}
}
if (flag == 0){
sort_arr[k++] = arr[i];
}
}
return j;
}
I see 3 problems with the code:
1. You should return k, not j. k is the size of sort_arr
2. You should loop until arr[i] / 2, not one less than that (see <= in code below)
3. You do not handle negative numbers. Change your loop to the following:
for (j = 2; j <= abs(arr[i])/2; j++)
Without the code that prints your values, I'm not sure exactly what you're looking for, but hopefully fixing these will fix your problem.
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This is what the code should do:
“Lift off in T minus 5 4 3 2 1 Blast-off!”
When I run it, it just keeps printing ''Sum = 5'' forever.
Code:
int main(void) {
int sum = 5;
int i;
printf("Lift off in T minus\n");
for (i = 0; i < 5; i=i+i) {
sum = sum - i;
printf("sum = %d\n",sum);
}
printf("Blast-off",sum);
return 0;
for (i = 0; i < 5; i=i+i) { // use i = i+1
sum = sum - i; //sum-- or sum = sum -1
printf("sum = %d\n",sum);
}
As initially i=0, so
i=i+i; //will be zero always, no increment.
And
sum = sum -1;
otherwise
i = 0 =>sum = sum - i; // = 5 as i=0
i = 1 =>sum = sum - i; // = 4 as i=1
i = 2 =>sum = sum - i; // = 2 as i=2
i = 3 =>sum = sum - i; // = -1 as i=3
Why not run the loop backwards?
for (i = 5; i > 0; --i) {
printf("i = %d\n",i);
}
This is simpler so the potential for bugs to creep in is reduced. Also, your final printf if malformed: you're missing a format specifier for sum.
Your specific problem: replace i=i+i with i=i+1 or something similar. (I prefer ++i).