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I got an if condition as a multiples of 5, i need to check if condition until a value <= 10000. My if statement looks like this
// in main function
if(value >=0 && value <16){
function(number,value);
}
else if(value >=5 && value <10){
value-=16;
function(number,value);
}
....
// function
int function(int n, int value){
return (n<<value)|(n>>(16-value))
}
Is there a better way to do this if statement. I am new to programming world and a bit curious to know how to do this.
Thanks in advance
You could use function pointers.
typedef void (*func)();
func fpointers[] = {func1, func2, func3}
int check = value / 5;
fpointers [check] ();
Put the amounts you're subtracting from value into an array, indexed by the multiple of 5 for each range:
int subtract[] = [2, 5, ...];
if (value > 0 && value < 5*(sizeof subtract/sizeof(*subtract))) {
value -= subtract[value/5];
functioncall(value);
}
If you need to evaluate the condition for multiples of 5 , I suggest you to use swich case
int check = value/5;
switch(check)
{
case 0: // 0 <= value < 5
// do things
break;
case 1 : // 5 <= value < 10
// do things ...
break;
.............
default:
break;
}
I got an if condition as a multiples of 5
To check if a number is a multiple of 5, use the modulus operator:
if (number % 5 == 0) ...
i need to check if condition until a value <= 10000
This sounds like you need a loop.
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I dont really understand i have written return in the end but still it gives error
CODE :
int factorial(int num)
{
int N;
if (num > 1)
{
N = (num * factorial(num--));
}
else
return N;
}
int main()
{
cout << factorial(5);
return 0;
}
ERROR : warning: control reaches end of non-void function [-Wreturn-type]
16 | }
Your issue is that you don't return anything. If you look at the flow of the program you can see that for num > 1 you do the factorial stuff and for num <= 1 you just return N. For num > 1 the return statement is never reached. This issue can be fixed by removing the else, BUT that leaves an other issue mentioned, namely that for num <= 1 N is never initialised. If you initialise it to 1 that should solve that, but as people pointed out you don't need N, you can do return num * factorial(num - 1); and simply return 1 for num <= 1. The final problem with your code is that you do num * factorial(num--). factorial(num--) will call factorial(num), when you would need factorial(num-1), because num-- is the post-decrement operator.
Other suggestions in the comments are good to heed as well, like implementing guards from integer overflow and the like.
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I'm trying to generate and print out 5 numbers from 1 to 5, but not in sequence. I'm using a self-written function, 'appearBefore' that will check whether the number has appeared before.
The function appearBefore will return '0' if the number has not appeared before, and '1' if the function has appeared before.
At the moment, the do-while loop doesn't get out of the loop even when 0 is returned. The program never ends. Any recommendations on what I can do?
EDITS - The downvotes sure comes fast. I have added the counter++, but it still does not work. Perhaps someone can advice on the inner-loop?
while (count < 5) {
repeat = 1;
do {
randomNumber = rand() % 4 + 1;
cout << randomNumber;
repeat = appearBefore(randomNumber);
cout << " " << repeat << endl;
} while (repeat == 1);
//Add the number into an array of numbers that have appeared before
checker[counterForChecker] = randomNumber;
counterForChecker++;
counter++;
}
This is the function appearBefore (the variables are global variables):
int appearBefore(int number) {
int x = 0;
int match = 0;
while (x < counterForChecker+1) {
if (checker[x] == number) {
match = 1;
break;
}
else {
match = 0;
}
x++;
}
return match;
}
You check for count < 5 while increasing counterForChecker.
Set the while condition to
while (counterForChecker < 5)
or increase the counter
counter++; // counterForChecker++;
(Assuming that counter++; actually says count++;...)
If x and k are positive, x % k is a number between 0 and k - 1.
So you have four possible values to choose from (1,2,3,4), and you're looping until you've found five unique values.
That will never end well.
To generate numbers from 1 to 5, use rand() % 5 + 1;
You'll want to change the variable count. This is now not done, so since the value does not change the loop will not end.
The loop is running while count < 5 and you never increment count.
Did you mean to use:
while (counterForChecker < 5)
If not, add:
++count;
at the end of the while loop
Use a debugger and step through the code, looking at the values of the variables as you do so. You'll learn more about how everything is working.
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I'm having like an assesment exercise.
So given a Number, for example 12345, I must find out the sum sequence of the digits of the given number (1 + 2 +3 + 4 +5) and then add to it the result (15), and repeat this till the sum sequence of the last number is a digit (in this case is 6).
Example : 12345 + 15 + 6 = 12366;
666 + 24 + 6 = 696;
I've been thinkig to store the digits in an array, but then I realized the array's size is static. Now I'm thinking to make a linked list, but I'm not really sure. Does it involve linked lists?
Just guide me to the right path. What should I use?
There's no magic needed here. Just do the obvious computation on integers:
int reduce(int n)
{
int result = 0;
while (n != 0) { result += n % 10; n /= 10; }
return result;
}
int your_problem(int n)
{
int result = n;
while (n >= 10) { n = reduce(n); result += n; }
return result;
}
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#include<iostream>
using namespace std;
int factorial(int x)
{
if(x == 1)
{
return 1;
}
else
{
return x*factorial(x-1);
}
}
int main()
{
cout<<factorial(5)<<endl;
}
I don’t get the part when the value reaches 1. Why doesn't the program print 1 as the output, because when 1 is reached it returns 1. Consider the below steps.
5*factorial(4)=5*4*factorial(3)=5*4*3*factorial(2)=5*4*3*2*factorial(1)
So now, when the x value becomes 1 and goes into if , condition becomes true and 1 is returned. So why doesn't it output 1? Is it that the value of the 5*4*3*2*factorial(1) is stored somewhere and the returned value of just gets multiplied with 5*4*3*2*1 and outputs 120?
Also please explain, what happens when we pass 0 instead of 5,how will it output 1? (0!=1)
it is exactly like you said:
5*factorial(4)=5*4*factorial(3)=5*4*3*factorial(2)=5*4*3*2*factorial(1)
So if it reaches 1 then this will be replaced by
5*factorial(4)=5*4*factorial(3)=5*4*3*factorial(2)=5*4*3*2*1
so the result of the last step goes into the second last step.... where 2*1 will be calculated... After that the 3rd last step gets the value of 2*1 = 2 and multiplies 3 on in and so on.
If you insert smaller or equal 0 into your function, you will get an endless recursion because if(0 == 1). But if you replace this code with
int factorial(int x)
{
if(x <= 1)
{
return 1;
}
else
{
return x*factorial(x-1);
}
}
Then also 0 and -numbers will work
The stack stores in some sense all pending operations. As soon as fact(2) gets 1 from the call, it multiplies it by 2 and returns 2. fact(3) receives that 2, multiplies it by 3 and returns 6. fact(4) receives that 6, multiplies it by 4 and returns 24. And so on...
Passing a 0 was not thought of in the current form of the program and will actually cause the program to crash (most likely). Changing the if(x==1) line to if(x==0) will fix this.
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What is the difference between this:
s = 0;
if (x > 0) s++;
if (y > 0) s++;
and this:
s = 0;
if (x > 0) s++;
else if (y > 0) s++;
Any help would be greatly appreciated.
When you write else if instead of if, program will not check the else if statement if x > 0, but when you write two if statements program will check both conditions, no matter if x > 0 or not.
In the first case the both conditions are checked because there are two different if statements.
In the second case the second condition is checked only if the first condition is evaluated to false.
Say x is 10 and y is 10. At the end of the first set of statements, s will be equal to 2. At the end of the second set of statements, s will be equal to 1.
The second example
s = 0;
if (x > 0) s++;
else if (y > 0) s++;`
will check for the y value only if x > 0 is false. The first example will execute the check regardless of x's value.