im sorry to have to post for something like this but i am desperate.
in what .h can the round() function be found? again, im sorry to have to make a new thread for something like this :/
Printable View
im sorry to have to post for something like this but i am desperate.
in what .h can the round() function be found? again, im sorry to have to make a new thread for something like this :/
while you're at it ( :) )
how can i, for example, define an array with an user-defined size?
as in, i need a proggy in which the user inputs the number of "digits" in the array, and then defines each
only thing that comes to mind is make something like:
1. tell the user to enter the # of values he wants the array to consist of.
2. tell the user to enter all the values, and each time he enters a value, lets say, i use a counter to count the # of values he used.
only problem is, how then do i arrange those values in the array?
ive tried declaring a variable n so the user lets the proggy know how many values he will use, but then i cant declare an array like w[n] cause it says:
[quote]expected constant expression
cannot allocate an array of constant size 0
'w' : unknown size]
:/ any help, id appreciate it :)
compiler?
but most likely you will need to use new and delete to allocate memory.
for example:
Code:int *array;
int size;
std::cin >> size;
array = new int[size];
using MS Visual c++
this is what i got so far:
Code:#include <iostream.h>
#include <math.h>
int main() {
int n;
cout << "Enter the number of values to process:\n";
cin >> n;
while(n<0){
cout << "Enter the number of values to process:\n";
cin >> n;
}
if(n==0)
cout << "No values to process.\n";
if(n>0){
cout << "First Value:\n";
cin >> w[0];
}
for(int i=1;i<n;i++){
cout << "Next Value:\n";
cin >> w[i];
}
return 0;
}
you haven't declared your array.
whoa
you helped :D
yeah i took it out to show the code: i didnt want peeps to mock me for my n00b skillz ;/
it worked :D :DCode:#include <iostream.h>
#include <math.h>
int main() {
int n, *w;
cout << "Entrar el numero de datos a procesar:\n";
cin >> n;
w = new int[n];
while(n<0){
cout << "Entrar el numero de datos a procesar:\n";
cin >> n;
}
if(n==0)
cout << "No hay datos para procesar.\n";
if(n>0){
cout << "Primer Numero:\n";
cin >> w[0];
}
for(int i=1;i<n;i++){
cout << "Proximo Numero:\n";
cin >> w[i];
}
return 0;
}
we're not here to mock you. we understand that you and others may be newbies. I'm still sortof a newbie. we're here to help.Quote:
Originally posted by vege^
yeah i took it out to show the code: i didnt want peeps to mock me for my n00b skillz ;/
I'm glad I could help :D
oh i almost forgot: where's the round() function :(
i mean, what #include do i have to do in order to use it :/
on your above code, add this line before return 0;
as for the round() fxn. i don't know. have you tried msdnCode:delete []w;
actually, there are some round fxns in math.h .
edit: i hope its what you're looking for. so #include <math.h>
to round a float/double to nearest int try.....
int x = floor( your_float_or_double + 0.5 );
That will round to nearest int. You can wrap that in its own function no problem.
the round function im looking for rounds a number to 2 decimal places
the professor gave us this example:
but when i try to make a function like thisCode:round(x*100)/100
it says something about double/int conversions, but even if i change it to double, when i summon the function in the program it still gives me the double/int error :/Code:int round(int x){
int r;
r=(x*100)/100;
return r;
}
any thoughts?
You can still use what i gave you with some creative multiplication and division no problem. Its easy to move a decimal point isnt it.
i am teh dumb :/
if i have:
how then would i round d to 2 decimal places using what you gave me? sorry im bothering so much :/Code:a=minValue(w,n)-.01;
b=maxValue(w,n)+.01;
d=(b-a)/5;