Programming Forums > Application Development > C++

tommclaughlan · Feb 1st, 2005, 1:10 PM

#include <cstdlib>
#include <iostream>
#include <stdint.h>
#include <io.h>
#include <math.h>
int main()


    {
    double phi = 5;

    for (long int n = 0; n < 5000; n=n+1)


        {
        phi=(1/phi)+1;
        }
    printf("Estimated PHI value: %f "),phi;
    scanf("press enter: %d");
    return 0;
}

i want this to return a value which is approximately phi (1.618033989...) but it just gives me "Estimated PHI value: 0.00000000"
any help would be appreciated

Ooble · Feb 1st, 2005, 2:13 PM

You have the printf statement messed up - phi is outside the brackets. You should have:

(Toggle Plain Text)

printf("Estimated PHI value: %f ", phi);

And BTW, what the hell is stdint.h?

Hockeyman · Feb 1st, 2005, 2:16 PM

You've got the variable outside the printf statement...
try:

(Toggle Plain Text)

printf("Estimated PHI value:%f ", phi);

Ooble · Feb 1st, 2005, 2:17 PM

Beat ya to it. :p

Hockeyman · Feb 1st, 2005, 2:32 PM

Ooble - by seconds =^)

Tomm - is that code part of a larger program? If not, you don't need most/all of those include statements...and I'm surprised that printf works without stdio.h (although I'm not sure what's in some of those libraries so it might be in there as well)

Ooble · Feb 1st, 2005, 3:12 PM

I believe it's in io.h as well.

tommclaughlan · Feb 2nd, 2005, 4:49 PM

i hope to expand this program to calculate many significant figures of phi, which is the solution to the equation x*x-x-1=0.
i'm not really an experienced programmer so i just learned from online tutorials and such so i don't really know what those include things are, etc.

suriyaa · Feb 3rd, 2005, 7:20 AM

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
void main()
{
double phi =5;
for( long int n=0;n<5000;n++) //n++ is same as n=n+1;
{
phi=(1/phi)+1
}
cout<<"estimated phi values is : "<<phi; \\printf("estimated phi value is :%f",phi);
getch();\\to make wait
}

L F Permenter · Feb 14th, 2005, 11:40 AM

// Phi, the Golden ratio
// We can calculate Phi without recursion or iteration,
// as it is one solution to a quadratic equation.
// phi = 1/2 = 1/2 *(SQRT(5)).

double PHI = 0.5 +pow(5.0,0.5)/2.0;

// print out in the normal manner

//

Quote:

Originally Posted by tommclaughlan

(Toggle Plain Text)

#include <cstdlib>
#include <iostream>
#include <stdint.h>
#include <io.h>
#include <math.h>
int main()


    {
    double phi = 5;

    for (long int n = 0; n < 5000; n=n+1)


        {
        phi=(1/phi)+1;
        }
    printf("Estimated PHI value: %f "),phi;
    scanf("press enter: %d");
    return 0;
}

i want this to return a value which is approximately phi (1.618033989...) but it just gives me "Estimated PHI value: 0.00000000"
any help would be appreciated

Feb 1st, 2005, 1:10 PM	#1
tommclaughlan Newbie Join Date: Jan 2005 Posts: 3 Rep Power: 0	calculating phi. problem (Toggle Plain Text) #include <cstdlib> #include <iostream> #include <stdint.h> #include <io.h> #include <math.h> int main() { double phi = 5; for (long int n = 0; n < 5000; n=n+1) { phi=(1/phi)+1; } printf("Estimated PHI value: %f "),phi; scanf("press enter: %d"); return 0; } #include <cstdlib> #include <iostream> #include <stdint.h> #include <io.h> #include <math.h> int main() { double phi = 5; for (long int n = 0; n < 5000; n=n+1) { phi=(1/phi)+1; } printf("Estimated PHI value: %f "),phi; scanf("press enter: %d"); return 0; } i want this to return a value which is approximately phi (1.618033989...) but it just gives me "Estimated PHI value: 0.00000000" any help would be appreciated

Feb 1st, 2005, 2:13 PM	#2
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	You have the printf statement messed up - phi is outside the brackets. You should have: (Toggle Plain Text) printf("Estimated PHI value: %f ", phi); printf("Estimated PHI value: %f ", phi); And BTW, what the hell is stdint.h? __________________ Me :: You :: Them

Feb 1st, 2005, 2:16 PM	#3
Hockeyman Programmer Join Date: Jan 2005 Location: Vancouver, Canada Posts: 60 Rep Power: 4	You've got the variable outside the printf statement... try: (Toggle Plain Text) printf("Estimated PHI value:%f ", phi); printf("Estimated PHI value:%f ", phi); __________________ ~ You know, Hobbes, some days even my lucky rocketship underpants don't help. ~ read my blog @ My Lucky Rocketship Underpants

Feb 1st, 2005, 2:17 PM	#4
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	Beat ya to it. :p __________________ Me :: You :: Them

Feb 1st, 2005, 2:32 PM	#5
Hockeyman Programmer Join Date: Jan 2005 Location: Vancouver, Canada Posts: 60 Rep Power: 4	Ooble - by seconds =^) Tomm - is that code part of a larger program? If not, you don't need most/all of those include statements...and I'm surprised that printf works without stdio.h (although I'm not sure what's in some of those libraries so it might be in there as well) __________________ ~ You know, Hobbes, some days even my lucky rocketship underpants don't help. ~ read my blog @ My Lucky Rocketship Underpants

Feb 1st, 2005, 3:12 PM	#6
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	I believe it's in io.h as well. __________________ Me :: You :: Them

Feb 2nd, 2005, 4:49 PM	#7
tommclaughlan Newbie Join Date: Jan 2005 Posts: 3 Rep Power: 0	i hope to expand this program to calculate many significant figures of phi, which is the solution to the equation x*x-x-1=0. i'm not really an experienced programmer so i just learned from online tutorials and such so i don't really know what those include things are, etc.

Feb 3rd, 2005, 7:20 AM	#8
suriyaa Newbie Join Date: Feb 2005 Posts: 2 Rep Power: 0	#include<iostream.h> #include<stdio.h> #include<conio.h> void main() { double phi =5; for( long int n=0;n<5000;n++) //n++ is same as n=n+1; { phi=(1/phi)+1 } cout<<"estimated phi values is : "<<phi; \\printf("estimated phi value is :%f",phi); getch();\\to make wait }

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Programming Forums > Application Development > C++

calculating phi. problem