not a homework question

[Three Headed Monkey]

write a program in "C" language that computes 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

I tried

#include <stdio.h>

int pow(int n)
{
int i,power;
power=n;
for(i=0;i<n;i=i +1)
power=power*pow er;
return power;
}

void main()
{
int result;
char ignore;
result= pow(9,pow(8,pow (7,pow(6,pow(5, pow(4,pow(3,pow (2,1))))))));
printf("\nresul t is %d", result);
printf("\nPress ENTER");
gets(&ignore);
}
but it does not work.

how to do that in "C" standard language?

I am using lcc-win32 compiler & windows 98.

help!

[WANG Cong]

On Wed, 12 Mar 2008 05:57:12 +0100，Three Headed Monkey wrote：

write a program in "C" language that computes
9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

I tried

#include <stdio.h>

int pow(int n)

Change the name please. C has a std library function with the same
name and it's prototype is:

double pow(double x, double y);

{
int i,power;
power=n;
for(i=0;i<n;i=i +1)
power=power*pow er;
return power;
}

void main()

main() is never void in C.

{
int result;
char ignore;
result=
pow(9,pow(8,pow (7,pow(6,pow(5, pow(4,pow(3,pow (2,1))))))));

Your _own_ pow only takes one parameter, why here it has two??

printf("\nresul t is %d", result);

I suspect it overflows, since 'result' is only an int.

printf("\nPress ENTER");
gets(&ignore);

gets() is considered harmful, NEVER use it.

}
but it does not work.

how to do that in "C" standard language?

I am afraid you can't, the result may be too big. You can choose
a non-standard libary that supports huge number operations,
e.g. gmp.

[Ian Collins]

Three Headed Monkey wrote:

write a program in "C" language that computes 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

I don't think you can in standard C, the result will be huge...

--
Ian Collins.

[Brice Rebsamen]

On Mar 12, 12:57 pm, Three Headed Monkey
<four_headed_mo [e-mail] rote:

write a program in "C" language that computes 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

I tried

#include <stdio.h>

int pow(int n)
{
int i,power;
power=n;
for(i=0;i<n;i=i +1)
power=power*pow er;
return power;

}

void main()
{
int result;
char ignore;
result= pow(9,pow(8,pow (7,pow(6,pow(5, pow(4,pow(3,pow (2,1))))))));
printf("\nresul t is %d", result);
printf("\nPress ENTER");
gets(&ignore);

}

but it does not work.

how to do that in "C" standard language?

I am using lcc-win32 compiler & windows 98.

help!

First, you may want to have the pow function take 2 arguments: int
pow(int a, int n) { ... }
Second, the pow function defined in math.h does the job for you,
except that it deals with doubles: float pow(double a, double n)
returns a to the power of n as a float. So if you want to deal with
integers you have to convert the result.
Finally you may want to use a loop to do this. It'd look like this:

#include <math.h>
#include <stdio.h>
int main(){
double res=1;
int n;
for( n=2; n<=9; n++ )
res = pow((double)n,r es);
printf("res=%f\ n",res);
return 0;
}

Don't forget to link with the math library when compiling (-lm)

However this might overflow, resulting in res reaching inf. You can
try using long double and powl... Or more complicated stuff. Any idea
of what the resulting number might be?

[Peter Nilsson]

Three Headed Monkey <four_headed_mo [e-mail] rote:

Subject: not a homework question

Your subject should reflect the nature of your problem,
not merely that you have one.

It's not homework questions we mind, rather it's homework
questions that are quoted verbatim without so much as an
attempt.

write a program in "C" language that computes
9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

#include <stdio.h>

int main(void)
{
printf("%d\n", 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) ) );
return 0;
}

If you mean exponentiation, then realise it's pretty big!
Just 5^(4^(3^(2^1))) alone yields a 183000+ digit number.
Raise 6 to the power of that and you have... big!

I am using lcc-win32 compiler & windows 98.

Ah well... the non-standard extension qfloat should knock
that over easily. ;)

--
Peter

[user923005]

On Mar 11, 9:57*pm, Three Headed Monkey <four_headed_mo [e-mail]>
wrote:

write a program in "C" language that computes 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

I guess that even LCC's qfloat data type will be too small.
Do you have any idea how many digits there are in that number?

[Richard Heathfield]

Three Headed Monkey said:

>
write a program in "C" language that computes
9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

#include <stdio.h>

int main(void)
{
printf("%d\n",
9^(8^(7^(6^(5^( 4^(3^(2^1)))))) ));
return 0;
}

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[user923005]

On Mar 11, 10:35*pm, Peter Nilsson <[e-mail] .auwrote:

Three Headed Monkey <four_headed_mo [e-mail] rote:

Subject: not a homework question

Your subject should reflect the nature of your problem,
not merely that you have one.

It's not homework questions we mind, rather it's homework
questions that are quoted verbatim without so much as an
attempt.

write a program in "C" language that computes
9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

* #include <stdio.h>

* int main(void)
* {
* * printf("%d\n", 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) ) );
* * return 0;
* }

If you mean exponentiation, then realise it's pretty big!
Just 5^(4^(3^(2^1))) alone yields a 183000+ digit number.
Raise 6 to the power of that and you have... big!

I am using lcc-win32 compiler & windows 98.

Ah well... the non-standard extension qfloat should knock
that over easily. ;)

I doubt it.
Maple with precision set to 10,000,000 digits overflowed:

Digits=10000000 ; y := 9.^(8.^(7.^(6.^ (5.^(4.^(3.^(2. ^1.)))))));

10 = 10000000

Error, (in evalf/power) argument too large

>

[Micah Cowan]

Micah Cowan <mi***@cowan.na mewrites:

Otherwise, you may be best-suited using the standard library's own
pow() function along with a floating point type (double would make
sense, given that's what pow() deals in). OTOH, if you happen to have
an implementation with <tgmath.h(and don't care to be portable to
them wot don't), you might opt for long double.

(Obviously, I wasn't thinking too clearly on the magnitude of this
number...)

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

[Doug Miller]

In article <12************ ******@aioe.org >, Three Headed Monkey <fo************ ****@yahoo.comw rote:

>
write a program in "C" language that computes 9^(8^(7^(6^(5^( 4^(3^(2^1)))))) )

Do you have even the slightest idea how large that number is? Did you perhaps
mean to type * instead of ^ ?

>
I tried

#include <stdio.h>

int pow(int n)
{
int i,power;
power=n;
for(i=0;i<n;i=i +1)
power=power*pow er;
return power;
}

Examine what you've written here a little more carefully, why don't you, and
see exactly what this function calculates. If you want to compute a^b, one
might suppose that you'd probably want a function that accepts both a and b as
input parameters.

>
void main()
{
int result;
char ignore;
result= pow(9,pow(8,pow (7,pow(6,pow(5, pow(4,pow(3,pow (2,1))))))));
printf("\nresul t is %d", result);
printf("\nPress ENTER");
gets(&ignore);
}
but it does not work.

A little more explanation would be helpful. Start with a complete description
of the manner in which it "does not work", including what you expect it to do
that it does not, and what it does that you expect it not to do.