C++ Program to Calculate X^Y

[way87]

Question:
Write a Program using Functions that calculates X^Y where both X and Y are doubles and X is positive. (Hints: Call the function pow() in <math.h>)

[sicarie]

So what have you tried? Where are you stuck?

[JosAH]

Question:
Write a Program using Functions that calculates X^Y where both X and Y are doubles and X is positive. (Hints: Call the function pow() in <math.h>)

I don't consider that a hint; I consider that a complete spoiler. If you don't recognize
a hint like that you wouldn't recognize a hint even if it were dancing on a piano
wearing a purple tutu singing "I am a hint; I am a hint".

kind regards,

Jos

[way87]

i jus started to do programming since last month and i still a newbie =(
i cant solve the question ,it's hard for me =(

[sicarie]

The experts on this site are more than happy to help you with your problems but they cannot do your assignment/program for you. Attempt the assignment/program yourself first and post questions regarding any difficulties you have or about a particular function of the code that you don't know how to achieve.

Please read the Posting Guidelines and particularly the Coursework Posting Guidelines.

Then when you are ready post a new question in this thread.

[way87]

is it ok?

#include <stdio.h>
int square(int y);
 
int main (void)
{
 int x;
 for (x = 0;x>=0;x++) {
 printf ("%d",square (x) );
 }
 
 printf("\n");
 
 return 0;
}
 
int square (int y)
{
 return y*y;
}

[kreagan]

Question:
Write a Program using Functions that calculates X^Y where both X and Y are doubles and X is positive. (Hints: Call the function pow() in <math.h>)

(define x x)
(define y y)
(pow x y)

hehehehe.

[sicarie]

tis is wat i hav done ,but it seems not correct =(

#include <stdio.h>
int square(int y);
 
int main (void)
{
 int x;
 for (x = 1;x>=0;x++) {
 printf ("%d",square (x) );
 }
 
 printf("\n");
 
 return 0;
}
 
int square (int y)
{
 return y*y;
}

Why aren't you using the easy method your teacher suggested? Why are you reinventing something that has already been created?

Also, why use a for loop? Was that a requirement as well? FYI - you declare them as ints, not doubles.

[kreagan]

tis is wat i hav done ,but it seems not correct =(

 for (x = 1;x>=0;x++) {
 printf ("%d",square (x) );
 }
}

X is equal than 1, continue while X is greater than 0, after each iteration, increase X by 1.

I hope you see atleast 1 problem with this for loop.

[way87]

is it ok?

#include <stdio.h>
int square(int y);
 
int main (void)
{
 int x;
 for (x = 0;x>=0;x++) {
 printf ("%d",square (x) );
 }
 
 printf("\n");
 
 return 0;
}
 
int square (int y)
{
 return y*y;
}

i still have no idea about doubles =(

[sicarie]

i still have no idea about doubles =(

http://www.cplusplus.com/doc/tutorial/variables.html

If you are having trouble with something that basic, we are going to have a difficult time helping you with your programs.

Ask to sit down and spend an hour with your teacher or TA going over variables, program structure, and control structures (loops), or you are going to have a LOT of trouble very quickly.

[way87]

#include <stdio.h>
#include <math.h>
 
int main ()
{
    double x = 1;
    double y = 1;
    double result;
 
    for ( x = 1; x <= 10; x++ ){
        printf("%f\n",result = pow(x,y) );
        }
 
 
    return 0;
}
 

i did it but the output is seems not correct:
1
2
3
4
5
6
7
8
9
10

[JosAH]

#include <stdio.h>
#include <math.h>

int main ()
{
double x = 1;
double y = 1;
double result;

for ( x = 1; x <= 10; x++ ){
printf("%f\n",result = pow(x,y) );
}


return 0;
}

i did it but the output is seems not correct:
1
2
3
4
5
6
7
8
9
10

That output is perfectly correct because for every x != 0, x^1 == x.

kind regards,

Jos

[sicarie]

#include <stdio.h>
#include <math.h>
 
int main ()
{
    double x = 1;
    double y = 1;
    double result;
 
    for ( x = 1; x <= 10; x++ ){
        printf("%f\n",result = pow(x,y) );
        }
 
 
    return 0;
}
 

i did it but the output is seems not correct:
1
2
3
4
5
6
7
8
9
10

Okay, let's trace through one of these

x=1
y=1

for x =1
while x<10 // while 1< 10
print pow(x,y) // print x^y -> print 1^1(1)
x++ // x=2
while x< 10 // while 2 < 10
print pow (x,y) // print x^y -> print 2^1 (2)
etc...

What you are printing out is correct math. Set y to something else and you will see it.

[way87]

it should not be correct..
becoz the output should be:
1
4
9
16
25
36
49
64
81
100

?

[sicarie]

it should not be correct..
becoz the output should be:
1
4
9
16
25
36
49
64
81
100

?

No, not the way you have it written. You do nothing to 'y' through the whole thing. So it is always 1. You need to increment y.

[way87]

im really stuck,i dunno how to fix it

[way87]

#include <stdio.h>
#include <math.h>

int main (void)
{
double x;
double y;
double result;

printf("Enter an integer for x\n");
scanf("%d",&x);

printf("Enter an integer for y\n");
scanf("%d",&y);

result = pow(x,y);

printf("x raised to power y is %f\n",result);

return 0;

}

it not works,i input 8 as x and 2 as y,the output is 1...

[oler1s]

First, please use CODE tags unless your code is something like one function call or some obviously readable block. Be on the safe side as a beginner and always use them.

Next, why declare x and y as doubles if you treat them as integers in your program? Shouldn't you rather treat them as doubles? For example, your scanf format specifier would be %f, not %d, right? (You should be verifying this with a C reference that you can easily google. Or type in "man scanf" in Google).