Can't do this question using while or for loop
Write a program that prints the powers of two from 2^0 to 2^10, i.e. the output of the program should look like this:
2^0 = 1
2^1 = 2
2^2 = 4...
2^10 = 1024
Note 1: Although you may want to use the pow function in this program, you are encouraged not to.
Note 2: This program does not require any input from the user.
Please can someone help me?
Thanks
To be clear, you're not allowed to use any loops?Originally Posted by Zuri786
If not, you can just code each equation in sequence.
It sounds like you are encouraged to write your own exponent function, but it's not mandatory (i.e. you are given a pre-existing alternative).
What have you written so far, and how are you stuck?
I haven't got the code on me right now but can you do me a sample of what you mean because i'm new to programming?
Thanks
We are here to help you with problems, not provide hand-outs. If you're new to programming, then you really need to try solving this on your own, and ask for help only when you get stuck. The programming assignments are only going to get more difficult, so now is a great time to start practicing and building up your skills.
Surely you have learned enough by now to declare variables and print their values. So start making an attempt to write this program, and if you get stuck, post your progress (in code tags) and explain exactly what you are stuck on.
outputCode:#include<stdio.h> /* A recursive function to get a^b Works only if a >= 0 and b >= 0 */ int pows(int a, int b) { if(b) return multiply(a, pows(a, b-1)); else return 1; } /* A recursive function to get x*y */ int multiply(int x, int y) { if(y) return (x + multiply(x, y-1)); else return 0; } /* driver program to test above functions */ int main() { printf("\n 2^0 = %d", pows(2,0)); printf("\n 2^1 = %d", pows(2,1)); printf("\n 2^2 = %d", pows(2,2)); printf("\n 2^3 = %d", pows(2,3)); printf("\n 2^4 = %d", pows(2,4)); printf("\n 2^5 = %d\n", pows(2,5)); printf("\n 2^10 = %d\n", pows(2,10)); return 0; }
Write you own Power without using multiplication(*) and division(/) operators - GeeksforGeeksCode:[userx@devuan powers]$ ./a.out 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^10 = 1024 [userx@devuan powers]$
but you'd better understand how it is really working, that is why I provided the link ...
how functions can call themselves, what values are being returned
putting printf in functions and looking at the values,
something like this to try and figure it out
I'm sure that teacher may know the answer is out there too. so you'd should make sure you're up to snuff on that.Code:/* A recursive function to get a^b Works only if a >= 0 and b >= 0 */ int pows(int a, int b) { printf("b = %d\n", b); if(b){ printf("%d\n", multiply(a, pows(a, b-1))); return multiply(a, pows(a, b-1)); } else return 1; }
it's hard being a teacher now days, the answers are on google.
Last edited by userxbw; 10-05-2017 at 08:25 PM.
It seems more likely that there should be a comma in the title.
Can't do this question using while or for loop
Vs
Can't do this question, using while or for loop
A simple loop and the << operator is all you need.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
