recursive function C program

Question

x and y are integers, the function f(x, y) = x^y needs to be calculated. Calculate the function f(x, y) recursively.

#include <stdio.h>
#include <stdlib.h>

int f(int x, int y) {
    if (x == 0 && y != 0) {
        printf("answer: 0\n");
        return 0;
    } else if (x != 0 && y == 0) {
        printf("result: 1\n");
        return 1;
    } else if (x > 0 && y == 1) {
        f(x, 1) == x;
        return x;
    } else if (x > 0 && y > 0) {
        printf("result: %d\n", x * f(x, y - 1));
        return x * f(x, y - 1);
    } else {
        y = -y;
        printf("result: %d\n", 1 / f(x, y));
        return  1 / f(x, y);
    }
}
int main() {
    int k, l;
    float result;

    printf("*****************ust alma*********************\n\n");
    printf("enter two number: ");
    scanf("%d\n%d", &k, &l);
    result = f(k, l);
    printf("girilen result: %d", result);

    return 0;
}

I am waiting for your help I can not do this lesson. Really hard for me.

I need calculate the function f (x, y) recursively., the codes are just examples but not working — has
– has, Commented Apr 26, 2020 at 20:09
f(x,1)==x; just discards the result of the function call and comparison - what were you intending to do here? — UnholySheep
– UnholySheep, Commented Apr 26, 2020 at 20:14
I have tried to reindent your code, return is missing for some else, some tests are always false. %d and result are not consistent. What is the definition on f ? — Ôrel
– Ôrel, Commented Apr 26, 2020 at 20:15

Ôrel · Accepted Answer · 2020-04-26 20:55:36Z

2

You have some issue on your type, with negative y your need to use double or float. You will get 0 with int I have simplified your function, you were closed

Not the . for 1 to use a double and not an int

#include <stdio.h>
#include <stdlib.h>


double f(int x, int y){
    if(y == 0) {
        return 1;
    } else if(y > 0) {
        return x * f(x, y - 1);
    } else {
        y = -y;
        return  1. / f(x, y);
    }
}
int main() {
    int k,l;
    float result;

    printf("*****************ust alma*********************\n\n");
    printf("enter two number: ");
    scanf("%d\n%d", &k, &l);
    result = f(k,l);
    printf("girilen result: %f\n", result);

    return 0;
}

edited Apr 26, 2020 at 20:55

answered Apr 26, 2020 at 20:35

Ôrel

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3 Comments

has Over a year ago

Perfect! Thank u very much!

nikoss Over a year ago

0,0 should return 1 actually

chqrlie Over a year ago

@nikoss: Thank you for this gem. Learn something new everyday :)

nikoss · Accepted Answer · 2020-04-26 20:42:29Z

0

#include <stdio.h>
#include<stdint.h>

double my_pow(ssize_t x,ssize_t y)
{
    if (y==0)return 1.0;
    else if (y==1)return x;
    else if (y<0) return 1.0/my_pow(x,-y);
    else return x*my_pow(x,--y);
}

int main(void) {
    printf("%.2lf\n",my_pow(2,-1));
}

answered Apr 26, 2020 at 20:42

nikoss

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11 Comments

chqrlie Over a year ago

The l in %.2lf is redundant and ignored. ssize_t is a non standard type.

nikoss Over a year ago

isnt ssize_t posix standard ?

chqrlie Over a year ago

ssize_t is defined in POSIX, but not in the C Standard. Why use ssize_t instead of int or long?

nikoss Over a year ago

because ssize_t is exactly a word size of the target machine but int or long might have alignment issues. I am writing rust normally and in rust we throw isize and usize everywhere possible so this is somewhat c equivalent of it @chqrlieforyellowblockquotes

chqrlie Over a year ago

Because, unlike scanf's, the printf conversion specifier for a double argument is %f, not %lf. float arguments are automatically converted to double when passed as extra arguments to vararg functions, so there is no need to distinguish float and double arguments.

|

chqrlie · Accepted Answer · 2020-04-26 20:58:55Z

0

Here is a compact solution:

#include <stdio.h>

double f(int x, int y) {
    return y ? y < 0 ? f(x, y + 1) / x : f(x, y - 1) * x : 1;
}

int main() {
    int x, y;
    printf("*****************ust alma*********************\n\n"
           "enter two numbers: ");
    if (scanf("%d%d", &x, &y) == 2)
        printf("girilen result: %f\n", f(x, y));
    return 0;
}

answered Apr 26, 2020 at 20:58

chqrlie

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Comments

abelenky · Accepted Answer · 2020-04-27 02:40:17Z

0

#include <stdio.h>

double f(int x, int y)
{
    return y<0 ? 1 / f(x, -y)  :
           y   ? x * f(x, y-1) : 1;
}

int main(void)
{
    printf("%.3f\n", f(4,0));  // To the 0-power: 1
    printf("%.3f\n", f(4,1));  // To the 1-power: X
    printf("%.3f\n", f(4,-2)); // To the negative power: 1/(X^Y)
    return 0;
}

Output

Success #stdin #stdout 0s 4472KB
1.0000
4.0000
0.0625

edited Apr 27, 2020 at 2:40

answered Apr 26, 2020 at 21:16

abelenky

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