Problems using pointers, i think

[guisantogui]

guys,

i'm newbie in C, them i'm trying to write a C code that calculates the fatorial, but when i'll compile the source i got this stack >

main.c: In function ‘fatorial’:
main.c:5:13: warning: comparison between pointer and integer
main.c:8:23: error: invalid operands to binary * (have ‘int **’ and ‘int **’)

I'm trying to use two methods, and call fatorial in main, follow the code:

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

void fatorial(int *number){
    if(&number == 1){
        return;
    }
    fatorial((&number-1) * &number);
}

float main(){
    int fat = 0;
    scanf("%d", &fat);
    fatorial(&fat);
    printf("%d", fat);
}

Thanks in advance.

[stahta01]

Just repeating my remarks because I think you missed the last two or more of my edits.

FYI: Most basic recursion functions do NOT require pointers to solve/implement.
FYI: Most basic recursion functions return a value.
Note: Factorial value of zero is the one. I think you code will have issues if passed zero as input.
Note: I think you code is likely to only work if passed the value of one.

Tim S.

[guisantogui]

I've changed my code, and i got this:

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

void fatorial(int *number){
    if((*number) == 1){
        return;
    }
    fatorial(((*number)-1) * (*number));
}

float main(){
    int fat = 0;
    scanf("%d", &fat);
    fatorial(&fat);
    printf("%d", fat);
}

it compiles, it runs, but when i pass the parameters (just one) and type enter it prints "Segmentation fault", some idea, how to fix this one?

[stahta01]

Decided this post help the user too much; so deleted content.

[jimblumberg]

I get several compile errors when I try to compile your code:

main.c||In function ‘fatorial’:|
main.c|8|error: passing argument 1 of ‘fatorial’ makes pointer from integer without a cast|
main.c|4|note: expected ‘int *’ but argument is of type ‘int’|
main.c|11|error: return type of ‘main’ is not ‘int’|
main.c||In function ‘main’:|
main.c|16|warning: control reaches end of non-void function|
||=== Build finished: 3 errors, 2 warnings ===|

Jim

[laserlight]

float main should be int main. I suggest that you implement the factorial function in this incomplete program:

#include <stdio.h>

int factorial(int number)
{
    /* complete this */
}

int main(void)
{
    int result = factorial(4);
    printf("%d\n", result);
    return 0;
}

When you have this down pat, then you can start thinking of what happens if you want to use a pointer parameter instead and change the return type to void.

[camel-man]

fatorial(((*number)-1) * (*number)) That takes an integer not a pointer to an integer.

[guisantogui]

I'm using gcc compiler, and i got this statement when i compiled the code.

I get several compile errors when I try to compile your code:

main.c:8:2: warning: passing argument 1 of ‘fatorial’ makes pointer from integer without a cast
main.c:4:6: note: expected ‘int *’ but argument is of type ‘int’

Jim

[laserlight]

guisantogui: accept my challenge in post #6. Until you prove that you can handle that, you have a snowball's chance in hell of getting this right.

[guisantogui]

Ok, let's do it, now!

guisantogui: accept my challenge in post #6. Until you prove that you can handle that, you have a snowball's chance in hell of getting this right.

[guisantogui]

HAHA, it was easy

#include <stdio.h>

int factorial(int number){
   if(number == 1){
       return number;
   }
   return factorial(number-1) * number;
}

int main(){
    int result = factorial(4);
    printf("%d\n", result);
    return 0;
}

But, i was trying to use pointers, to learn about that, do you have some "challenge" using pointers laserlight?

[Salem]

Very good - now count how many temporary integers are created when you calculate factorial(4)

Now think how many integers exist when you try it with your int* approach.

[guisantogui]

Ok, here we go!

Very good - now count how many temporary integers are created when you calculate factorial(4)

Now think how many integers exist when you try it with your int* approach.

[guisantogui]

It was harder than another challenge, but i got this (i think), follow the code:

#include <stdio.h>

int factorial(int number, int *cont){
    *cont = *cont+1;
    if(number == 1){
        return number;
    }
    //printf("%d",cont);
    return factorial(number-1, cont) * number;
}

int main(){
    int cont = 0;
    int val;
    scanf("%d", &val);
    int result = factorial(val, &cont);
    printf("%d\n%d\n", result, cont);
    return 0;
}

Very good - now count how many temporary integers are created when you calculate factorial(4)

Now think how many integers exist when you try it with your int* approach.

Then, do you have another challenge?

[laserlight]

Then, do you have another challenge?

If you are taking this as a challenge, then you did not actually meet the challenge

So, complete the factorial function of this program without changing the function's signature or return type:

#include <stdio.h>

void factorial(int *number)
{
    /* complete this */
}

int main(void)
{
    int result = 4;
    factorial(&result);
    printf("%d\n", result);
    return 0;
}