Thread: negative factors of an Integer

Hello Guys, this is my first of many many coming posts.

Having this code. i got negative numbers, does anyone know why?.

Code:

#include <stdio.h>

int factors(long int b);

int main () {
  long int num = 12345678987;
  long int ret;
  ret = factors(num);

  return 0;
}

int factors(long int b) {
    long int i;
    
    for( i=1; i <= 12345678987 ; i++ ){
        if( b % i == 0 ){
            printf( "Value is: %d\n", i);
        }
    }
}

Output
Value is: 87558007
Value is: 242072137
Value is: 262674021
Value is: 726 216 411
Value is: -179740967
Value is: -539222901

Are you getting any warnings when you compile?

Code:

main.c||In function 'main':|
main.c|6|warning: integer constant is too large for 'long' type|
main.c|6|warning: overflow in implicit constant conversion|
main.c||In function 'factors':|
main.c|16|warning: integer constant is too large for 'long' type|
main.c|16|warning: comparison is always true due to limited range of data type|
main.c|18|warning: format '%d' expects type 'int', but argument 2 has type 'long int'|

The maximum size of a long int is 32 bits. Therefore the maximum value for a (signed) long int is 2³¹ - 1 = 2147483647.

Try using type "long long" instead.

Originally Posted by Matticus

The maximum size of a long int is 32 bits.

On a 32-bit machine, perhaps.

Originally Posted by algorism

On a 32-bit machine, perhaps.

That limitation is directly from the standard, Section 5.2.4.2.1:

Originally Posted by c11-draft

- maximum value for an object of type long int
LONG_MAX +2147483647 // 2³¹ − 1

I may have mis-interpreted the standard

Their implementation-defined values shall be equal or greater in magnitude (absolute value) to those shown, with the same sign.

[edit]
Yes, you are correct.

I did a little change to the code and now i got a btter result, You got me, i'm not a C programmer, i have been doing this for just 2 or 3 days, but i want to start with simple math programs.
I read my factoring routine is very inefficient, Can you share a better routine?, i'd like to see your point of view of how to attack the problem here.
Thanks in advance.

Code:

#include <stdio.h>

long int factors(long int b);

int main () {
  long int num = 12345678987;
  factors(num);

  return 0;
}



long int factors(long int b) {
     long int i;
    
    for( i=1; i <= b ; i++ ){
        if( b % i == 0 ){
            printf( "Value is: %ld\n", i);
        }
    }
}

BTW, this is the warning i got

Code:

gcc -Wall sample_factor.c
sample_factor.c: In function ‘factors’:
sample_factor.c:22:1: warning: control reaches end of non-void function [-Wreturn-type]
 }

Originally Posted by Matticus

I may have mis-interpreted the standard
[edit]
Yes, you are correct.

Actually, now that I think about it, you're right as far as portability goes. However, gcc -Wall -W -pedantic gives me no warnings when compiling his program on my machine.

The previous program I posted has a bug in that it doesn't generally print the final factor (final value of n). Also, it has an inefficiency in not being bound by the square root of n (which is why it took so long for large primes). Those are fixed below as well as changing it use unsigned long and making it take input from the command line.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>  // sqrtl
#include <errno.h>

typedef unsigned long UL;

void factor(UL n) {
    printf("%lu: ", n);

    for ( ; n % 2 == 0; n /= 2)
        printf("2 ");

    UL sq = (UL)sqrtl((long double)n);
    for (UL i = 3; i <= sq; ) {
        if (n % i == 0) {
            printf("%lu ", i);
            n /= i;
            sq = (UL)sqrtl((long double)n);
        }
        else
            i += 2;
    }

    if (n > 1)
        printf("%lu", n);
    putchar('\n');
}

int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("Usage: myfactor NUMBER...\n");
        exit(EXIT_FAILURE);
    }

    for (int i = 1; i < argc; i++) {
        char *end = NULL;
        errno = 0;
        UL n = strtoul(argv[i], &end, 10);
        if (errno == ERANGE)
            fprintf(stderr, "%s: number out of range of unsigned long\n",
                    argv[i]);
        else if (*end)
            fprintf(stderr, "%s: invalid number\n", argv[i]);
        else if (n != 0)
            factor(n);
    }

    return 0;
}