Computing the nth number of the fibonacci sequence, where n is entered at command line

I want to write a program to compute the nth number of the fibonnacci sequence, which i had done using printf and scanf. But I was hoping to change my program so that the sequence number is entered at the command li开发者_Go百科ne rather than entered when prompted by the program. This is what i've come up with. It compiles, but then it crashes when i run it... not sure why. Any suggestions would be appreciated.

This is a program to compute the nth number of the fibonnacci code using iteration. I have written it as such: You must enter the number of the sequence you wish to compute at the command line argv[1]. The program then takes this command line argument and uses it in the while loop, and also prints this number.

#include <stdio.h>


int main( int argc, char**argv ) {
int fib[3] = {0,1};
int counter = 0;
  printf("The %dth Fibonacci number is:\n", atoi(argv[1]));
while ( counter < atoi(argv[1]) ) {

    fib[2] = fib[0] + fib[1];
    fib[0] = fib[1];
    fib[1] = fib[2];
    counter++;
}
printf("%d\n", fib[0]);
getchar();
  return 0;
}

Check if the user actually passed an argument:

int main( int argc, char**argv ) {
    if (argc < 2) {
        printf("Usage: %s number\n", argv[0]);
        return 1;
    }
    ...
}

If he didn't, argv[1] is null, and you'll crash

The program should check if a command line argument is present:

if (argc < 2)
{
    printf ("usage:  %s n\n  where n is a positive integer\n", argv[0]);
    return 1;
}

If no argument is supplied, it will probably crash.

= = = = = edit = = = = =

I don't see any cause for a crash once the bug above is fixed. This works okay:

#include <stdio.h>
int main( int argc, char**argv )
{
        int fib[3] = {0,1};
        int counter = 0;
        if (argc < 2)
        {
                printf ("usage: %s number\n", argv[0]);
                return 1;
        }

        printf("The %dth Fibonacci number is:\n", atoi(argv[1]));
        while ( counter < atoi(argv[1]) )
        {

                fib[2] = fib[0] + fib[1];
                fib[0] = fib[1];
                fib[1] = fib[2];
                counter++;
        }
        printf("%d\n", fib[0]);
        return 0;
}


[wally@zenetfedora ~]$ ./a.out 
usage: ./a.out number
[wally@zenetfedora ~]$ ./a.out 3
The 3th Fibonacci number is:
2
[wally@zenetfedora ~]$ ./a.out 4
The 4th Fibonacci number is:
3
[wally@zenetfedora ~]$ ./a.out 5
The 5th Fibonacci number is:
5
[wally@zenetfedora ~]$ ./a.out 6
The 6th Fibonacci number is:
8
[wally@zenetfedora ~]$ ./a.out 7
The 7th Fibonacci number is:
13
[wally@zenetfedora ~]$ ./a.out 8
The 8th Fibonacci number is:
21
[wally@zenetfedora ~]$ ./a.out 9
The 9th Fibonacci number is:
34
[wally@zenetfedora ~]$ ./a.out 10
The 10th Fibonacci number is:
55

What number did you test it with? Because an int won't be able to hold anything higher than the ~50th Fibonacci number. I had to do something like this in my Algorithms class and we had to find the 239th Fibonacci number which was 64202014863723094126901777428873111802307548623680 Which is something an int nor a long can hold correctly.I don't think that would make it crash but just giving a heads up if you need to find a large Fibonacci number.

#include<stdio.h>

#include<stdlib.h>

int main(int c,char *v[])

{

int *numberOfTerms;

int x,y,z,i,count;

if(c==1)

{

printf("provide number of terms as command line argument");

return 0;

}

numberOfTerms=(int *)malloc(sizeof(int));

*numberOfTerms=atoi(v[1]);

x=0;

y=1;

if(*numberOfTerms==0) return 0;

if(*numberOfTerms<=2)

{

for(i=0;i<*numberOfTerms;i++) printf("%d\n",i);

return 0;

}

printf("%d\n",x);

printf("%d\n",y);

count=2;

while(count<*numberOfTerms)

{

z=x+y;

printf("%d\n",z);

x=y;

y=z;

count++;

}

}

If you want to print the series upto a certain number specified as command line argument, then go for this program:

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

int main(int c,char *v[])
{
    int *number;
    int x,y,z;
    if(c==1)
    {
        printf("provide number upto which febonacci is to be printed (greater than 2)");
        return 0;
    }
    number=(int *)malloc(sizeof(int));
    *number=atoi(v[1]);
    x=0;
    y=1;
    printf("%d\n",x);
    printf("%d\n",y);
    z=x+y;
    while(z<=*number)
    {
        printf("%d\n",z);
        x=y;
        y=z;
        z=x+y;
    }
}