Binary conversion code Segmentation fault - c

Question

I am new to programing in C (only 2 weeks into it). I am unable to figure out why my code is throwing a segmentation fault. I am able to get the program to work if I set long int num equal to a static number. However I need the program to be able to accept user input from the command line (not once the program is running)
Example:
./binary 7
should output
The binary number for 7 is: 111
I have tried using strcpy(num, argv[0]) but that also throws errors when compiling.
#include<stdio.h>
#include <string.h>
#include<math.h>
void decToBinary(long int num) // Function Definition
{
long int remainder[50];
int i=0;
int length=0;
printf("The binary number for %d is: ",num);
while(num > 0)
{
remainder[i]=num%2; // does the mod function
num=num/2; // Divides original number by 2
i++; // Increases count for the upcoming for-loop
length++; // Increases length display digits
}
for(i=length-1;i>=0;i--) // Prints out the binary number in order (ignoring the previous 0's)
{
printf("%ld",remainder[i]);
}
printf("\n"); // Adds a new line after the binary number (formatting)
}
//================================================================================================
int main(char argc, char* argv[])
{
long int num; //HOW DO I TAKE ARGV[0] AND MAKE IT A USEABLE VARIABLE???
printf("Enter the decimal number: "); //TEMPORARY until problem above is solved
scanf("%ld",&num); //TEMPORARY until problem above is solved
decToBinary(*num); // Calling decToBinary function
return 0; // Program terminated successfully
}

Answer 1

Hint1: If you want to change num's value inside decToBinary() function, you have to pass its address in it. So you should call it like this:
decToBinary(&num);
Then your function's prototype would look like this:
void decToBinary(long int *num)
So you have to modify the function's body properly too. I guess that's what causes the Segmentation Fault.
Hint2: argc stands for arguments count, so its type isn't char but int.
As #Kerrek SB pointed out on his comments, scanf(3) is a function with a return value of type int. It's considered wise to check the return value and handle any possible error that may occur.
argv[0] is the name of the executable. If you want to use the first argument from your command line, it's stored in argv[1], and it's of type char *. So if you want to use it as a number, you can use one of these functions of stdlib.h.

Answer 2

Something like (does not handle some error situations)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
void decToBinary(long int num) // Function Definition
{
long int remainder[50];
int i=0;
int length=0;
printf("The binary number for %ld is: ", num);
while(num > 0)
{
remainder[i]=num%2; // does the mod function
num=num/2; // Divides original number by 2
i++; // Increases count for the upcoming for-loop
length++; // Increases length display digits
}
for(i=length-1;i>=0;i--) // Prints out the binary number in order (ignoring the previous 0's)
{
printf("%ld",remainder[i]);
}
printf("\n"); // Adds a new line after the binary number (formatting)
}
//================================================================================================
int main(int argc, char* argv[])
{
if (argc > 1)
{
long int mylong = atol(argv[1]);
decToBinary(mylong); // Calling decToBinary function
} else
{
int errno = 0;
long int num;
int res;
printf("Enter the decimal number: "); //TEMPORARY until problem above is solved
res = scanf("%ld",&num); //TEMPORARY until problem above is solved
if (res != EOF)
{
decToBinary(num); // Calling decToBinary function
} else
if (errno != 0)
{
perror("scanf");
} else
{
printf("No number found in input\n");
}
}
return 0; // Program terminated successfully
}
** I'm building this on Ubuntu 14.04 **
Note that decToBinary(*num) is wrong. You would only use the dereference operator on a pointer not an int or any non pointer type.
A pointer holds an address in memory and when you dereference it you are telling your program to read at that address. If you were dereferencing an int value of for example 10, you are telling the program to read a value at address 10. This is a very low address value and will be outside of the memory range that your programme is allowed to access and so it will be a segmentation fault.
Firstly you must pay attention to compiler warnings. If applicable use the -Werror and possibly -Wextra flags when you build your executable.
e.g
gcc -g -Werror -Wextra myfile.c -o myfile
This is a good opportunity to learn to use a debugger to find your segfault, if you are serious about C programming its essential knowledge. Note above the -g flag which builds debug symbols into your executable (without the symbols the debugger wont know the function names). You can then use gdb (or another debugger as applicable) and launch your executable, then run it and get a back trace to tell you when the segfault happens.

Answer 3

A simple method of extracting the argv[1] parameter to its constituent bits and displaying those bits...
// following eliminates any nonnumeric char from input
// by stopping at first nonnumeric char
long int value = atol(argv[1]);
int position = sizeof( long int ) -1;
// following loop could be modified/prefixed with a loop to skip leading 0's
for( ; position >= 0; position-- )
{
printf( "%c, ", ('0' + ( value >> position) & 1) );
}
printf( "\n" );