Convert integers into binary

Question

I've made a very simple program that takes integer number arguments either with plus/minus sign or without them, and print the number in binary format.

binary.c

#include <stdio.h>

void printBinary(int);
int toInt(int*, char[]);
int handleProgram(char[]);

int main(int argc, char *argv[])
{   
    return handleProgram(argv[1]);
}

int handleProgram(char param[])
{
    int number, state;

    if( toInt( &number, param ) != -1 )
    {
        printBinary(number);

        state = 0;
    }
    else
    {
        printf("\tmain()::Exiting program...\n");

        state = -1;
    }

    return state;
}


void printBinary(int number)
{
    for (int i = 31; i >= 0; i--)
    {
        if (number >> i & 1 == 1)
            putchar('1');
        else
            putchar('0');

        if (i != 0)
        {
            if (i % 4 == 0)
                putchar(' ');

            if (i % 16 == 0)
                printf(": ");
        }
    }

    printf("\n");
}

int toInt(int *num, char param[])
{
    *num = 0;
    int sign = 1;
    int i = 0;

    if(param == NULL)
    {
        printf("\ttoInt()::No parameter, please enter a valid number!\n");

        return -1;
    }

    if(param[i] == '-')
    {
        sign = -1;
        i++;
    }
    else if( param[i] == '+' )
    {
        i++;
    }

    // check that a number exist
    if(param[i] == '\0')
    {
        printf("\ttoInt()::No number detected, please enter a valid number\n");

        return -1;
    }

    for(; param[i] != '\0'; i++)
    {
        if(param[i] >= '0' && param[i] <= '9')
            *num = 10* *num + param[i] - '0';
        else
        {
            printf("\ttoInt()::Invalid parameter, please enter a valid number!\n");

            return -1;
        }
    }

    *num *= sign;

    return 0;
}

Sample output: positive and negative cases

D:\C\chapter2>gcc binary.c -o binary.exe

D:\C\chapter2>binary 3
0000 0000 0000 0000 : 0000 0000 0000 0011

D:\C\chapter2>binary +3
0000 0000 0000 0000 : 0000 0000 0000 0011

D:\C\chapter2>binary -3
1111 1111 1111 1111 : 1111 1111 1111 1101

D:\C\chapter2>binary +3s
        toInt()::Invalid parameter, please enter a valid number!
        main()::Exiting program...

D:\C\chapter2>binary -3s
        toInt()::Invalid parameter, please enter a valid number!
        main()::Exiting program...

D:\C\chapter2>binary s
        toInt()::Invalid parameter, please enter a valid number!
        main()::Exiting program...

D:\C\chapter2>binary +
        toInt()::No number detected, please enter a valid number
        main()::Exiting program...

D:\C\chapter2>binary -
        toInt()::No number detected, please enter a valid number
        main()::Exiting program...

D:\C\chapter2>binary
        toInt()::No parameter, please enter a valid number!
        main()::Exiting program...

Any ideas for a better approach (where to fix, what to do about negative cases, any more functions need)?

Answer 1

Functionality

1. Why 31? Why not 15 or 63 or 42? If code is to handle 32-bit integers, then use long as that type is at least 32-bit, rather than int, which can be as small as 16. Alternatives int32_t, int_least32_t. Even better: re-code for the largest integer type intmax_t.

   // Why 31?
   for (int i = 31; i >= 0; i--)
   

2. Right shifting a negative number leads to implementation defined behavior. Consider unsigned instead.

   // ... (int number)
   ... (unsigned number)
   ... 
   if (number >> i & 1 == 1)
   

3. Good that code allows a leading '+'.

4. Pedantic point. int overflow is undefined behavior and that is the result of the following with argv[1] == "-2147483648", a string that should "work".

   *num = 10* *num + param[i] - '0';
   

5. Avoid using printf(some_string). Should some_string contain a %, maybe due to a code update, that will invoke UB as printf() expects the first arg to be a format string.

   // printf(": ");
   fputs(": ", stdout);
   // or (2nd choice)
   printf("%s", ": ");
   

6. Do not use argv[1] unless it is valid

   int main(int argc, char *argv[]) {   
     if (argc > 1) return handleProgram(argv[1]);
     else ...  
   

7. Prefer terminating error messages to go to stderr

   // printf("\ttoInt()::No number detected, ...
   fprintf(stderr, "\ttoInt()::No number detected, ...
   

8. Overflow not detected in toInt().

Style

1. Avoid naked magic numbers. Consider

   // for (int i = 31; i >= 0; i--)
   #define INT_WIDTH  32
   for (int i = INT_WIDTH - 1; i >= 0; i--)
   

2. int toInt() returns only 2 different int. Consider bool.

Minor

1. See little value in excessive in vertical spacing throughout code. Example: The need for a blank line between buys little clarity for the price of a line.

   printBinary(number);
   
   state = 0;
   

2. Consider parens ()

   // if (number >> i & 1 == 1)
   if ((number >> i) & 1 == 1)
   // or simply
   if ((number >> i) & 1)
   

Answer 2

Two things about your code.

First, try putting all of your constants in macros. This will make your code more readable and less change-error prone.

Second, you assume that all of your input integers are signed. But if the user doesn't want a signed int and wants to convert an unsigned int to binary he would get one major difference: The int range is two times smaller.

I suggest you ask for input like s* if * is a signed number or u* if * is unsigned. Also change / add a function to handle this case.