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円')
 {
 printf("\ttoInt()::No number detected, please enter a valid number\n");
 return -1;
 }
 for(; param[i] != '0円'; 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)?

• Ezequiel Tolnay: If your aim is writing elegant code making use of existing standard libraries, then the function strtol (in stdlib) is the way to go (the man page includes a comprehensive example on how to use it), and can reduce your code by half. Even though your code doesn't require IO performance, it's good practice using a buffer for constructing your output and print only once rather than on multiple little chunks.
• Levent Divilioglu: I want to build up everything like reinventing the wheel. That's why I didn't used stdlib functions. I've used to code in C in the field however after 3 years, I want to go deep to be a better programmer thus started from the page no:1 of the Kernighan's book.

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.

• THE AMAZING: I would also specify signed long ints
• MarcusJ: I disagree. macros are hard to debug at best, and it clutters up the code. I personally only use macros for values I want to be editable.