COMP 1917 Computing 1
Session 2, 2016

Tutorial Solutions - Week 4

1. Write a program that prints the numbers from 1 to 100. But for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz". Use void fizzbuzz(int i) to handle your printing.

   #include <stdio.h>
   #include <stdlib.h>
   void fizzbuzz(int num);
   
   int main(){
       int i = 1;
       
       while(i < 100){
           fizzbuzz(i);
           i++;
       }
       
       return 0;    
   }
   
   void fizzbuzz(int num){
       if((num % 3 == 0) && (num % 5) == 0){
           printf("FizzBuzz\n");
       }else if(num % 3 == 0){
           printf("Fizz\n");
       }else if(num % 5 == 0){
           printf("Buzz\n");
       }else{
           printf("%d\n",num);
       }
       return;
   }
   

2. Write your own versions of these character functions

   int isdigit( int ch )
   {
      return( ch >= '0' && ch <= '9' );
   }
   
   int islower( int ch )
   {
      return( ch >= 'a' && ch <= 'z' );
   }
   
   int toupper( int ch )
   {
      if( ch >= 'a' && ch <= 'z' ) {
         return( ch + 'A' - 'a' );
      }
      else {
         return( ch );
      }
   }
   

3. Convert from Binary to Decimal:

   10010101

   method 1 - using powers of 2:

   128 + 16 + 4 + 1 = 149
   

   method 2 - repeated multiplication:

     0 x 2 + 1 =   1
     1 x 2 + 0 =   2
     2 x 2 + 0 =   4
     4 x 2 + 1 =   9
     9 x 2 + 0 =  18
    18 x 2 + 1 =  37
    37 x 2 + 0 =  74
    74 x 2 + 1 = 149
   

4. Convert from Decimal to Binary:

   186

   method 1 - using powers of 2:

   186 -128 = 58  -> 1
    58 < 64       -> 0
    58 - 32 = 26  -> 1
    26 - 16 = 10  -> 1
    10 -  8 =  2  -> 1
     2 <  4       -> 0
     2 -  2 =  0  -> 1
     0 <  1       -> 0
   

   so 186₁₀ = 10111010₂

   method 2 - using repeated division

   186 / 2 = 93 r 0
    93 / 2 = 46 r 1
    46 / 2 = 23 r 0
    23 / 2 = 11 r 1
    11 / 2 =  5 r 1
     5 / 2 =  2 r 1
     2 / 2 =  1 r 0
     1 / 2 =  0 r 1
   

   reading upwards, answer = 10111010

5. Convert from Hexadecimal to Binary:

   B7A3₁₆

   B = 1011
   7 = 0111
   A = 1010
   3 = 0011

   so B7A3₁₆ = 1011011110100011₂

6. Convert from Binary to Hexadecimal:

   1100111101010010

   1100 = C
   1111 = F
   0101 = 5
   0010 = 2

   so 1110111101010010₂ = CF52₁₆

7. Use Two's Complement to represent:

   -19

   • Represent 19 in binary: 00010011
   • Negate it: 11101100
   • Add 1: 11101101 is -19.

8. Write a C program which reads a number in binary format, one character at a time, from standard input, and then prints the number in decimal format, to standard output.

   int main( void )
   {
     int num = 0;
     int ch;
   
     ch = getchar();
   
     // scan binary digits from left to right
     while( ch == '0' || ch == '1' ) {
       num = 2 * num + ( ch - '0' );
       ch = getchar();
     }
   
     printf("Value = %d\n", num );
   
     return 0;
   }
   

9. Now modify your program so that it prompts the user to enter a base (between 2 and 10) followed by a number expressed in that base, computes the value of that number and prints the result in decimal format, e.g.

   Enter base: 7
   Enter number: 5342
   Value = 1892
   

   How would you modify your program to handle bases larger than 10?

   int main( void )
   {
     int base;
     int num=0;
     int ch;
   
     printf("Enter base: ");
     scanf("%d",&base);
   
     printf("Enter number: ");
   
     // skip the newline and any additional spaces
     ch = getchar();
     while( ch == '\n' || ch == ' ' ) {
       ch = getchar();
     }
   
     while(( ch >= '0' && ch <= '9' )||( ch >= 'A' && ch <= 'Z' )) {
       if( ch >= '0' && ch <= '9' ) {
         num = base*num + ( ch - '0');
       }
       else {
         num = base*num + ( ch + 10 - 'A');
       }
       ch = getchar();
     }
   
     printf("Value = %d\n", num );
   }
   

10. Any questions about Assignment 1.

11. (if time permits) Write a C program which reads a number in decimal format, from standard input, and then prints the number in binary format, one character at a time, to standard output.

    int main( void )
    {
      int num;
      int power2=1;
      int ch;
    
      scanf("%d", &num );
    
      // find the largest power of 2 that is not larger than num
      while( power2 <= num/2 ) {
        power2 = power2 * 2;
      }
    
      // now compute the binary digits from left to right
      while( power2 > 0 ) {
        if( num >= power2 ) {
          printf("1");
          num = num - power2;
        }
        else {
          printf("0");
        }
        power2 = power2 / 2;
      }
      printf("\n");
    
      return 0;
    }