Number systems and review of C++ syntax

1. Number systems and review of C++ syntax In base 10 all digits are between 0 and 9 and can be interpreted as 419 = 9 x 1 + 1 x 10 +4 x 100 = 9 x 100 +1 x 101 + 4 x 102 23= 2 x 1 + 3 x 10 = 3 x 100 +2 x 101 In base 2 all digits are either 0 or 1 and can be interpreted as 10111 = 1 x 1 + 1 x 2 + 1x 4 + 0 x 8 + 1 x 16 =1 x 20 +1 x 21 + 1 x 22 +0 x 23 +1 x 24 = 23( base 10) In base 16 we have the digits 0 though 9 and 10 is A, 11 is B , 12 is C, 13 is D, 14 is E and 15 is F and can be interpreted as 1A3= 3 x 16 0 +10 x 161 + 1 x 162 = 3 + 10 x 16 + 256 = 419(base10) 17 = 7 x 16 0 +1 x 161 = 7 + 16 = 23 (base 10)

2. Base 10 Base 2 Base 16 • 1 1 • 10 2 • 11 3 • 100 4 • 101 5 • 110 6 • 111 7 • 1000 8 • 1001 9 • 1010 A • 1011 B • 1100 C • 1101 D • 1110 E • 1111 F • 1000 10

3. What would be the output from #include <iostream.h> #include <iomanip.h> int main() { cout << 23 << " in base 16 is " << setiosflags(ios::hex) << 23 << endl; return 0; } ______________________________________ 23 in base 16 is 17

4. Can we do our own conversions? 419%10 = 9 419% 16 = 3 419/ 10 = 41 419/16 = 26 41%10 = 1 26%16 = 10 = A(base 16) 41/10 = 4 26/16 = 1 4%10 = 4 1%16 = 1 4/10 = 0 1/16 = 0 23%10 = 3 23%16 = 7 23%2 = 1 23/10 = 2 23/16 = 1 23/2 = 11 2%10 = 2 1%16 = 1 11%2 =1 2/10 = 0 1/16 = 0 11/2 = 5 5%2 = 1 5/2 = 2 2%2 = 0 2/2 = 1 1%2 =1

5. Program fragment to do binary conversions- first cut int dec; dec=23; while (dec >0) { cout << dec%2; dec= dec/2; } What is produced by the code above? What’s wrong? How do we fix it?

6. Binary conversion in correct order int dec, i=0, bin[100]; dec=23; while (dec >0) { bin[i]=dec%2; //store stuff in array bin i++; dec= dec/2; } for ( int j =i-1; j>=0; j--) cout << bin[j]; //print bin in reverse order cout << endl;

7. Hex conversion- first cut int dec, i=0, hex[100]; dec=23; while (dec >0) { hex[i]=dec%16; //store stuff in array hex i++; dec= dec/16; } for ( int j =i-1; j>=0; j--) cout << hex[j]; //print hex in reverse order cout << endl; What is the output? What would the output be if dec=26?

8. Hex conversion- version with array int dec, i=0,hex[100]; char hexchar[16]={'0','1','2','3','4','5','6','7','8','9', 'A','B','C','D','E','F'}; dec=26; while (dec >0) { hex[i]=dec%16; //store stuff in array hex i++; dec= dec/16; } for ( int j =i-1; j>=0; j--) cout << hexchar[hex[j]]; //print hex in reverse order cout << endl;

9. int dec=27, i=0,hex[100]; char let; while (dec >0) { hex[i]=dec%16; //store stuff in array hex i++; dec= dec/16; } for ( int j =i-1; j>=0; j--) if (hex[j]<10) cout <<hex[j]; //these are numbers else { let='A'+hex[j]-10; //use fact that lettershave cout <<let; //consecutive rep. } cout << endl;

10. Hex conversion with switch int dec=26, i=0,hex[100]; while (dec >0) { hex[i]=dec%16; //store stuff in array hex i++; dec= dec/16; } for ( int j =i -1; j>=0; j--) if (hex[j] <10) cout <<hex[j]; else switch(hex[j]) { case 10: cout <<'A'; break; case 11: cout <<'B'; break; case 12: cout <<'C'; break; case 13: cout <<'D'; break; case 14: cout <<'E'; break; case 15: cout <<'F'; break; }

11. Binary conversion in function void dectobin( int dec); int main() { dectobin(23); return 0;} void dectobin( int dec) {int i=0, bin[100]; while (dec >0) { bin[i]=dec%2; //store stuff in array bin i++; dec= dec/2; } for ( int j =i-1; j>=0; j--) cout << bin[j]; //print bin in reverse order cout << endl; }

12. Binary conversion with recursion void dectobin( int dec); int main() { int dec=6; dectobin(dec); return 0; } void dectobin(int dec) { if (dec==0) //base case return; else { dectobin(dec/2); //calling self cout <<dec%2; //putting cout after call gets reverse order } }

13. Homework Fri: Review chapter 2 in Deitel for assessment test to determine who should not be in this class and how much to review. Tues. Read Appendix C and hand in 23, 24, 26, 29 Wednesday: Write a recursive function that receives a decimal input parameter and prints out its hexadecimal equivalent. Try out your function on the decimal number 27.