(2 7 0) + (2 6 1) + (2 5 1) + (2 4 0) + (2 3 1) + (2 2 1)+(2 1 0)+(2 0 1)

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1 BITS PILANI, DUBAI CAMPUS DUBAI INTERNATIONAL ACADEMIC CITY, DUBAI FIRST SEMESTER COURSE : COMPUTER PROGRAMMING (CS F111) COMPONENT : Tutorial# 1 SOLUTIONS DATE : 27-AUG-2017 Q1. Represent the following in 8 bits binary, signed binary and 1 s & 2 s Compliment form a) -20 Start with +20 in 8 bits take 1 s Complement Take 2 s Complement b) 15 Start with +15 in 8 bits For positive numbers the signed binary only will be the 1 s C and 2 s C c) -38 Start with +38 in 8 bits take 1 s Complement Take 2 s Complement Q2. Find the decimal equivalent for the following 2 s Compliment binary number a) The MSB (Most Significant Bit or left most bit) is 1 therefore the decimal equivalent will be a ve number. b) The MSB (Most Significant Bit or left most bit) is 0 therefore the decimal equivalent will be a +ve number. (2 7 *0) + (2 6 *1) + (2 5 *1) + (2 4 *0) + (2 3 *1) + (2 2 *1)+(2 1 *0)+(2 0 *1) CS F111 COMPUTER PROGRAMMING Page 1 of 11

2 (-2 7 *1) + (2 6 *1) + (2 5 *1) + (2 4 *0) + (2 3 *1) + (2 2 *1)+(2 1 *0)+(2 0 *1) = +109 = -20 c) The MSB (Most Significant Bit or left most bit) is 0 therefore the decimal equivalent will be a +ve number. (2 5 *0) + (2 4 *1) + (2 3 *0) + (2 2 *0)+(2 1 *1)+(2 0 *1) = +19 Q3. Perform the following arithmetic operation and check for the overflow a) The operation will be performed as 23+(-15) The 2 s C binary equivalent of 23 is The 2 s C binary equivalent of -15 is Adding these two will result in (overflow is not there, but carry forward is there) b) The operation will be performed as (-16)+(-37) The 2 s C binary equivalent of -16 is The 2 s C binary equivalent of -37 is Adding these two will result in (overflow is not there, but carry forward is there) CS F111 COMPUTER PROGRAMMING Page 2 of 11

3 Q4. Using 16 bits, what is the range of signed integers that can be represented in each of the following cases a) Sign magnitude -(2 15-1) to (2 15-1) i.e to b) 1's Complement -(2 15-1) to (2 15-1) i.e to c) 2's Complement -(2 15 ) to (2 15-1) i.e to Q5. Perform the following conversions a) ( )2 = ( 0614 )8 = ( 396 )10 b) (1A03)16 = ( )2 = ( )8 Q6. Represent the following into IEEE floating point representation a) This is a ve number so the sign bit will be 1 The binary equivalent to 19 is The binary equivalent to.75 is 11 So the representation will look like After normalization it will become * 2 4 After adding the biased exponent * The binary equivalent of exponent 131 is So the IEEE representation will be b) This is a +ve number so the sign bit will be 0 The binary equivalent to 397 is The binary equivalent to.5 is 1 So the representation will look like After normalization it will become * 2 8 CS F111 COMPUTER PROGRAMMING Page 3 of 11

4 After adding the biased exponent * The binary equivalent of exponent 135 is So the IEEE representation will be Q7. Draw the flow chart for the following a) Arithmetic operations (+, -, *, /) CS F111 COMPUTER PROGRAMMING Page 4 of 11

5 b) Reverse a given number Answer 8 CS F111 COMPUTER PROGRAMMING Page 5 of 11

6 int i=8, j=5; float x=0.5, y = -1.5; float c=2.0; int d=6; void reinit() i=8, j=5; x=0.5, y = -1.5; c=2.0; d=6; d = 2*i/5+4*j-3%i+j-2; = (2*8)/5+(4*5)-(3%8)+5-2; = 16/ ; *, % have higher precedence than +, - = ; left right associativity = 23 i=8, j=5, x= , y= , c= , d=23 c = (i-3*x)/(y+2*j)/x y; = (8-3*0.5)/(-1.5+2*5)/0.5 - (-1.5) = (8-1.5)/( )/ = -6.5/-8.5/ = (-6.5/-8.5)/ = /0.5 = = (assuming default precision of %f) i=8, j=5, x= , y= , c= , d=6 d = ++i + j-- * ++c / y--; = 9 + 5* 3 / -1.5 (i, c are incremented and used, whereas j,y are used used and incremented ), = / -1.5 = 9 + (-10) = -1 i=9, j=4, x= , y= , c= , d=-1 c = j i * (-d/x); = 5 7 * (-6/0.5) (6 is type promoted to float) = 5-7 * ( ) = = i=7, j=4, x= , y= , c= , d=6 Answer 9 (Valid means no Compile Time Errors) int i=8, j=5; float x=0.5, y = -1.5; float c=2.0; int d=6; void reinit() i=8, j=5; x=0.5, y = -1.5; c=2.0; d=6; CS F111 COMPUTER PROGRAMMING Page 6 of 11

7 d = x*(i%j)/y-d; Valid ( both operands to % are int) d = x*(i+j)%y-d; Invalid ( 2 nd Operand to % is float) c = x*i%j/y-d; Valid ( both operands to % are int) c = x*i+j%y-d; Invalid ( 2 nd Operand to % is float ) c = x++ * d++ / --y; Valid ( Itsjust right ) x= (x+i)++; Invalid ( lvalue required for increment operand) x = x+i++; Valid ( treated as x + i++ as ++ has higher precedence to + ) c = x+++++i; Invalid (treated as ((x++)++) + i, So lvalue required for increment operand) c = x+++i++; Valid ( treated as x++ + i++) Answer 10 int a = 5, b = 0, c = -4, d = 3, e = 8, f =-10; a + f > 0; rel op have lesser precedence to +, so this is treated as (a + f) > 0 ) = -5-5 > 0 is FALSE c + d == a + b % 3; c + d = = 7 a + b % 3 = % 3 = = 5 7 == 5 is FALSE b; b is FALSE. ( zero is treated as FALSE ) b (a + c) && (e - f); This expression is treated as b ( (a+c) && (e-f) ) as && has higher precedence to. b CS F111 COMPUTER PROGRAMMING Page 7 of 11

8 b is FALSE. ( zero is treated as FALSE ) ( (a+c) && (e-f) ) is evaluated as i) (a + c) = = -1 (TRUE) ii) (e f) = = 18 (TRUE) so ( (a+c) && (e-f) ) is TRUE finally, FALSE TRUE is TRUE d > (a + e++) (d == c + f) &&!(a < c); This expression is treated as (d > (a + e++) ) ( (d== (c+f)) &&!(a<c) ) i) (d > (a + e++) ) is 3 > (5 + 8) is FALSE (value of e used is 8 and then incremented to 9) ii) ( (d== (c+f)) &&!(a<c) ) is evaluated as i) (d== (c+f)) is 3 == is FALSE ii)!(a<c) is not evaluated (because for && if one operand is FALSE entire expression is FALSE) so overall FALSE FALSE is FALSE b > 1; 0 > 1 is FALSE b++ > 1; 0 > 1 is FALSE (value of b used is 0 and then incremented to 1) ++b > 1; 1 > 1 is FALSE (value of b used is incremented to 1 and then used ) Answer 11 int a = 5, b = 0, c = -4, d = 3, e = 8, f =-10; void reinit() a = 5, b = 0, c = -4, d = 3, e = 8, f =-10; a *= b + (c *= 4) e++; is evaluated as a = a * (b + (c *= 4) e++); Further *= is right to left associative,.. c *= 4 is evaluated first as c = c * 4 is c = -4 * 4 is 16 a = a * ( b ); ( value of e used is 8 the then incremented to 9) = 5 * ( ) = 5 * -24 = -120 (a=-120, b=0, c=-16, d=3, e=9, f=-10) c = (a /= b+d); is evaluated as c = a = a / (b+d); = is Right to Left associative. a = 5 / (0+3) = 5/3 = 1 CS F111 COMPUTER PROGRAMMING Page 8 of 11

9 c = a So c = 1 (a=-120, b=0, c=-16, d=3, e=9, f=-10) d = (a > b)? a : b; = (5 > 0) is TRUE so d = a i.e. d = 5; (a=5, b=0, c=-4, d=5, e=8, f=-10) c = (a > b)? ((b > c)? a : b) : c; = (5 > 0) is TRUE so ((b > c)? a : b) is evaluated and assigned to c ((b > c)? a : b) = (0 > -4) is TRUE so 5 (value of a) is the value of the expression. So c = 5 (a=5, b=0, c=5, d=3, e=8, f=-10) Answer 12 int a = 10, b = 3, e = 0, f=10; int aa, bb, cc; char c = 'A', d = 'B'; float x = 10, y = 2.5, z = 0.5; float xx, yy, zz = 3; void reinit() a = 10, b = 3, e = 0, f=10; c = 'A', d = 'B'; x = 10, y = 2.5, z = 0.5, zz = 3; e = c++ + a%3; ( value of c used is 'A' (ascii 65) and then incremented to 66 ) = %3 = = 66 printf ( "c=%d, d=%d, e=%d\n", c,d,e); c=66, d=66, e=66 d = c++ + a%3; printf ( "c=%d, d=%d\n", c,d); c=66, d=66, e=0 c = (c >= 'A' && c <= 'Z')? (c-'a'+'a'): c; printf ( "c=%c\n", c); c='a' (converting to lower case) xx = a/b+y+z/f; printf ( "xx=%f\n", xx); xx= aa = a/b+y+z/f; CS F111 COMPUTER PROGRAMMING Page 9 of 11

10 printf ( "aa=%d\n", aa); aa=5 xx = (float)a/b+y+z/f; printf ( "xx=%f\n", xx); xx= aa = (float)a/b+y+z/f; printf ( "aa=%d\n", aa); aa=5 xx = a/(float)b+y+z/f; printf ( "xx=%f\n", xx); xx= aa = a/(float)b+y+z/f; printf ( "aa=%d\n", aa); aa=5 Answer 13. Examine whether the following statements are true or false 1) True 2) False (sizeof() is an operator) 3) True Answer To calculate SI and CI #include <math.h> /* To calculate */ float principle, intrate, nyears; /* Inputs */ float simpleinterst, compoundinterest; /* Read Inputs */ simpleinterest = principle * intrate * nyears; compoundinterst = /* I forgot the formula */ /* extract money */ 2. /* To reverse a number */ int num, revnum; printf ("Pl enter no. to be reversed : " ); scanf("%d", &num); /* for negative nos?? */ revnum = 0; for(; num; revnum = (revnum *= 10) + num % 10, num /= 10 ) ; printf ("Reversed num = %d\n", revnum ); CS F111 COMPUTER PROGRAMMING Page 10 of 11

11 /* Better clarity revnum = 0; while ( num > 0 ) int lastdigit = num % 10; revnum = revnum * 10 + lastdigit; num /= 10; */ 3. To print the word corresponding the digit /* To reverse a number */ int digit; printf ("Pl enter the digit : " ); scanf("%d", &digit); switch (digit) case 0 : printf ("Zero\n"); break; case 1 : printf ("One\n"); break; case 2 : printf ("Two\n"); break; case 3 : printf ("Three\n"); break; case 4 : printf ("Four\n"); break; case 5 : printf ("Five\n"); break; case 6 : printf ("Six\n"); break; case 7 : printf ("Seven\n"); break; case 8 : printf ("Eight\n"); break; case 0 : printf ("Nine\n"); break; default : printf ("Illegal digit\n"); break; /* or digittoword[] = "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" ; if ( digit >= 0 && digit <= 9 ) printf ("%s\n", digittoword[digit] ); else printf ("Illegal digit\n ); */ ***END*** CS F111 COMPUTER PROGRAMMING Page 11 of 11

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