Download CSC107 F06 Midterm #1 – Exam Questions - Prof. Matthew A. Hertz and more Exams Computer Science in PDF only on Docsity!
Problems:
- [10 * 1 points] State whether each of the following names is a:
- GOOD VARIABLE NAME -or-
- GOOD SYMBOLIC CONSTANT NAME -or-
- STYLISTICALLY POOR VARIABLE NAME -or-
- RESERVED WORD -or-
- OTHER ILLEGAL VARIABLE NAME
(a) [1 points] counter good variable name
(b) [1 points] 3141519 stylistically poor variable name
(c) [1 points] sigma2nd good variable name
(d) [1 points] voidableCost good variable name
(e) [1 points] 2ndInput other illegal variable name
(f) [1 points] GRAV CONSTANT good symbolic constant name
(g) [1 points] powerOf2 good variable name
(h) [1 points] UNION good symbolic constant name
(i) [1 points] floor good variable name
(j) [1 points] costIn$ other illegal variable name
- [5 * 4 points] Compute the value and datatype of each of the following expressions. If an expression is “boolean”, write down the result as “zero” or “non-zero”. You must show your work to earn partial credit.
(a) [4 points] toupper(‘b’ + 3) ⇒ toupper(‘e’) ⇒ Value is ‘E’ which is a char
(b) [4 points] 3 / ceil(3.0 + 0.1) ⇒ 3 / ceil(3.1) ⇒ 3 / 4 ⇒ Value is 0 which is an int
(c) [4 points] 7 + (^6) * sqrt(9) ⇒ 7 + 6 * 3.0 ⇒ 7 + 18.0 ⇒ Result is 25.0, a double
(d) [4 points] fabs(-3) / 2 ⇒ 3.0/2 ⇒ Value is 1.5 which is a double
(e) [4 points] !isupper(‘a’) || isalnum(toupper(‘.’))⇒ !”false” || isalnum(‘.’) ⇒ “true” || “false” ⇒ Result is “true” (non-zero), a “boolean”
- [25 total points] For this problem, write the C statements (not full programs) that perform the actions desired. You should assume that all variables have been declared and are appropriately defined.
(a) [15 points] Uses a loop to compute and print out the first 15 powers-of-two (e.g., 2^0 , 2^1 , 2^2 , · · ·, 2^14 )
for (power = 0; power < 15; power++) { printf(‘‘%.0lf\n’’, pow(2, power)); }
(b) [10 points] If the variable counter is greater than 5 and less than 10, assign the variable inRange the value 1. Otherwise, inRange should be set to 0.
if ((counter > 5) && (counter < 10)) { inRange = 1; } else { inRange = 0; }
- [2 * 10 points] Generate a trace and show the output for the following snippets of C code. If a loop will not terminate, show the first 3 iterations of the loop and state that the loop is infinite.
(a) [10 points] for (i = 0, j = 1; i < j; i++) { switch(i) { case 2: printf(‘‘Two’’); case 1: printf(‘‘Two’’); case 0: printf(‘‘One’’); case 3: printf(‘‘Three\n’’); break; case 4: printf(‘‘Four’’); break; default: printf(‘‘Something else\n’’); break; } } line# i j Output 0 0 1 1 4 One 5 (on same line as above) Three 6 12 1 0
(b) [10 points] int i = -1, k = 20; do { printf(‘‘%d\n’’, k); i += 1; } while(i = 0); line# i k Output 0 -1 20 1 2 20 3 0 4 0
