E2 – Fundamentals, Functions & Arrays
Please refer to announcements for details about this exam. Make sure you fill the information below to avoid not being graded properly;
Last Name |
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First Name |
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Student ID # |
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Here is the grading matrix where the TA will leave feedback. If you scored 100%, you will most likely not see any feedback J
Question |
# points |
Feedback |
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Max |
Scored |
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1 |
Tracing |
3 |
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2 |
Testing |
2 |
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3 |
Refactoring |
2 |
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4 |
Debugging |
3 |
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To help you fill the trace table in question #1, we start with an example of a small recursive program & provide you with the trace table we expect you to draw.
This program implements a power function recursively. We do not have local variables in either function, thus making the information in each activation record a bit shorter.
1 #include <stdio.h>
2 #include <stdlib.h>
3
4 int pwr(int x, int y){
5 if( y == 0 ) return 1;
6 if( y == 1 ) return x;
7 return x * pwr(x, y-1);
8 }
9
10int main(){
11 printf(“%d to power %d is %dn”, 5, 3, pwr(5,3));
12 return EXIT_SUCCESS;
13}
Please note the following about the table below;
– We only write down the contents of the stack when it is changed, i.e. when we enter a function or assign a new value to a variable in the current activation record.
– When we write the contents of the stack, we write the contents of the whole stack, including previous activation records.
– Each activation record is identified by a bold line specifying the name of the function & the parameters passed to it when it was invoked.
– It is followed by a bullet list with one item per parameter or local variable.
– New activation records are added at the end of the contents of the previous stack
Line # |
What happens? |
Stack is |
10 |
Entering main function |
main’s activation record – No local vars / parameters |
11 |
Invoking function pwr as part of executing the printf |
Stack is the same, no need to repeat it |
4 |
Entering pwr function with arguments 5 & 3 |
main’s activation record – No local vars / parameters pwr(5,3) activation record – x is 5 – y is 3 |
5 |
Testing if y is 0 à false |
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6 |
Testing if y is 1 à false |
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7 |
Invoking pwr(5,2) as part of return statement |
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4 |
Entering pwr function with arguments 5 & 2 |
main’s activation record – No local vars / parameters pwr(5,3) activation record – x is 5 – y is 3 pwr(5,2) activation record – x is 5 – y is 2 |
5 |
Testing if y is 0 à false |
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6 |
Testing if y is 1 à false |
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7 |
Invoking pwr(5,1) |
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4 |
Entering pwr function with arguments 5 & 1 |
main’s activation record – No local vars / parameters pwr(5,3) activation record – x is 5 – y is 3 pwr(5,2) activation record – x is 5 – y is 2 pwr(5,1) activation record – x is 5 – y is 1 |
5 |
Testing if y is 0 à false |
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6 |
Testing if y is 1 à true |
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6 |
Return value x which is 5 |
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7 |
Back from invocation of pwr(5,1) with result 5.
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main’s activation record – No local vars / parameters pwr(5,3) activation record – x is 5 – y is 3 pwr(5,2) activation record – x is 5 – y is 2 |
7 |
Return the result * x = 5 * 5 = 25 |
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7 |
Back from invocation of pwr(5,2) with result 25 |
main’s activation record – No local vars / parameters pwr(5,3) activation record – x is 5 – y is 3 |
7 |
Return the result * x = 5 * 25 = 125 |
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11 |
Back from invocation of pwr(5,3) with result 125 |
main’s activation record – No local vars / parameters |
11 |
Printf now displays “5 to the power 3 is 125 |
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12 |
return EXIT_SUCCESS from main function |
Stack is empty |
We are going to trace the following program, i.e. simulate in your head how it would execute on a computer. To help you, a trace table is provided for you to fill. Unlike exam E1, our focus here is not only on keeping track of the values of each variables but also the activation records pushed on the program’s execution stack.
1 #include <stdio.h>
2 #include <stdlib.h>
3
4 int mystery(int v1, int v2){
5 if( (v1 < 1) || (v2 < 1) ) return 0;
6 if( v1 < v2 ) return mystery(v1, v2-1)+1;
7 if( v2 < v1 ) return mystery(v1-1, v2)+1;
8 return mystery(v1-1, v2-1);
9 }
10
11int main(){
12 int tmp = 0;
13 tmp = mystery(3,2);
14 printf(“result is %dn”, tmp);
15 return EXIT_SUCCESS;
16}
Feel free to add / remove rows if necessary
Line # |
What happens? |
Stack is |
11 |
Entering main function |
main’s activation record – |
12 |
Define & initialize tmp |
main’s activation record – tmp is 0 |
… |
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You are going to write tests for the following program.
Its requirements are to
– Take an integer array data of SIZE elements
– Take a positive, non-null, integer value n
– If the value is null or negative, the program should not alter the array
– If it is positive, each element in the array should be shifted right by n positions
– If an element is pushed past the end of the array, we keep pushing it as if the end of the array connected to its start. Our array is a kind of “ring”.
Your objective is to write tests which will guarantee
– The program conforms to the requirements; the program below might or might not, your tests need to be able to determine this
– All possible execution paths have been tested
– Your program does not feature any of the novice errors discussed in the textbook / videos / …
1 // all arrays in this program will have same size
2 #define SIZE 3
3
4 void rotate(int data[], int n){
5 int index = 0;
6 int tmp[SIZE];
7
8 // copying data into tmp array
9 for(index = 0 ; index < SIZE ; index++){
10 tmp[index] = data[index];
11 }
12
13 for(index = 0 ; index < SIZE ; index++){
14 next = (index + n) % SIZE;
15 data[next] = tmp[index];
16 }
17}
Feel free to add / remove rows if necessary
Test # |
Inputs’ Values |
Expected Results |
Explanations; What did you use this test for? Why is it not redundant with others? |
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data |
n |
data |
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0 |
1 |
2 |
0 |
1 |
2 |
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Refactor the following code; i.e. improve its quality without modifying its behavior;
– Use meaningful names for variables, parameters & functions
– Provide proper documentation as required in the PAs
– Provide proper indentation & programming style similar to the examples from the textbook, videos & PAs
– Remove useless code
– Simplify program
– Improve performance
You will provide your version in the “Your Modified Version” subsection which is already pre-filled with the original. Then, for each improvement you applied, use the table in the “Summary” subsection to explain what you did & why you did it.
1 int mystery(int v1, int v2){
2 if( (v1 < 1) || (v2 < 1) ) return 0;
3 if( v1 < v2 ) return mystery(v1, v2-1)+1;
4 if( v2 < v1 ) return mystery(v1-1, v2)+1;
5 return mystery(v1-1, v2-1);
6 }
1 int mystery(int v1, int v2){
2 if( (v1 < 1) || (v2 < 1) ) return 0;
3 if( v1 < v2 ) return mystery(v1, v2-1)+1;
4 if( v2 < v1 ) return mystery(v1-1, v2)+1;
5 return mystery(v1-1, v2-1);
6 }
Feel free to add rows to, or remove rows from, this table as needed;
What did you modify |
How did you modify it? |
Why did you modify it? |
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The following function has all sorts of problems in implementing the requirements. We need you to list the errors you found in the table below along with how you would fix them. You will then provide the fixed version of the function. Here is the documentation for the interleave function. You will use this information as requirements;
– Role
§ Modifies the array result so it contains alternated elements from d1 & d2
§ Example – if d1 = {1,3,5} & d2 = {2,4,6} then result = {1,2,3,4,5,6}
– Parameters
§ d1 & d2 arrays of SIZE integers
§ result array of SIZE*2 integers
§ No need to pass the size of the arrays, we expect SIZE to have been globally #define’d
– Return Values
§ n/a
1 // arrays passed as d1 / d2 have the following size
2 // array passed as result will always be 2*SIZE
3 #define SIZE 3
4
5 void interleave(int d1[], int d2[], int result[]){
6 int n=0;
7 for( ; n <= SIZE ; n++){
8 result[n] = d1[n];
9 result[n+1] = d2[n];
10 }
11}
Identify each bug you find & provide the lines # where it happens. In addition, explain what the problem was, then how you fixed it. If the same bug appears at different line, just enter it one time in the table but specify all the lines where it happens.
Bug # |
Lines # |
Problem you identified |
How you fixed it |
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Apply all the fixes you listed in the table to the code below to make it your fixed version. Please note that if a bug is fixed below but not in the table, or the other way around, you won’t get credit for it. You need to have both the bug and its explanations in the table and have it fixed below.
1 // arrays passed as d1 / d2 have the following size
2 #define SIZE 3
3
4 void interleave(int d1[], int d2[], int result[]){
5 int n=0;
6 for( ; n <= SIZE ; n++){
7 result[n] = d1[n];
8 result[n+1] = d2[n];
9 }
10}
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