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Unformatted text preview: LAB 3
C Programming Techniques
……Pointers, Bitwise Operations and More Rajesh Rajamani
ME 4231
Department of Mechanical Engineering
University Of Minnesota POINTERS p points at k the address of k is stored in p k is called the referent The expression *p stands for the referent
If k is the referent of the pointer p, then m=*p;
means the same thing as m = k;
Similarly *p = n; means the same thing as k =n; 1 POINTERS Example
int
int
int
int *p;
k =17;
m;
*pt = NULL; // An int pointer variable, initialized to NULL p = &k;
m = *p + 2;
*p = m;
pt = p; POINTERS AND ARRAYS Arrays The name of an array is the same as the address
of its zero’th element array and &array[0] both contain the address of
the zero’th element
An array can be passed as an argument to a
function simply by passing the name of the array
Passing an array is like passing a pointer Changes in the array made in the function are
reflected back in the main program 2 TASKS IN LAB Task 1
Modify the sorting program from Lab 1 to make it
modular Write a function called sort that accepts an integer array,
sorts it, and returns the sorted array to the main function.
Call this function from the main program. In addition, write
a function called print_array that accepts an integer array
and prints it out in this format:
The array is [0 1 2 3 4 5 6 7 8 9]
Use this function to print out the array before and after
sorting. ORDER ARRAY ELEMENTS Determine i th element, i =0,1,2, ….,n i =0 i =1 Find smallest number among elements i , i +1, i +2,….,n
Place the smallest number in the i th element This can be done using 2 “for loops” i =0,1,2,3,…..,n
j =i+1,i+2,i+3,…..,n i j =i+1,i+2,….,n Compare i th
element with j
th element and
swap if j th
element is
smaller 3 BIT OPERATIONS
Bitwise operation Symbol AND
OR
EXCLUSIVEOR
NOT
Shift right
Shift left &

^
~
>>
<< BIT OPERATIONS “&” AND Operator A = 0xa2; B = 0x34; C = A&B;
A=
1010 0010
= 0xa2
B=
0011 0100
= 0x34
A&B =
0010 0000
= 0x20 “” OR operator A=
B=
AB = 1010 0010
0011 0100
1011 0110 = 0xa2
= 0x34
= 0xb6 “^” XOR operator A=
B=
A^B = 1010 0010
0011 0100
1001 0110 = 0xa2
= 0x34
= 0x96 4 BIT OPERATIONS “<<” Left shift operator
Assume A is a short int, i.e. a 16bit integer.
A = 0xa234 = 1010 0010 0011 0100 A<<4 = 0x2340 = 0010 0011 0100 0000 A<<2 = 0x88d0 = 1000 1000 1101 0000 Vacancies in the right most bits are filled with zeroes. BIT OPERATIONS “>>” Right shift operator:
A = 0xa234 =
A>>4 = 0x0a23 = 1010 0010 0011 0100
0000 1010 0010 0011 Vacancies in the left most bits are filled with zeroes. 5 TASKS IN LAB Task 2
Write a function called bit_merge that accepts 2 four
digit hexadecimal numbers (16 bits each) and
combines their bits into a single 32 bit number as
follows: void main()
{
unsigned short hn1 = 0x1234;
unsigned short hn2 = 0xabcd;
int result;
result = bit_merge(hn1,hn2);
printf(“\n Merging 0x%x and 0x%x results in \n”, hn1, hn2, result);
} TASKS IN LAB Task 2 The output should look like:
merging 0x1234 and 0xabcd results in 0xabcd1234
To do this requires a left shift of hn2 by 16 bits and then a
bitwise OR of hn1 with the leftshifted hn2.
Remember that hn2 needs to be left shifted into a 32 bit
variable, or all the data will be lost when shifted. Be careful
of data types (int versus unsigned short), and think about
what is going on at a bit level. 6 TASKS IN LAB Task 3
Write a function that accepts a decimal number
between 0 and 255 and prints it out on the screen
using 8 binary digits for representation.
void binary_print(short a)
{
// Print an error message if a is not between 0 and 255
// Find LSB and print on screen
// Find Next MSB and print on screen
// ….
// Find MSB and print on screen
} TASKS IN LAB Task 4
Parity method for error detection Assume that your data sequence has 7 bits and an
eighth bit is added to achieve even parity.
Write a program that accepts an 8bit word (or
equivalently, a 2 digit hex number) and does an even
parity check on the input data.
If the input data violates even parity, print out an
error message 7 ...
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