Lab46 Wiki

information

lab46:~$ screen 
lab46:~$ irssi
[(status)] /server irc
[(status)]/join #data
lab46:~/src/data$ svn add example.c
A         example.c
lab46:~/src/data$ svn commit -m "Submitted example.c"
Adding         data/example.c
Transmitting file data .....
Committed revision 40.
lab46:~/src/data$
  1 #include<stdio.h>
  2
  3 int main()
  4 {
  5    int number = 10;//int variable number
  6    int *pointer = &number;//int pointer pointing to the address of number
  7    printf("number is: %d\n",number);
  8    printf("number is stored at 0x%x\n", &number);
  9    printf("pointer points to 0x%x\n", pointer);
 10    printf("The value of pointer is: %d\n", *pointer);//dereferencing a pointer
 11
 12    return 0;
 13 }
lab46:~/src/data$ ./pointer
number is: 10
number is stored at 0x594780d4
pointer points to 0x594780d4
The value of pointer is: 10
lab46:~/src/data$
#include<stdio.h>
#include<stdlib.h>
int main()
{
   int array1[10];//static allocation of an array of 10 elements
   int *array2;//declaration of an integer pointer
   array2 = (int*)  malloc(sizeof(int)*10);//dynamically allocating memory for 10 integers
   int i,j = 0;

   for( i = 0; i < 10; i++)//This loop displays the address of the first element of the array
   {                       //and then increments i by one to move to the next location
      printf("Element %d is stored at 0x%x\n", i , &array1[i]);
   }
   printf("\n");
   for( i = 0; i < 10; i++)//This next loop uses pointer arithmetic to display the memory addresses
   {
      printf("Element %d is stored at 0x%x\n", i , array2);
      array2++;
   }

   return 0;
}
lab46:~/src/data$ ./example
Element 0 is stored at 0x2b5e9a0
Element 1 is stored at 0x2b5e9a4
Element 2 is stored at 0x2b5e9a8
Element 3 is stored at 0x2b5e9ac
Element 4 is stored at 0x2b5e9b0
Element 5 is stored at 0x2b5e9b4
Element 6 is stored at 0x2b5e9b8
Element 7 is stored at 0x2b5e9bc
Element 8 is stored at 0x2b5e9c0
Element 9 is stored at 0x2b5e9c4

Element 0 is stored at 0xfa4010
Element 1 is stored at 0xfa4014
Element 2 is stored at 0xfa4018
Element 3 is stored at 0xfa401c
Element 4 is stored at 0xfa4020
Element 5 is stored at 0xfa4024
Element 6 is stored at 0xfa4028
Element 7 is stored at 0xfa402c
Element 8 is stored at 0xfa4030
Element 9 is stored at 0xfa4034
lab46:~/src/data$
#include<iostream>
using namespace std;
int main()
{
   int **num1;//declaration of a double pointer
   int *num2;//declaration of a single pointer
   int num3 = 0;//setting value of num3 to 0
   num2 = &num3;//assigning num3's address to num2
   num1 = &num2;//assigning the address of the single pointer to the double pointer

   //now **num1, *num2 and num3 all = 0
   cout << "**num1 = " << **num1 << endl;
   cout << "*num2 = " << *num2 << endl;
   cout << "num3 = " << num3 << endl;
   return 0;
}
int x = 4;
void* q = &x;
int* p = q;  /* void* implicitly converted to int*: valid C, but not C++ */
int i = *p;
int j = *(int*)q; /* when dereferencing inline, there is no implicit conversion */
int exampleFunc(int x, int y)
int (*exampleFunc)(int x, int y)
int number;//allocates 4 bytes for the int variable number
char ch;//allocates two bytes for char variable ch
double db;//allocates 8 bytes for double variable db
int *number = NULL;
char *ch = NULL;
double *db =NULL;

number = malloc(sizeof(int));
ch = malloc(sizeof(char));
db = malloc(sizeof(double));
int *number;
char *ch;
double *db;

number = new int;
ch = new char;
db = new double;
free(number);
free(ch);
free(db);
delete number;
delete ch;
delete db;
struct studentType
{
    string studentFName;
    string studentLName;
    int testScore;
    char grade;
};
studentType student;//struct variable declaration
//StructVariableName.membername
//in our example. to print the students last name you write this line
cout << "The students last name is " << student.studentLName << endl;
#include "linkedlist.h"
#include <stdlib.h>

struct node {
   char value;
   struct node *next;
   struct node *prev;
};
typedef struct node Node;

void append(Node** end, char input)
{
      Node* tmp;
      tmp = *end;

      tmp = (Node *) malloc (sizeof(Node));
      (*end) -> next = tmp; // tack new node onto end of list
      tmp -> prev =(*end); // new node points to current end
      (*end) = (*end) -> next; // advance end to new end
      (*end) -> value = input; // put input in node
      (*end) -> next = NULL; // nothing beyond end
}
struct node1 {
   char value;//1 byte

};
struct node2 {
   char value;//1 byte
   int number;//4 bytes
};
typedef struct node1 Node1;
typedef struct node2 Node2;
#include<stdio.h>
int main()
{

   char v1 = 0;
   int v2 = 0;

   printf("char is %d bytes\n", sizeof(v1));
   printf("int is %d bytes\n", sizeof(v2));
   printf("node1 is %d bytes\n", sizeof(Node1));
   printf("node2 is %d bytes\n", sizeof(Node2));

   return 0;                                                                                                                                                  
}
lab46:~/src/data$ ./sizelab46:~/src/data$ ./size
char is 1 bytes
int is 4 bytes
node1 is 1 bytes
node2 is 8 bytes
lab46:~/src/data$
struct node1 {
   char value;
   struct node1 *tmp1;
};
struct node2 {
   char value;
   int number;
   struct node2 *tmp1;
   struct node2 *tmp2;
   struct node2 *tmp3;
};
typedef struct node1 Node1;
typedef struct node2 Node2;
lab46:~/src/data$ ./size
char is 1 bytes
int is 4 bytes
node1 is 16 bytes
node2 is 32 bytes
lab46:~/src/data$
#include<stdio.h>
struct node1 {
   int value;
   struct node1 *tmp1;
};
typedef struct node1 Node1;
int main()
{
   int array[10000];
   int i;
   for (i = 0; i < 10000; i++)
   {
      array[i] = i;
   }

   printf("node1 is %d bytes\n", (sizeof(Node1)*10000));
   printf("array is %d bytes\n", (sizeof(array[0])*10000));

   return 0;
}
lab46:~/src/data$ ./size
node1 is 160000 bytes
array is 40000 bytes
lab46:~/src/data$
#include<stdio.h>
#include<stdlib.h>

int main()
{
   int x = 0;
   int a,b,c,d,e,f,g,h,i,j;
   int iterations = 0;
   while (x == 0)
   {

      a = (int)malloc(sizeof(int));
      b = (int)malloc(sizeof(int));
      c = (int)malloc(sizeof(int));
      d = (int)malloc(sizeof(int));
      e = (int)malloc(sizeof(int));
      f = (int)malloc(sizeof(int));
      g = (int)malloc(sizeof(int));
      h = (int)malloc(sizeof(int));
      i = (int)malloc(sizeof(int));
      j = (int)malloc(sizeof(int));
      iterations = iterations + 10;
      printf("iterations: %d\n",iterations);
   }
   return 0;
}
lab46:~$./malloc
iterations: 1
.
.
.
.
.
.
iterations: 38220610
iterations: 38220620
iterations: 38220630
iterations: 38220640
iterations: 38220650
iterations: 38220660
iterations: 38220670
Killed
lab46:~/src/data$
lab46:~$ ps
USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND
dschoeff  1152  0.0  0.0  13660    12 pts/7    SNs  Sep16   0:00 /bin/bash
dschoeff  1155  0.0  0.1  42544  1428 pts/7    SN+  Sep16   1:59 irssi
dschoeff  3547  0.0  0.0  13632   864 pts/47   SNs  21:42   0:00 -bash
dschoeff 19715  0.0  0.1  13624  1488 pts/45   SNs  23:43   0:00 -bash
dschoeff 20561  9.0 80.0 1054040 836352 pts/47 SN+  23:49   0:32 ./malloc
dschoeff 21328  0.0  0.0   8584  1028 pts/45   RN+  23:55   0:00 ps u
lab46:~$ 

Table of Contents

Part 1

Entries

September 5th, 2011

September 18th, 2011

September 27th, 2011

September 30th, 2011

DATA Topics

Version Control

Pointers

arrays, pointer arithmetic

pointers to pointers

Null pointer

Void pointers

Function pointers

Static allocation vs. Dynamic allocation

Memory allocation (malloc(), new)

Memory De-allocation (free(), delete)

structures(structs/records)

Structure pointers

DATA Objectives

Write a program that uses linked lists

Method

Measurement

Analysis

Experiments

Experiment 1

Question

Resources

Hypothesis

Experiment

Data

Analysis

Conclusions

Experiment 2

Question

Resources

Hypothesis

Experiment

Data

Analysis

Conclusions

Experiment 3

Question

Resources

Hypothesis

Experiment

Data

Analysis

Conclusions