Mark: 5 revision(s)

2008-11-03T05:19:06Z

5 revision(s)

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== Task 1 ==

This task demonstrated that all pointers are 4 bytes irrespective of what they are pointing to.

'''Code'''
<pre>#include <stdio.h>

int main(void)
{
double *ip;
printf("The size of the pointer is = %d", sizeof(ip));
return 0;
}
</pre>

'''Windows Output:'''

<pre>The size of the pointer is = 4</pre>

'''Linux Output:'''

<pre>The size of the pointer is = 4</pre>

----
== Task 2 ==

When using ''long'' predefined for each variable the answers are as expected. When using a single ''long'' predefinition for the substraction however you get the expected answers divided by 4.

'''Code'''
<pre>#include <stdio.h>

int main(void)
{
int x;
int y;
printf("The adress of x = %p\nThe adress of y = %p\nThe"
" difference of the adresses are = %ld\n", &x, &y, (long)&x- (long)&y);
printf("The adress of x = %p\nThe adress of y = %p\nThe"
" difference of the adresses are = %ld\n", &x, &y, (long)(&x- &y));

return 0;
}
</pre>

'''Windows Output:'''

<pre>The adress of x = 0x22ccc4
The adress of y = 0x22ccbc
The difference of the adresses are = 4
The difference of the adresses are = 1
</pre>

'''Linux Output:'''

<pre>The adress of x = 0xff81072c
The adress of y = 0xff810728
The difference of the adresses are = 4
The difference of the adresses are = 1
</pre>

----
== Task 3 ==
This task demonstrates that if you initialise a value to the '''(n+x)th''' position of an array, where the array is '''n''' long and x is any integer greater than one, the value initialised will be written outside the array and may overwrite other variables in memory.

'''Code'''
<pre>#include <stdio.h>

int main(void)
{

int x;
char arr[4];
int y;

x=0;
y=0;
arr[4] = 10;
printf("The adress of x = %p\nThe adress of y = %p\nThe"
" difference of the adresses are = %ld\n", &x, &y, (long)&x- (long)&y);
printf("adress of arr = %p\n"
"adress of arr[4] = %p\n"
"value of arr+4 = %d\n",
&arr, &arr[4], arr[4]);
printf("x =%d\ny= %d", x, y);
return 0;
}
</pre>

'''Windows Output:'''

<pre>The adress of x = 0x22ccc4
The adress of y = 0x22ccbc
The difference of the adresses are = 8
adress of arr = 0x22ccc0
adress of arr[4] = 0x22ccc4
value of arr+4 = 10
x =10
y= 0
</pre>

'''Linux Output:'''

<pre>The adress of x = 0x10010b64
The adress of y = 0x10010b68
The difference of the adresses are = -4
adress of arr = 0xffc4073c
adress of arr[4] = 0xffc40740
value of arr+4 = 10
x = 0
y = 0</pre>
----

== Task 4 ==
This time the memory adresses are in a different sectors altogether and hence have vastly different values.
Also making x and y global variables enables you to change the character array without it overwriting the x or y variables, this is because these variables are in different memory sectors.
----

== Task 5 ==
p1 is the adress of q
p2 is the adress of r
----

== Task 6 ==
This task demonstrates the different sectors of memory used under different situations.

'''Code'''
<pre>#include <stdio.h>

char *local_str(){
char s[8] = "1234567";
return s;
}

char* local_str2(){
char s[8] = "abcdefg";
return s;
}

char *malloc_str(){
char *s = malloc(8);
strcpy(s, "hijklmn");
return s;
}

char *static_str(){
static char s[8] = "tuvwxyz";
return s;
}

int main (void)
{
char *sp;

sp = local_str();
printf("sp = %p(%s)\n", sp, sp);

sp = local_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

sp = static_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

sp= malloc_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

return 0;

}
</pre>

'''Windows Output:'''

<pre>sp = 0x22cc90(567)
sp = 0x22cc90(efg)
sp = 0x402000(tuvwxyz)
sp = 0x10601a0(hijklmn)
</pre>

'''Linux Output:'''

<pre>sp = 0xffb0f704( þîÐ►)
sp = 0xffb0f704(abcdefg)
sp = 0x10010cf8(tuvwxyz)
sp = 0x10011008(hijklmn)
local_str = 0x100004dc</pre>

----

== Task 7 ==
This task demonstrates the adresses the variables are stored within a structure. It also shows that the structure itself has a NULL adress. This is because its not actually storing anything, its just C's way of representing an structure and its not really a physical memory structure.

'''Code'''
<pre>#include <stdio.h>
struct
{
char my_char;
short my_short;
int my_int;
long my_long;
float my_float;
double my_double;

}my_struct;

int main (void)
{
printf("&my_struct = %p\n", my_struct);
printf("offsets:\n"
"my_char: %ld\n"
"my_short: %ld\n"
"my_int: %ld\n"
"my_long: %ld\n"
"my_float: %ld\n"
"my_double: %ld\n",
(long)&my_struct -(long)&my_struct.my_char,
(long)&my_struct-(long)&my_struct.my_short,
(long)&my_struct-(long)&my_struct.my_int,
(long)&my_struct-(long)&my_struct.my_long,
(long)&my_struct-(long)&my_struct.my_float,
(long)&my_struct-(long)&my_struct.my_double);

return 0;
}
</pre>

'''Windows Output:'''

<pre>&my_struct = 0x0
offsets:
my_char: 0
my_short: -2
my_int: -4
my_long: -8
my_float: -12
my_double: -16
</pre>

'''Linux Output:'''

<pre>&my_struct = 0xff917710
offsets:
my_char: 0
my_short: -2
my_int: -4
my_long: -8
my_float: -12
my_double: -16
</pre>

----
== Task 8 ==
An union stacks the variables on top of each other. All the variables within the structure share the same adress
----

== Task 9 ==
This task demonstrates that if you free a malloced pointer it still points at the adress it was pointing to before but the memory is now unalocated.

'''Code'''
<pre>
#include<stdio.h>

int main (void)
{
char *sp1, *sp2, *sp3;

sp1 = malloc( 10 );

printf("sp1 = %p\n"
"sp2 = %p\n"
"sp3 = %p\n\n",
sp1, sp2, sp3);

sp2 = malloc( 10 );

printf("sp1 = %p\n"
"sp2 = %p\n"
"sp3 = %p\n\n",
sp1, sp2, sp3);

free( sp1 );

printf("sp1 = %p\n"
"sp2 = %p\n"
"sp3 = %p\n\n",
sp1, sp2, sp3);

sp3 = malloc(10);

printf("sp1 = %p\n"
"sp2 = %p\n"
"sp3 = %p\n\n",
sp1, sp2, sp3);

return 0;
}
</pre>

'''Windows Output:'''

<pre>gcc malloc.c -o malloc.exe && ./malloc.exe
sp1 = 0x1060198
sp2 = 0x611021a0
sp3 = 0x0

sp1 = 0x1060198
sp2 = 0x10701b0
sp3 = 0x0

sp1 = 0x1060198
sp2 = 0x10701b0
sp3 = 0x0

sp1 = 0x1060198
sp2 = 0x10701b0
sp3 = 0x1060198
</pre>

'''Linux Output:'''

<pre>malloc.c: In function 'main':
malloc.c:7: warning: incompatible implicit declaration of built-in function 'mal
loc'
sp1 = 0x10011008
sp2 = 0xfffe39c4
sp3 = 0x1

sp1 = 0x10011008
sp2 = 0x10011018
sp3 = 0x1

sp1 = 0x10011008
sp2 = 0x10011018
sp3 = 0x1

sp1 = 0x10011008
sp2 = 0x10011018
sp3 = 0x10011008</pre>

----
== Task 10 ==
The functions are stored near the global variables in the actual code sector.

'''Code'''
<pre>#include <stdio.h>

char *local_str(){
char s[8] = "1234567";
return s;
}

char* local_str2(){
char s[8] = "abcdefg";
return s;
}

char *malloc_str(){
char *s = malloc(8);
strcpy(s, "hijklmn");
return s;
}

char *static_str(){
static char s[8] = "tuvwxyz";
return s;
}

int main (void)
{
char *sp;

sp = local_str();
printf("sp = %p(%s)\n", sp, sp);

sp = local_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

sp = static_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

sp= malloc_str();
local_str2();
printf("sp = %p(%s)\n", sp, sp);

printf("local_str = %p\n\n", local_str);

return 0;
}
</pre>

'''Windows Output:'''

<pre>sp = 0x22cc90(567)
sp = 0x22cc90(efg)
sp = 0x402000(tuvwxyz)
sp = 0x10201a0(hijklmn)
local_str = 0x401050
</pre>

'''Linux Output:'''

<pre>sp = 0xffb0f704( þîÐ►)
sp = 0xffb0f704(abcdefg)
sp = 0x10010cf8(tuvwxyz)
sp = 0x10011008(hijklmn)
local_str = 0x100004dc</pre>

----

SE250:lab-2:sbas046 - Revision history

Mark: 5 revision(s)