SE250:lab-2:shua066
Lab2
Q 1
The question is use the 'sizeof()' operator to determine the size of a point.
int main(){
int *ip;
int *dpt;
printf("%d %d\n",sizeof(ip),sizeof(dpt));
return 0;
}
and i got
4 4
the diff names of the pointer , the same size.
I try to use the other types I got the answers all the same.
4
so the sizeof a pointer is "4".
and I got that I can declare the variavles first as well
double x=10; double *dpt=&x;
Q2
to find the address of two pointer and find out the diff between these address.
int x;
int y;
printf("&x=%p\n &y=%p\n diff=%ld\n,&x,&y,(long)(&x-&y));
I got the outcome is
&x=0x22ccc4 &y=0x22ccc0 diff=1
I change
printf("&x=%p\n &y=%p\n diff=%ld\n,&x,&y,(long)&x-(long)&y);
result
&x=0x22ccc4 &y=0x22ccc0 diff=4
the same address of x and y ,but not the same as diff,because the first one is the diff of two address ,so it is one byte. and the second one is the diff of two numbers,so it is 4.
Q3
int x; char arr [4]; int y;
check the size of the array ,it is
printf("%d\n", sizeof(arr[4]));
result
1
printf("&arr=%p\n", &arr);
result
&arr = 0x22ccc0
printf("arr+4=%p, &arr+4=%p, &arr[4]=%p\n",arr+4,&arr+4,&arr[4]);
result
arr+4=0x22ccc4, &arr+4=0x22ccd0, &arr[4]=0x22ccc4
the "arr+4" and "&arr[4]" are the same.
Next ,I change the size of the array from 0 to 10.
the address of x add of y diff the value is 0 0x22ccbc 0x22cc9c 8 the value is 1 0x22ccc4 0x22ccbc 2 the value is 8 0x22ccc4 0x22ccb4 4
the value is 1&2,3,4 are the same the value is 0&3,5,6,7,9,10 are the same looks like the number play around of that address, it is the compiler give the arrary the space.
Set the value of x&y =0,and arr[4]=10. result
10 0
Q4
code
int main(){
int g_x;
int g_y;
printf("&g_x=%p,&g_y=%p,diff =%ld\n", &g_x,&g_y,(long)(&g_x-&g_y));
printf("&g_x=%p,&g_y=%p,diff =%ld\n", &g_x,&g_y,(long)&g_x-(long)&g_y);
return 0;
}
result
&g_x=0x22ccc4,&g_y=0x22ccc0,diff =1 &g_x=0x22ccc4,&g_y=0x22ccc0,diff =4
same as Q2
code
int main(){
int g_x;
char arr[4];
int g_y;
printf("&g_x=%p,&g_y=%p,diff =%ld\n", &g_x,&g_y,(long)(&g_x-&g_y));
printf("&g_x=%p,&g_y=%p,diff =%ld\n", &g_x,&g_y,(long)&g_x-(long)&g_y);
printf("size of arr[4] is %d\n",sizeof(arr[4]));
printf("arr+4=%p, &arr+4=%p, &arr[4]=%p\n",arr+4,&arr+4,&arr[4]);
return 0;
}
result
the same as Q2
result
the address of g_x add of g_y diff the value is 0 0x22ccbc 0x22cc9c 8 the value is 1 0x22ccc4 0x22ccbc 2 the value is 8 0x22ccc4 0x22ccb4 4
the result of 0&3,5,6,7,9,10 are the same, the result of 1&2,4 are the same. it is not the same as in Q2
set the value x ,y to 0,and arr[4]to 10
result
10 0
Q5
code
int main(){
int *p1,*p2;
{int q; p1 = &q;}
{int r; p2 = &r;}
printf("%d %d\n",p1,p2);
return 0;
}
result
2280636 2280632
Q6
code
char *local_str(){
char s[8]="0123456";
return s;
}
char *local_str2(){
char s[8]="abcdefg";
return s;
}
char *static_str(){
static char s[8]="tuvwxyz";
return s;
}
char *malloc_str(){
char *s=malloc(8);
strcpy(s, "hijklmn");
return s;
}
int main(){
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();
loca_str2();
printf("sp=%p(%s)\n",sp,sp);
return 0;
}
result
gcc lab.c -o lab&& ./lab.exe lab.c: In function `local_str': lab.c:3: warning: function returns address of local variable lab.c: In function `local_str2': lab.c:7: warning: function returns address of local variable /cygdrive/c/Users/shua066/AppData/Local/Temp/ccabFLmD.o:lab.c:(.text+0x112): undefined reference to `_loca_str2' collect2: ld returned 1 exit status
Q7
code
struct {
char my_char;
short my_short;
int my_int;
long my_long;
float my_float;
double my_double;
}my_struct;
result
gcc task7.c -o task7&& ./task7.exe &my_struct = 0x0 offsets: my_char: 0 my_short: -2 my_int: -4 my_long: -8 my_float: -12 my_double: -16
Q8
change the struct to union
result
gcc task7.c -o task7&& ./task7.exe &my_struct = 0x0 offsets: my_char: 0 my_short: 0 my_int: 0 my_long: 0 my_float: 0 my_double: 0