What will happen when typecasting 8 bit pointer to 16 bit pointer?

[vigneshmck]

Hi all,
I have one unsigned char array[2] of 8 bit.it contains value of same type.i tried to assign to unsigned short(16 bits) pointer by type casting.now how will be the value stored in a pointer.please refer this example code.

unsigned char t[2];
unsigned short *p;
t[0]=1;
t[1]=2;
*p=(unsigned char *) *t;
 

how will be the value stored in the memory while reading if starting address of p is 0x 04000000

[weaknessforcats]

Well, t is a char* so *t is a char. The typecast tells the compiler to consider it to be a pointer to an unsigned char. The compiler knows this is phony but does what you say because you must know something the compiler doesn't.

If the value in the *t is 25 then that's the value of your unsigned char*.

After a cast you, and not the compiler , is responsible for correct values.

[kudos]

Pointers, it seems to be surrounded with a certain mystique.

First, a pointer is something that points to the memory in your computer. The memory is an array.

for instance;

int a[] = {1,2,3,4,5};

An array right, this array is placed somewhere in memory.

void main(){
int a[] = {1,2,3,4,5};
printf("%d\n",a);
}
 

You probably get some warnings, and the output is a weird number. The value is the location in the memory where the array a is placed.

Now, lets prove that this is a pointer!

void main(){
int a[] = {1,2,3,4,5};
printf("%d\n",*a);
}
 

The value is 1, so this is the same as

void main(){
int a[] = {1,2,3,4,5};
printf("%d\n",a[0]);
}
 

Now, we need to be a bit careful...becau se there is two things here! The address of the element (i.e., the weird value) and the content of that address which is 1. One more example:

void main(){
int a[] = {1,2,3,4,5};
int *b;
b = a;
printf("%d\n",b);
printf("%d\n",*b);
}
 

Here we see the difference between the address of the pointer and the content of that pointer. Now....here is where you want to be careful. Typecasting the value of the pointer, for instance 1 fits inside an 8 bit value.

void main(){
int a[] = {1,2,3,4,5};
printf("%d\n",(char*)*a);
}
 

Here we tell, lets turn the value 1 into an 8 bit pointer. But, we try ask, what address does this pointer point to...well, it points to the memory address 1. We do not try to actually read what is placed at memory location 1. Ofcourse, we could try...

void main(){
int a[] = {1,2,3,4,5};
char *b = (char*)*a;
printf("%d\n",(char*)*a);
printf("%d\n",*b);
}
 

Did it segfault? I think this part of the memory would be used by the bootloader.

[vigneshmck]

thanks now i got something understood