Thread: Bit confused with this logic

I have seen several places like this

Code:

int main(int argc, char *argv)
{
  char value=12;
  int *ptr = (int *)(&value); 
}

Complicated statements like structure in place of int means what? Why do you typecast?

please help me to understand the use of the above statement. Will * and & not cancel out? Another statement like

Code:

int *ptr = (int *)(&value) and int *ptr = (int *) &value; both are same.

Please help.

(int *) is a cast - It's more so for the intentional convention from char* to int*

& is "address of"

I see you're asking this KIND of question in multiple places, so read my answer on your other recent question. What I'll say of the first example you post here is that it is likely going to crash because there's a fundamental error, which I can get to in the other answer (I'll be posting it shortly).

In the statement:

Code:

int *ptr = (int *)&value;

You are getting the "address of" the variable "value" and initializing the pointer ptr with it. The (int *) is a casting (a conversion of types), so the compiler will not complain...

But, beware: You are assigning the adress of an 8 bit type to a pointer which will access 32 bits (pointer to int). If you do:

Code:

*ptr = 0;

In this case, the * is the OPERATOR of indirection (will take the address stored in ptr and access the data stored there), but will deal with 4 bytes, instead of a single one. It can cause segmentation fault or corrupt the contents of another variable or the stack.