在c中发现指针的混乱

Question

#include<stdio.h>

int main(){
    int a = 320;
    char *ptr;
    ptr =( char*)&a;
    printf("%d",*ptr);
    return 0;
}

I just wanted to know what is the use (char*) ? What if I used (int*) instead of that?

Answer 1

Your variable is of type integer ie int a = 320; Your pointer is of type character ie char *ptr;

character pointer can only points to character type variable, but you in your code want to point a integer variable so how would that be possible? By this line of code: ptr =( char*)&a; what it doing is type casting your pointer varble explicitly or forcefully (whatsoever you might want to call), by making it character type.

Answer 2

Let's change the program a bit:

int main() {
    int a = 0x12345678;   // put hexadecimal value 12345678 into a
    char *charptr = (char*)&a;
    int *intptr = &a;                
    printf("%x\n", *charptr);   // %x prints in hexadecimal
    printf("%x\n", *intptr);
    return 0;
}

On a little endian system the output will be :

78
12345678

• &a is the memory address of the a variable.
• charptr is a pointer to char
• intptr is a pointer to int

The second output 12345678 is obvious, but the first output 78 is not quite as obvious.

On a little endian system the int a in memory looks like this:

+----+
| 78 |  <---- both 'charptr' and 'intptr' point here
+----+
| 56 |
+----+
| 34 |
+----+
| 12 |
+----+

The char (or byte) at the address &a is 0x78, therefore if you dereference the charptr pointer you get 0x78 .

On a big endian system we have this:

+----+
| 12 |  <---- both 'charptr' and 'intptr' point here
+----+
| 34 |
+----+
| 56 |
+----+
| 78 |
+----+

and the first output would be 12 instead of 78 .

---

In this line

char *charptr = (char*)&a;

the (char*) is called a cast, and it tells the compiler to consider &a as a pointer to char .

Without the cast:

char *charptr = &a;

you will get a compiler warning such as incompatible types - from 'int *' to 'char *' , because you are trying to assign a pointer to int to a pointer to char .

If you write this:

char *charptr = (int*)&a;

then you will also get a compiler warning incompatible types - from 'int *' to 'char *' . Anyway in that case the (int*) cast is useless because &a is already a pointer to int .

Answer 3

calling a variable ptr doesn't make it a pointer. if you want it to be a pointer it has to be declared with a * ie int *ptr; or char *ptr;

Casting a pointer such as &a to type char is just silly as they're totally different things. Using (int *) wouldn't change the type of &a so you'd not need to put it in, but then your compiler would show you that ptr is totally the wrong type for storing the value of &a in.