How to see address of a structure in printf

c++pointersstruct

I have a function which returns address as following

struct node *create_node(int data)
{
        struct node *temp;
        temp = (struct node *)malloc(sizeof(struct node));
        temp->data=data;
        temp->next=NULL;
        printf("create node temp->data=%d\n",temp->data);
        return temp;
}

where struct node is

struct node {
        int data;
        struct node *next;
};

How can I see in printf("") the address stored in temp?

UPDATE
If I check the adressed in gdb the addresses are coming in hex number format i.e.
0x602010 where as same address in printf("%p",temp) is coming in a different number which is different from what I saw in gdb print command.

Best Solution

Use the pointer address format specifier %p:

printf("Address: %p\n", (void *)temp);

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

Difference between a Structure and a Union

With a union, you're only supposed to use one of the elements, because they're all stored at the same spot. This makes it useful when you want to store something that could be one of several types. A struct, on the other hand, has a separate memory location for each of its elements and they all can be used at once.

To give a concrete example of their use, I was working on a Scheme interpreter a little while ago and I was essentially overlaying the Scheme data types onto the C data types. This involved storing in a struct an enum indicating the type of value and a union to store that value.

union foo {
  int a;   // can't use both a and b at once
  char b;
} foo;

struct bar {
  int a;   // can use both a and b simultaneously
  char b;
} bar;

union foo x;
x.a = 3; // OK
x.b = 'c'; // NO! this affects the value of x.a!

struct bar y;
y.a = 3; // OK
y.b = 'c'; // OK

edit: If you're wondering what setting x.b to 'c' changes the value of x.a to, technically speaking it's undefined. On most modern machines a char is 1 byte and an int is 4 bytes, so giving x.b the value 'c' also gives the first byte of x.a that same value:

union foo x;
x.a = 3;
x.b = 'c';
printf("%i, %i\n", x.a, x.b);

prints

99, 99

Why are the two values the same? Because the last 3 bytes of the int 3 are all zero, so it's also read as 99. If we put in a larger number for x.a, you'll see that this is not always the case:

union foo x;
x.a = 387439;
x.b = 'c';
printf("%i, %i\n", x.a, x.b);

prints

387427, 99

To get a closer look at the actual memory values, let's set and print out the values in hex:

union foo x;
x.a = 0xDEADBEEF;
x.b = 0x22;
printf("%x, %x\n", x.a, x.b);

prints

deadbe22, 22

You can clearly see where the 0x22 overwrote the 0xEF.

BUT

In C, the order of bytes in an int are not defined. This program overwrote the 0xEF with 0x22 on my Mac, but there are other platforms where it would overwrite the 0xDE instead because the order of the bytes that make up the int were reversed. Therefore, when writing a program, you should never rely on the behavior of overwriting specific data in a union because it's not portable.

For more reading on the ordering of bytes, check out endianness.

Best Solution

Related Solutions

C++ – How to set, clear, and toggle a single bit

Setting a bit

Clearing a bit

Toggling a bit

Checking a bit

Changing the nth bit to x

Difference between a Structure and a Union

Related Question