CODEWITHSUNDEEP
cformat-specifiers
poorvank
poorvank

Reputation: 7612

%p Format specifier in c

How are the specifiers %p and %Fp working in the following code?

void main() 
{
    int i=85;

    printf("%p %Fp",i,i);

    getch();   
}

I am getting the o/p as 0000000000000055 0000000000000055

Upvotes: 20

Views: 118403

Answers (6)

Imtiaj Sreejon
Imtiaj Sreejon

Reputation: 63

%p is for printing an address but you need to use the ampersand i.e. & operator before i to get the address of i

So to get the address of variable i the correct format will be:

void main() 
{
    int i=85;

    printf("%p %Fp", &i, &i);

    getch();   
}

if you don't use & then u will just get the value contained in i which in this case is 85 or 55 in hex

Upvotes: 1

Saurabh Rana
Saurabh Rana

Reputation: 3548

Addition to what @Myforwik said

%p is for printing a pointer address.

%Fp is probably used to format a FAR pointer which is of the form --> (0x1234:0x5678)

and 85 in decimal is 55 in hexadecimal.

I hope its okay now.

References : http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html http://www.winehq.org/pipermail/wine-devel/2005-March/034390.html

Upvotes: 3

reegan vijay
reegan vijay

Reputation: 161

It's purpose is to print a pointer value in an implementation defined format. The corresponding argument must be a void * value.

And %p is used to printing the address of a pointer the addresses are depending by our system bit.

Upvotes: 3

user657862
user657862

Reputation: 406

Here is the compilation output from my machine:

format.c:7:5: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat]

format.c:7:5: warning: format ‘%F’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat]

so there are warnings but it does compile and the output is: 0x55 0.000000p

I am surprised you aren't getting a p at the end. Are you sure code and output matches? I guess it isn't impossible for the address of i to also be 0x0...055..but something looks wrong here.

btw: the typical usage of %p would be to print an address i.e. &i as opposed an int

Upvotes: 2

David Christo
David Christo

Reputation: 876

If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.

In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.

And of course, 55 in hex is 85 in decimal.

Upvotes: 29

Myforwik
Myforwik

Reputation: 3588

%p is for printing a pointer address.

85 in decimal is 55 in hexadecimal.

On your system pointers are 64bit, so the full hexidecimal representation is: 0000000000000055

Upvotes: 2

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