Thread: Assigning negative number to unsigned variables.

Code:

#include<stdio.h>
main()
{
        unsigned char a = -125;
        unsigned int b = -125;
        unsigned long c = -125;
        unsigned short d =-125;
        printf(" unsigned char a = %d sizeof a = %d \n", a,sizeof(a));                                                      
        printf("Unsinged int b = %d sizeof b = %d \n",b,sizeof(b));                                                         
        printf("Unsinged long c = %d sizeof c = %d \n",c,sizeof(c));                                                        
        printf("Unsigned short d = %d sizeof d = %d \n",d,sizeof(d));                                                       
}


Output:
---------

$ ./a.out                                                                                          
 unsigned char a = 131 sizeof a = 1 
Unsinged int b = -125 sizeof b = 4 
Unsinged long c = -125 sizeof c = 4 
Unsigned short d = 65411 sizeof d = 2 

$

How does the internal way representation differs from char to int,short,long?

Could anybody answer?

-Mahesh.

First post, therefore you get a warning. You aren't using code tags. Now go press edit and fix your post. Read the forum guidelines if you don't understand [code] tags.


You aren't using the correct format specifier for printf when you try to display unsigned numbers. %u is unsigned. It doesn't matter how it's represented if you display it incorrectly.



Quzah.

125 in hex is 7D, in binary 0111 1101. To make it negative, you invert each bit and add 1

The way numbers are stored internally is implementation-defined, but the way you have described is the most common (I think it's called ones complement or something).

Not only do you use %u for unsigned variables, as Quzah said, but there's the possiblity that a long is larger than an int. To print an unsigned long int, use

Code:

%lu

Flipping all the bits and adding one gives the two's complement value.
With one's complement, you don't add the extra one in after flipping all the bits.

The advantage of two's complement over one's complement is that the subtraction A - B can be achieved by simply adding the two's complement of B to A without any "end around carry", and also there is no redundant value for zero (a binary representation of all ones across the whole variable represents -0 in one's complement).

Keep in mind that all chars and shorts will be cast to int when passed to printf. Likewise all floats will be passed as doubles.

Originally Posted by King Mir

Keep in mind that all chars and shorts will be cast to int when passed to printf. Likewise all floats will be passed as doubles.

Where are you finding this information, because I'm not seeing that? They're all passed as pointers, probably as void pointers. So I doubt the conversion to a double / int on everything. Do you have the reference point in the standard that covers this?


Quzah.

Originally Posted by quzah

They're all passed as pointers, probably as void pointers.

Huh? Parameters to a vararg passed as void pointers?

I'm pretty sure the standard says absolutely nothing on the topic, but passing them as pointers would be kind of ... pointless.

Nah, by the typical x86 calling convention, they're indeed passed as simple 32-bit integers.
Not so sure about the float/double thing though.

Originally Posted by CornedBee

Huh? Parameters to a vararg passed as void pointers?

I'm pretty sure the standard says absolutely nothing on the topic, but passing them as pointers would be kind of ... pointless.

My bad. I was thinking scanf. I blame talking on the phone and typing at the same time.


Quzah.

>Do you have the reference point in the standard that covers this?
Section 6.5.2.2, Function calls.