C Programming - Bitwise Operators - Discussion

Answer to Why everyone taking for char 2bytes?

we are not taking char as 2 bytes.
Let's picture the real scenario, the memory stored in the register of RAM.
for example, suppose 'i' is stored at 0x00 as shown in the below:

Address values
0x06 0000 0000
0x05 0000 0000
0x04 0000 0000
0x03 0000 0000
0x02 0000 0000
0x01 0000 0000
0x00 1000 0000 <----- i = 0x80;

now if we right shift the value of 'i', then MSB(most significant bit) will shifted to the next register* as shown below,

Address values
0x06 0000 0000
0x05 0000 0000
0x04 0000 0000
0x03 0000 0000
0x02 0000 0000
0x01 0000 0001
0x00 0000 0000 <----- i = 0x00;

It should show 0 if we read the value of 'i' as a character(%C) which is NULL(it will print NULL which is not a visible character like space).
But here we are reading the value of 'i' as integer(%d) so it will read the 4 bytes, which are,

0x03 0000 0000
0x02 0000 0000
0x01 0000 0001
0x00 0000 0000

In single line 00000000 00000000 00000001 00000000 <----- i = 0x100 = 256.

@Gopi Krishna, If we right shift the variable then MSB goes to next register's LSB, not the same one in C language.

One you are describing is the case of the microcontroller/processes @Kunal.

Hello all,

We are representing 128 in 2 byte because of the followings :

Case 1 - sizeof i=1 \\i is of character data type here.
Case 2 - sizeof 0x80=2 \\ (for 16 bit OS).
Case 3 - sizeof 128=2 \\(for 16 bit OS).
Case 4 - sizeof 'A'=2 \\unsigned char i = 'A';

Case 4 also requires 2 byte to be stored in the memory. So all the R-value(constants) of character take 2 byte from the memory.

Character constants like '0x80' are stored in memory in their corresponding ASCII value,as ASCII values are integer they requires 2 byte of memory.

#include<stdio.h>
int main()
{
unsigned char i = 0x80;
printf("%d\n", i<<1);
printf("%d\n",sizeof 0x80);
printf("%d\n",sizeof i);
return 0;
}

Output
256
2
1

Thanks.

Some are having doubt that why everyone having explanation assuming it is 2 byte?

First clear in you mind that we are not converting 1 byte of character into 2 byte.

Because if we do this then we have to write i = i<<2; but we are just printing a integer value using integer format specifier, but not changing the value of i.

After this program try this.

#include<stdio.h>
int main()
{
unsigned char i = 0x80;
printf("%d\n", i<<1);
printf("%d\n",i);
return 0;
}

I am just want to show that value of i is not changing.

You will get the output as:

256 as value of integer after left shift and 128 as the value of the i.

So don't confuse between both.

if char i=0,if we use %c than it should print 0 on stdout.

But if we use %d it should print the equal ASCII value of that char.if it is 0 than 48 should print.memory will take on datatype but on formatspecifier provided.any way char should take 1byte to store,for signed char the range from -128 to +127.1 bit used for sign representation. For unsign char no loss of bit ranges from 0 to 255.256 not valid in char type.
in microcontroller suppouse 8051 all registers 8-bit othar dptr(16bit) if accumulator a regester value exceeds 8 bit then it will generate the carry.

Dear sutendra Mirajkar

int main()
{
unsigned char i = 0x80;
i=i<<1;// i is char so 0.
0x80 in binary 1000 0000
after letf shift(i<<1) is 0000 0000 .That value store in i char(1byte 00000000)
printf("%d\n", i);// out put is 0
return 0;
}
//////////////////////////////////////

our result in inter thats y i take int var; i<<1 value is store int var. The output is 256
int main()
{
unsigned char i = 0x80;
int var;
var=i<<1;
printf("%d %d %d\n", var);
return 0;
}

@Swathi:
to conevrt hex to binary u need to take each digit of hex value and write its binary equivalent.
For eg:-
if hex = 88 , its binary would be
bin = 1000 1000
since binary equivalent of 8(decimal) is 1000.

example2:-
if hex = AF , its binary equi is
bin = 1010 1111
as binary equi of 'A'(in decimal system its 10) 1010 and that of 'F'(in decimal its 15) is 1111

Hope this will help u.
Take care :)

Hello everyone in the syntax of function printf:

printf("format specifier",variables);

In which it would expect the datatype which format specifier specifies not which variable specifies.

Example: printf("%d\n",i<<0x80); will take 2 bytes because format specifier is %d instead if you write.

printf("%c\n",i<<0x80) it will take only one byte and output will be NULL.

First you have look what is value of i.
Value of i is 0x80
It means value in Hexadecimal.

So,
Write it's binary in four bit for each digit
For zero(0) 0000.
For eight(8) 1000.

Finally you get this binary in 16-bit 0000 0000 10000 0000.

Now solve this expression i<<1.
Shift one(1) bit after shifting you get this binary
0000 0001 0000 0000 and print it %d(decimal value) i.e 256.

Hello All,
Simply,

unsigned char i = 0x80;
printf("%d\n", i<<1);

Produces 256 value as the output of this operation hasn't been stored back in i variable. i variable still contains 0x80 value.

unsigned char i = 0x80;
i=i<<1;
printf("%d\n", i);

This which will produce 0 output.

Anyone found any other thing, please tell me.

@sumit %d %c %f are used for printing values in int, char, float respectively and do not do any datatype conversions. Study type casting first then you ll b more clear about your concepts. +whenever any binary operators like +, -. /, * etc are used b/w different data types it. It promotes or demotes the datatype accordingly and then perform operations.