C Programming - Bitwise Operators - Discussion

Yes it is correct @Abhinav but when we consider that and tried to shift by 1 then answer goes wrong. Please anyone tell how I convert this hex number into binary.

@Abhinav.

No, its not correct. 0X80 is a hexadecimal number, and for hexadecimal number you have to convert nibble by nibble.

For example:

0X80 --------> two nibbles 8 0.

1000 0000.

0X 75 -------> 7 5.

0111 0101.

Although, you are right if it were decimal number.

For example:

80 ----------> 01010000.

"0X" makes it hexadecimal.

#include<stdio.h>

int main()
{
unsigned char i = 0x80;
int a=i<<1;
unsigned long b=a;
unsigned char c=b;
printf("%d\n",c);
b=b>>1;
c=b;
printf("%d\n",c);
return 0;
}

Try this.

The answer should be zero. I checked it. Unsigned char takes only single byte.

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.

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.

// 1

#include<stdio.h>

int main()
{
unsigned char i = 0x80;
printf("%d\n", i<<1);
return 0;
}
Here output 256.

// 2
#include<stdio.h>

int main()
{
unsigned char i = 0x80;
i=i<<1;
printf("%d\n", i);
return 0;
}
Here output 0.

* The diff is assignment of variable i.

#include<stdio.h>

int main()
{
unsigned char i = 0x80;
i=i<<1;
printf("%d\n", i);
return 0;
}

In this case, also typecast operation is performed but final data is stored in 8 bit so the output is 0.

Kindly explain this program clearly.

Please, someone clearly explain to me what does the bitwise operator<< actually does.