C Programming - Bitwise Operators - Discussion

Size of char is 1 byte.
So it have 8 bits. The equivalent of 0*80 is 1000 0000.
If I do left shift then it is 0000 0001 so answer is 1.

This is my thinking. Can any one explain me correct?

#include<stdio.h>

int main()
{
char i=0x04;
char k=0x80;
printf("%d",i<<6);
printf("%d",k<<1);

return 0;
}

Output of first print statement is 256 while for other it is -256 how?

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.

Here x080 can be written as 0000 0000 0101 0000. Is it true?

Step 1 :

First convert the Hexadecimal into decimal
0x80 ---> 16*8=128.

Step 2 :

Then use bit-wise operation left shift.
128*pow(2,1).
Here 1 indicates the value after left shift.

@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.

Kindly explain this program clearly.

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