C Programming - Bitwise Operators - Discussion

Char should overflow to become zero instead its becoming 256.

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.

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.

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.

Kindly explain this program clearly.