C Programming - Bitwise Operators - Discussion

if were a larger type (e.g., int), 0x80 << 1 would be 256 (binary 100000000).
But the unsigned char is 8-bit, so the overflow bit is discarded, leaving 00000000 (0).
So, I think answer is 0

How 0x80 is converted into binary?

i=0x80 means( 0000 0000 1000 0000).

i<<1means (0000 0001 0000 0000)= 256.

0*80 means 0000 0000 1000 0000.

Then, one time left shift(i<<1):0000 0001 0000 0000
According to position 1 is at 8th position, then to convert binary to decimal, 2^8=256.

Please someone explain this solution.

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

Kindly explain this program clearly.

#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.

// 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.

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.