C Programming - Bitwise Operators - Discussion

Try this it will be all clear.

#include<stdio.h>

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

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.

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.

Not a single explanation was useful. If anyone knows correct explanation, please help us. Thank you.

What will be the answer if it was printf("%x",i<<1); ?

Here it is given unsigned char that's why we have taken 2 byte.

The hex representation of 80 is 1000 0000.

i = 0000 0000 1000 0000.

Left shift by 1.

We get 1= 0000 0001 0000 0000 = 256 as result.

What is the meaning of "return 0"?

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.

int 0x86 means it is in hexadecimal So the decimal equivalent of 86 is 128. Left shift of 128 is 256 because left shift by 1 of a number is equivalent to multiplying it with 2.

@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.