C Programming :: Bitwise Operators - Discussion - IndiaBIX

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C Programming - Bitwise Operators - Discussion

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Discussion Forum : Bitwise Operators - Find Output of Program (Q.No. 5) Bitwise Operators 5.What will be the output of the program? #include<stdio.h> int main() { unsigned char i = 0x80; printf("%d\n", i<<1); return 0; } 0 256 100 80 Answer: Option Explanation: No answer description is available. Let's discuss. Discussion:64 comments Page 1 of 7. Newest

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Namu said: 5 years ago How 0x80 is converted into binary? (3) Jasber said: 5 years ago i=0x80 means( 0000 0000 1000 0000).i<<1means (0000 0001 0000 0000)= 256. (5) Keziahtabraham said: 6 years ago 0*80 means 0000 0000 1000 0000.Then, one time left shift(i<<1):0000 0001 0000 0000According to position 1 is at 8th position, then to convert binary to decimal, 2^8=256. (1) Rajani R said: 6 years ago Please someone explain this solution. ANUPA A said: 7 years ago Please, someone clearly explain to me what does the bitwise operator<< actually does. Vaishnavi said: 7 years ago Kindly explain this program clearly. Sandeep Kumar said: 7 years ago #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. (8) Nagendra said: 8 years ago // 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. (4) Gopi.v said: 8 years ago 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) Mayur said: 8 years ago 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 values0x06 0000 00000x05 0000 00000x04 0000 00000x03 0000 00000x02 0000 00000x01 0000 00000x00 1000 0000