## not understanding decimal to binary with bitwise operators in c++

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// Step 1: You call a function defined as `void function(int a)` // which accepts one parameter of type int and returns a void. void function(int a) { // Step 2: This function checks if a is zero or some other value. // If it is zero, nothing happens and the function returns. // If it is non-zero, we proceed inside the if block. if(a){ // Step 3: we do two things here. // First we right-shift whatever value is stored in a. // Right-shifting means if we have e.g. a=8, which is 1000 in binary, // then rightshifting 8 would be 0100 or 4. Basically it is division by 2. // This continues until we have 0000. // Secondly, this function will call itself then. This is called recursion. // So the "void function(4)-> void function(2) -> void function(1)" calls will run. function(a>>1); // Here we will bitwise AND with 1. // So, 0001 & 0001 = 1 // 0010 & 0001 = 0 // 0100 & 0001 = 0 // 1000 & 0001 = 0 // And this is what you will see on the terminal as output: 1000 // The 1 gets printed first, because the inner most call in the recursion // finishes first and the call stack unwraps back up. cout<<(a&1); } }

**not understanding decimal to binary with bitwise operators in c++ ,** // Step 1: You call a function defined as `void function(int a)` // which accepts one parameter of type int and returns a void. void function(int a) Algorithm for Decimal to Binary using Bitwise Operator : Initialize ‘mask’ variable with 32768 [1000 0000 0000 0000] Perform ANDing of two number’s (i.e Given variable and Mask variable ) Check whether the Result of ANDing is 0 or not , if Yes Display 0 otherwise Display 1; Right shift mask variable by 1 [0100 0000 0000 0000]

Recursive function

void function(int a){ if(a){ // if a is not 0, continue calling recursively function(a>>1); // right shift a one step i.e. divide by 2, then call ourselves cout<<(a&1); // bitwise & with a i.e print if a is odd or even. } }

Example

function(4); a = 4, a>>1 == 2 function(2); a = 2, a>>1 == 0 function(0); will end recursion cout << (2&1) will print 0 since 0x10 & 0x01 == 0x00 cout << (4&1) will print 0 since 0x20 & 0x01 == 0x00

**C Program to Convert Decimal to Binary using Bitwise AND operator ,** C Program to Convert Decimal to Binary using Bitwise AND operator To convert the Decimal Number into Binary , Check First MSB bit of number , if it is 1 then Explanation of Decimal to Binary Program : As mask variable is not equal to 0 so once again while loop will be executed and in the next It is clear that, this C program will Convert Decimal to Binary using Bitwise Operator. For Example: If the Decimal number 27 is given as input, the program will print the binary equivalent as 11011 The below given C program is used to Convert Decimal to Binary using Bitwise Operator.

void function(int a) // function declaration { if(a) // a != 0? { function(a>>1); // recursive call to function with a shifted right one bit cout<<(a&1); // print out least significant bit of a } }

This function prints out a binary representation of a number given, e. g. a initially 13:

f(13) // 13 == 0b1101 { f(6) // 0b1101 >> 1 == 0b0110 { f(3) // 0b0110 >> 1 == 0b0011 { f(1) // etc { f(0) { } // does nothing std::cout << (1 & 1); // 1, most significant bit of original 13 } std::cout << (3 & 1); // 1 } std::cout << (6 & 1); // 0 } std::cout << (13 & 1); // 1 } 1101 in total

Signature is dangerous, though; wheras it is implementation defined if the shift is logic (always 0 shifted in at the left) or arithmetic (shift in current sign bit again), *usually* (question is C, but applies for C++ as well) arithmetic shift is implemented for signed values.

So you would end up in an endless recursion if you pass a negative value to the function. Better:

void function(unsigned int a); // ^^^^^^^^

For *unsigned* values, logic and arithmetic shift are the same...

Side note: if you pass 0 initially to the function, you don't get any output at all (either variant).

**C program to convert decimal to binary using bitwise and, shift ,** To know what is U , read documentation about 'integer constant'. Quote: i cannot understand the code although had tried to understand this code Let's take an example of using "bitwise AND" on 7 and 11 to understand and see how it works. The binary value of 11 is 00001011 and the binary value of 7 is 00000111. The bitwise AND operator just performs the bitwise operation on each bit. 00001011 & 00000111----- 00000011 → 3 in decimal system

**Program for Decimal to Binary Conversion,** Given a decimal number as input, we need to write a program to convert the given decimal number into equivalent binary number. C++ program to convert a decimal Note that bitwise operators work faster than arithmetic operators used above. C++ Decimal to binary conversion can also be done without using arrays. 35 = 00100011 (In Binary) Bitwise complement Operation of 35 ~ 00100011 _____ 11011100 = 220 (In decimal) Twist in bitwise complement operator in C Programming. The bitwise complement of 35 (~35) is -36 instead of 220, but why? For any integer n, bitwise complement of n will be -(n+1). To understand this, you should have the knowledge of 2's

**C Program to Convert Decimal to Binary using Bitwise Operator,** Related Other Popular C Programs. * Write a c program to check whether a given number is prime number or not? * Write a C program to find the In C, the following 6 operators are bitwise operators (work at bit-level) The & (bitwise AND) in C or C++ takes two numbers as operands and does AND on every bit of two numbers. The result of AND is 1 only if both bits are 1. The | (bitwise OR) in C or C++ takes two numbers as operands and does OR on every bit of two numbers. The result of OR

**Decimal to Binary conversion using bitwise shift operator C++ program,** Size of an integer is assumed to be 32 bits. We will use the bitwise operator "AND" to perform the desired task. We right shift the original number by 31, 30, 29, .. The following table lists the Bitwise operators supported by C. Assume variable 'A' holds 60 and variable 'B' holds 13, then − & Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) = 12, i.e., 0000 1100 | Binary OR Operator copies a bit if it exists in either

##### Comments

- For negative a, the value of a >> 1 is implementation-defined.
- You need to google a few things. "Left and right shift", "Bitwise AND and OR" and "Recursion".
- You need to target what you do not understand in this function: The bitwise operations (
`<<`

and`&`

)? The recursion of`function`

? Something else? *'First we right-shift'*- that's perhaps ambivalent; if had to rewrite the function solely based on this explanation, I'd do`a >>= 1; f(a); std::cout << (a&1)`

.- @OP, To avoid the problems, mentioned for the case of negative numbers, I would change
`if(a)`

by`if (a>0)`

. - @Aconcagua, Are you referring to "6.2.2 Evaluation Order [expr.evaluation]"? I always thought that the operation would be performed first, before the function call continues. Is it not so? If not, then apparently, I have learned something new today.
- Your understanding is fine, the point is: your wording could be misunderstood. Sure the value first is shifted right,
*then*passed to the function - to be more precise:*the result*of the shift is passed - wheras the operand`a`

remains unchanged.*'First we right-shift'*I'd understand, in contrast to, as modifying the operand as well, just as in`f(a >>= 1)`

(note the additional`=`

). - Apart from: It is totally legal to right-shift 0 (no UB at all)!
`0 >> 1`

just remains 0...