Is the Integer Division in Javascript the same as Math.floor(x) for x greater than or equal to 0, or is x | 0 equivalent?

After examining the examples below, it appears that Math.floor(x) is the same as x | 0 when x >= 0. Is this accurate? And if so, why is that the case? (or how is x | 0 computed?)

x = -2.9; console.log(Math.floor(x) + ", " + (x | 0));   // -3, -2
x = -2.3; console.log(Math.floor(x) + ", " + (x | 0));   // -3, -2
x = -2;   console.log(Math.floor(x) + ", " + (x | 0));   // -2, -2
x = -0.5; console.log(Math.floor(x) + ", " + (x | 0));   // -1, 0
x = 0;    console.log(Math.floor(x) + ", " + (x | 0));   //  0, 0
x = 0.5;  console.log(Math.floor(x) + ", " + (x | 0));   //  0, 0
x = 2;    console.log(Math.floor(x) + ", " + (x | 0));   //  2, 2
x = 2.3;  console.log(Math.floor(x) + ", " + (x | 0));   //  2, 2
x = 2.9;  console.log(Math.floor(x) + ", " + (x | 0));   //  2, 2
x = 3.1;  console.log(Math.floor(x) + ", " + (x | 0));   //  3, 3

This trick can be handy for executing integer division in Javascript: (5 / 3) | 0 instead of using Math.floor(5 / 3).

Answer №1

When using bitwise operators, numbers are converted to a 32-bit sequence. This means that the suggested alternatives will only be effective with positive signed 32-bit floats, which are numbers ranging from 0 to +2,147,483,647 (2^31-1).

Math.floor(2147483646.4); // 2147483647
2147483646.4 | 0; // 2147483647
// however…
Math.floor(2147483648.4); // 2147483648
2147483648.4 | 0; // -2147483648

Another distinction to make is that if x is not a number, the outcome of x | 0 may differ from the result of Math.floor(x).

Math.floor(NaN); // NaN
NaN | 0; // 0

Apart from these considerations, the result should be comparable to that of Math.floor() when dealing with positive numbers.

For more examples and performance tests, visit: http://jsperf.com/rounding-numbers-down

Answer №2

According to the specifications outlined in the ECMAScript document, section 11.10 delves into the realm of Binary Bitwise Operators:

Understanding the intricacies
In the scenario where A : A @ B, with @ representing one of the bitwise operators listed prior in the specifications, the process of evaluation unfolds in the following manner:
1. Begin by determining the reference value of A and store it as lref.
2. Proceed to calculate the actual value by obtaining lval through GetValue(lref).
3. Repeat the same process for B by first evaluating B and storing the result as rref.
4. Obtain rval by evaluating the value of rref.
5. Convert lval to a 32-bit signed integer, known as lnum.
6. Perform the same conversion for rval, giving you rnum.
7. Finally, utilize the bitwise operator @ to operate on lnum and rnum, producing the result as a signed 32 bit integer.

The procedure followed for calculating x | y is as follows: First, both x and y are converted to Int32 before the | operator is applied to them.

Answer №3

When working with bitwise operations in JavaScript, it's important to remember that the numbers are limited to 32 bits, so any float values will first be converted to integers.

For example, in the expression "3.9" | 0 = 3, it appears as though the parseInt function is being invoked behind the scenes.

Answer №4

When the vertical bar is used as a bitwise-or operator, interesting things happen. While the binary representation of zero consists of all zeros, the expression x|0 technically does nothing. However, to evaluate it, the operands must both be integers. Therefore, the floating-point number x needs to be converted into an integer first. This conversion process involves removing the decimal part, leading to situations where for certain values of x greater than or equal to zero, x|0 is equivalent to Math.floor(x).

It's important to note that the outcome can vary based on the size and sign of the internal integer type. For instance, you might observe:

2147483648|0     == -2147483648     // 0x80000000
Math.pow(2,32)|0 == 0               // the lowest 32 bits are all 0

Answer №5

The operation (x | 0) eliminates the decimal portion of x, leading to the following true relationship:

x | 0 = (x < 0 ? -1 : 1) * Math.floor(Math.abs(x)) ;

Interestingly, x >> 0 produces the same outcome as x | 0, therefore:

x >> 0 = x | 0 = (x < 0 ? -1 : 1) * Math.floor(Math.abs(x)) ;

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