How to get the bit length of an integer in C++? [duplicate] - c++

Question

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Minimum number of bits to represent a given `int`
(9 answers)
Closed 4 months ago.
This question is not a duplicate of Count the number of set bits in a 32-bit integer. See comment by Daniel S. below.
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Let's say there is a variable int x;. Its size is 4 bytes, i.e. 32 bits.
Then I assign a value to this variable, x = 4567 (in binary 10001 11010111), so in memory it looks like this:
00000000 00000000 00010001 11010111
Is there a way to get the length of the bits which matter. In my example, the length of bits is 13 (I marked them with bold).
If I use sizeof(x) it returns 4, i.e. 4 bytes, which is the size of the whole int. How do I get the minimum number of bits required to represent the integer without the leading 0s?

Answer 1

unsigned bits, var = (x < 0) ? -x : x;
for(bits = 0; var != 0; ++bits) var >>= 1;
This should do it for you.

Answer 2

Warning: math ahead. If you are squeamish, skip ahead to the TL;DR.
What you are really looking for is the highest bit that is set. Let's write out what the binary number 10001 11010111 actually means:
x = 1 * 2^(12) + 0 * 2^(11) + 0 * 2^(10) + ... + 1 * 2^1 + 1 * 2^0
where * denotes multiplication and ^ is exponentiation.
You can write this as
2^12 * (1 + a)
where 0 < a < 1 (to be precise, a = 0/2 + 0/2^2 + ... + 1/2^11 + 1/2^12).
If you take the logarithm (base 2), let's denote it by log2, of this number you get
log2(2^12 * (1 + a)) = log2(2^12) + log2(1 + a) = 12 + b.
Since a < 1 we can conclude that 1 + a < 2 and therefore b < 1.
In other words, if you take the log2(x) and round it down you will get the most significant power of 2 (in this case, 12). Since the powers start counting at 0, the number of bits is one more than this power, namely 13. So:
TL;DR:
The minimum number of bits needed to represent the number x is given by
numberOfBits = floor(log2(x)) + 1

Answer 3

You're looking for the most significant bit that's set in the number. Let's ignore negative numbers for a second. How can we find it? Well, let's see how many bits we need to set to zero before the whole number is zero.
00000000 00000000 00010001 11010111
00000000 00000000 00010001 11010110
^
00000000 00000000 00010001 11010100
^
00000000 00000000 00010001 11010000
^
00000000 00000000 00010001 11010000
^
00000000 00000000 00010001 11000000
^
00000000 00000000 00010001 11000000
^
00000000 00000000 00010001 10000000
^
...
^
00000000 00000000 00010000 00000000
^
00000000 00000000 00000000 00000000
^
Done! After 13 bits, we've cleared them all. Now how do we do this? Well, the expression 1<< pos is the 1 bit shifted over pos positions. So we can check if (x & (1<<pos)) and if true, remove it: x -= (1<<pos). We can also do this in one operation: x &= ~(1<<pos). ~ gets us the complement: all ones with the pos bit set to zero instead of the other way around. x &= y copies the zero bits of y into x.
Now how do we deal with signed numbers? The easiest is to just ignore it: unsigned xu = x;

Answer 4

Many processors provide an instruction for calculating the number of leading zero bits directly (e.g. x86 has lzcnt / bsr and ARM has clz). Usually C++ compilers provide an intrinsic for accessing one of these instructions. The number of leading zeros can then be used to calculate the bit length.
In GCC, the intrinsic is called __builtin_clz. It counts the number of leading zeros for a 32 bit integer.
However, there is one caveat about __builtin_clz. When the input is 0, then the result is undefined. Therefor we need to take care of this special case. This is done in the following function with (x == 0) ? 32 : ..., which gives the result 32 when x is 0:
uint32_t count_of_leading_0_bits(const uint32_t &x) {
return (x == 0) ? 32 : __builtin_clz(x);
}
The bit length can then be calculated from the number of leading zeros:
uint32_t bitlen(const uint32_t &x) {
return 32 - count_of_leading_0_bits(x);
}
Note that other C++ compilers have different intrinsics for counting the number of leading zero bits, but you can find them quickly with a search on the internet. Here is How to use MSVC intrinsics to get the equivalent of this GCC code? for an equivalent with MSVC.

Answer 5

The portable modern way since C++20 should probably use std::countl_zero, like
#include <bit>
int bit_length(unsigned x)
{
return (8*sizeof x) - std::countl_zero(x);
}
Both gcc and clang emit a single bsr instruction on x86 for this code (with a branch on zero), so it should be pretty much optimal.
Note that std::countl_zero only accepts unsigned arguments though, so deciding how to handle your original int parameter is left as an exercise for the reader.