c++ math/hex question

Sorry if this sounds a bit 'noobish' but I don't know c++ and I was looking at some code for the perlin noise, trying to figure it out when I came across

& 0x7fffffff

and

(n << 13) ^ n;

and I have no idea what those mean. I do know that 0x7fffffff is hex, but I don't know what the & means Any help?

The first expression is a bitmask. It basically clears out the upper bit of the number, or equivalently computes a mod of the number over 2^31.

The second expression is a shift, followed by an xor. In arithmetic it would be the same thing as multiplying n by 2^13, then flipping all the bits that are common between the shifted version and itself. Its purpose in the perlin noise code is to compute a procedural spatial hash of the x,y coordinates so that they can be used to seed the noise generator.

This same type of technique is used in a lot of procedural content to create dynamically varying psuedo random number generators that vary deterministically in space. These hashes are basically meant to be complicated, difficult to predict, nearly random functions and are usually discovered with a mixture of theory, guess work and experimentation. As a result, I would not recommend thinking too hard about why it is using that specific formula in this case.

& is a bit wise AND operator in C++

The bitwise AND operator (&) compares each bit of the first operand to the corresponding bit of the second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.

     11111111 11110000  
   & 00000000 01100011
     _________________
     00000000 01100000

<< is bit wise left shift operator in C++

The << operator shifts its first operand left by a number of bits given by its second operand, filling in new 0 bits at the right.

    0 1 0 1 0 1 1 0 << 2
    _____________________
    0 1 0 1 0 1 1 0 0 0

^ is Ex-Or operator in C++

The bitwise-exclusive-OR operator compares each bit of its first operand to the corresponding bit of its second operand. If one bit is 0 and the other bit is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.

     0 1 0 1 0 1 1 0
   ^ 0 0 1 1 0 0 1 0
   ___________________  
     0 1 1 0 0 1 0 0

So, & 0x7fffffff sets bit 31 of the 32-bit integer to zero and leaves other bits with values they had.

(n << 13) ^ n n is left-shifted with 13 and the result is XORed with n.

& is bitwise AND operator. So & 0x7fffffff stripping off the signed bit from an 32 bit integer. That means, it makes the leftmost bit of int zero.

And in the second one, first n is left-shifted with 13 using << operator which is left-shift operator, and the result of this expression used in XOR operation with n itself.

& 0x7fffffff means set bit 31 of the (presumed) 32-bit integer to zero, preserving the values of bits 0 through 30; IOW, make the 32-bit integer be positive; IOW, take the abs val of the integer.

(n << 13) ^ n; means shift n left by 13 bits, then xor that shifted value with the original value of n.

HTH

The symbols &, <<, and ^ are called operators. There is a Wikipedia page that lists C/C++ operators so you can learn about the names of those operators and then search Google for more information.

The 0x7fffffff is an integer written in hex (base 16) notation. The 0x prefix means it is hex.

The n is probably the name of an integer variable defined a few lines above the line you quoted.

The parentheses serve to group expressions together, so in your example they guarantee that n gets shifted left by 13 bits BEFORE it gets XOR'd with n. Without the parentheses, the order of operations would be determined by the operator precedence, (which in your example happens to give the same result).