what does this operator stands for ^=

[RubyS]

I was reading some paper regarding galios field and a snippet of code was given as shown below

for i = 0:7
for j = 0:7
result [i+j] ^= a[j] & b

in this i m not able to understand the operator "^=".Is this some reduction operator..kindly help

 

[BradtheRad]

In BASIC there is a shorthand expression that goes:

result += a

In longhand it means:

result = result + a

There's a chance your code uses a similar type of shorthand.

The '^' operator normally means 'raise to the power of'.

 

[keith1200rs]

Assuming it is C, it is a standard operator https://en.m.wikipedia.org/wiki/Operators_in_C_and_C++

Keith

 

[alexan_e]

I was reading some paper regarding galios field and a snippet of code was given as shown below

Code of which language?

I haven't seen for i = 0:7 in C but maybe the XOR operator is the same.

 

[keith1200rs]

^ is also XOR in Verilog.

Keith