问题
When I do this: count = ++count; Why do i get the warning - The assignment to variable count has no effect ? This means that count is incremented and then assigned to itself or something else ? Is it the same as just ++count ? What happens in count = count++; ? Why don't I get a warning for this ?
回答1:
count++ and ++count are both short for count=count+1. The assignment is built in, so there's no point to assigning it again. The difference between count++ (also knows as postfix) and ++count (also known as prefix) is that ++count will happen before the rest of the line, and count++ will happen after the rest of the line.
If you were to take apart count=count++, you would end up with this:
count = count;
count = count+1;
Now you can see why postfix won't give you a warning: something is actually being changed at the end.
If you take apart count=++count, you would end up with this:
count = count+1;
count = count;
As you can see, the second line of code is useless, and that's why the compiler is warning you.
回答2:
Breaking the statement up you are basically writing:
++count;
count = count;
As you can see count=count does nothing, hence the warning.
回答3:
the ++ operator is a shortcut for the following count = count + 1. If we break your line count = ++count it responds to count = count+1 = count
回答4:
To expand a little, count++ is postfix. It takes place after other operations so if you did something like
int a = 0, b = 0;
a = b++;
a would be 0, b would be 1. However, ++count is prefix if you did
int a = 0, b = 0;
a = ++b;
then a and b would both be 1. If you just do
count++;
or
++count;
then it doesn't matter, but if you are combining it with something else, it will
来源：https://stackoverflow.com/questions/11639525/the-assignment-to-variable-has-no-effect