C calculation problem, C how to calculate a little more thank you

It's simple: you can't write it in C++, I write it in C.

**As follows. #include ""

void main()

int x=20,w=30,y;

printf("%d",~x&w);

The result is 10 is and, is the negation.

Except for 1+1=1, the rest are equal to zero.

Replace 1 with 0 and 0 with 1

Calculate the lower 8 bits;

20 is 0001 0100

30 is 0001 1110

20 is 1110 1011

o(∩_o...Ha ha.

x=20 in memory is 0000 0000 0000 0010w = 30 in memory is 0000 0000 0000 0011, then x is 1111 1111 1111 1101 in &w is 0000 0000 0000 0001 (that is, 10), so y is 10

I don't have a compiler at home to test it.

I don't remember if the order of operation of sum & is left to right or right to left, anyway this is the idea :)

Bitwise operations are not supported in C++, so X is wrong!

Put it in the denominator, bring out a, and then reduce it.

- -a =1;a=1; b + = 3;b=4; (a) +b + = 4; -c = 3;c=3;

-a) +b + c = 1; (b + = 1) => (b = b + 1=5)

So the answer is a=1 b=5 c=3

The yes or symbol you just asked, non-zero equals 1, e.g. 3|4=1, your question is a bit wrong, -a + b++ c is a whole, and the four operators have the highest priority, |It should be ranked last, and if you don't understand, you can ask again, thank you.

--c preceding --, immediately changes the value of c, which is 1b++ and posterior ++, in this equation, the value of b is still 3 and does not change, and b will become 4 after the operation after the equal sign

The --, in front of c immediately changes the value of c, which is 3, so it becomes.

b+=1+3-3 where b has become 4.

i.e. b=4; b+=1;So b is 5

What should be noted here is the difference between the front ++ and the rear ++, you can try to replace b++ with (++b) and the final result is 6.

b += --a + b++-c);Add parentheses to compile in vc++ 2005.

i.e. b=b+--a+b++-c).

4 There is --a, b++, c in the above equation, so after execution, a=1, b=5, c=3

It is important that you understand the priority of the operator: the self-incrementing operator" plus and minus the operator, so the program execution process is like this: first a self-subtract 1 and then a==1;Then b adds 1 at this time b==4, c subtracts 1 at this time c==3, and finally b+=1+4-3, the leftmost b==3 in this process, when this process is executed, b=3+(1+4-3)=5;

It's detailed.

6 * 103 4 rounded.

133 % 4 remainder.

Then subtract 153

The result is 3. (a=1,a++,b=1,a&&b++) is a comma expression, and its value is =a&&b++=1, so the value of the whole expression is a+1=3

Something completely meaningless.,Why don't you study it?。。。

Could it be that the computer output you a garbled code that day, and you still have to study the composition of this garbled code, embarrassment.

If you want to ask why, ask the guy who designed the C language.

This kind of program is meaningless, and the results will vary depending on the compiler.

int a=2

char b='a'Your original ** is wrong, then a+b+2 5= 99

This involves the automatic conversion of C++ types, for comparison: 2 5 = 0; = ;

The division of two shapes is still an integer, so it is 0 after rounding;

So a+b+2 5= 99