a b c b c in what order is it calculated,

As mentioned above, +- is the same priority, ||&&& is the same priority, and +- has a higher priority than ||&& has a high priority, so (b+c) (b-c) is counted first, and a |b+c, if the value of this formula is true, then it will be calculated with the next formula (&&b-c), otherwise, the value of the whole formula is 0

To solve this problem, we need to know their priority, first count the high priority, and then count the low priority. In this formula, +- is the same priority, ||&&& is the same priority, and +- has a higher priority than ||&&& has a higher priority, and the same priority is calculated from left to right.

The arithmetic operator takes precedence over the logical operator, and the combination of + and - is from left to right.

Naturally, b+c, b-c are calculated separately first, and then by priority, &&& is higher than ||Seems like it should.

First &&& then ||, but due to &&& and ||. in cIt's special.

The order of their calculations is strictly from left to right.

So the right one is first||Rear &&

First, calculate a, b+c, b-c

If a=a, b=b+c, c=b-c

Just the next step to judge.

b&&c if the result is r

Finally, a||r

over:)

First +, then -, then &&, and finally ||

Take a look at the Visual C++ Language Reference - Operator Precedence and Associate article.

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Please do the divination test:

Summary. The first is to remove the parentheses, the parentheses in front of it: a + c - a + b) are merged according to the algorithm:

A + C - A - B is then calculated in order from left to right: (A - A) +C - BSince A - A is equal to zero, we can simplify to 0 + C - B, and the final result is C - B, so the result of (A+C)-(A+B) is C - B.

a+c)-(a+b).

Okay. The first is to remove the parentheses, the parentheses in front of the world: a + c - a + b) The root segment is merged according to the family return algorithm:

A + C - A - B is then calculated in order from left to right: (A - A) +C - BSince A - A is equal to zero, we can simplify to 0 + C - B, and the final result is C - B, so the result of (A+C)-(A+B) is C - B.

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The first is to remove the parentheses, the parentheses in front of the world: a + c - a + b) The root segment is combined according to the family return algorithm: a + c - a - b and then calculated in order from left to right:

A - a) +c - bSince a - a is equal to zero, we can simplify to 0 + c - b and the final result is c - b, so the result of (a+c)-(a+b) is c - b.

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!(a+b)+c-1&&b+c/2

Because the && operator is a short-circuit algorithm, it is judged first! Whether the result of (a+b)+c-1 is 0, if it is 0, then there is no need to calculate the b+c 2 on the right, even if b+c 2 has a higher priority, it can be verified, modified to! (a+b)&&b+(c=22));So that the rightmost assignment operator should be evaluated first, but as long as !

a+b)=0, the sentence c=22 on the right will not be executed, so the value of c will not be changed.

If! If the result of (a+b)+c-1 is 1, it is calculated in the following order, attention! The order of (a+b)+c-1 will not be affected.

Calculate a+b in parentheses first

And then calculate! (a+b)

Calculate again! (a+b)+c-1, judge here! (a+b)+c-1 is 1, if 1, the expression on the right is executed, ie.

Calculate C2 first

Then calculate b+c 2

Finally put it again! (a+b)+c-1 and b+c2.

Make a b=b laugh c=c a=k

then a=kbb=kc

c=ka additive.

a+b+c=k(a+b+c)

a+b+c)(k-1)=0

a+b+c=0 or k-1=0

If a+b+c=0, then (a+b+c) c=0 if k-1=0, k=1, then a=b=c, then touch the surplus (a+b+c) missing c=3c c=3

So (a+b+c) c=0 or 3