So you think you know the priority of operators in c++?

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Closed 13 years ago.

我只读了一部法典,它不仅汇编了,而且带来了预期结果(X是分类值):

int y = (int)(0.5 * x * x + + + 0.6 * x + 1.2);

我只想说什么发生,我必须说这是一个有趣的经营者问题。如果不汇编该方案,以下业务的结果是什么?

int a = 1;
int b = 2;
int z = a + + + b;
int z1 = a + - + b;
int z2 = a + - - b;
int z3 = a - - + b;
int z4 = a - - - b;
int z5 = a +- b;

我仍然有一个问题,尽管:标准是否提供这种结果,还是具体汇编?

Explanation: Because the + and - operators have spaces between them, the "+ + +" sequence is not compiled as "++ +", but as unary operators on the right member. So

int y = (int)(0.5 * x * x + + + 0.6 * x + 1.2);

实际:

int y = (int)(0.5 * x * x + 0.6 * x + 1.2);

这是预期的结果。

因此,

z  = a + + + b = a + + (+b) = a + (+b) = a + b = 3;
z1 = a + - + b = a + - (+b) = a + (-b) = a - b = -1;
z2 = a + - - b = a + - (-b) = a + (+b) = a + b = 3;
z3 = a - - + b = a - - (+b) = a - (-b) = a + b = 3;
z4 = a - - - b = a - - (-b) = a - (+b) = a - b = -1;
z5 = a +- b = a + (-b) = a - b = -1;

Answer 1

z = 3
z1 = -1
z2 = 3
z4 = 3
z5 = -1

Without compiling, i d say it s a simple math operator combining
I ve learned it with this phrase (translated from Brazilian portuguese):

Different signs = subtract
Equal signs = add

So

int z = a + + + b; //  +  +  +  =  +  , the resulting  +  +  +  = + again.
int z1 = a + - + b; //  +  +  -  =  -  , the resulting  -  +  +  = - again.
int z2 = a + - - b; //  +  +  -  =  -  , the resulting  -  +  -  = +.
int z3 = a - - + b; //  -  +  -  =  +  , the resulting  +  +  +  = + again.
int z4 = a - - - b; //  -  +  -  =  +  , the resulting  +  +  -  = -.
int z5 = a +- b; //simple "different signals = subtract" :)

Answer 2

回答标题中提出的问题:

So you think you know the priority of operators in c++?

我认为,没有人会这样做,更重要的是,我无意学习。在读书时,每当文件没有保证时,我将使用括号。我需要记住更多的重要事情。

Answer 3

[source ] 普通+/-约束比附加/减量+/-和准权利紧缩。因此......

int a = 1;
int b = 2;
int z = a + + + b; // equivalent to a + (+(+b))
int z1 = a + - + b; // equivalent to a + (-(+b))
int z2 = a + - - b; // equivalent to a + (-(-b)) = a + b
int z3 = a - - + b; // equivalent to a - (-(+b)) = a + b
int z4 = a - - - b; // equivalent to a - (-(-b)) = a - b
int z5 = a +- b; // equivalent to a + (-b) = a - b

Answer 4

为了回答你的问题,答案并不具体。它既未界定实施,也未界定明确的行为。

So you think you know the priority of operators in c++?

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