# Operator precedence

Nancy Gu

Greenhorn

Posts: 6

posted 15 years ago

- 0

Hello, everybody:

What is the eveluating order and results for these questions?

Pls expain it for me. Thanks.

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int n = 7;

n<<=3;<br /> n = n & n + 1 | n + 2 ^ n + 3;<br /> n >>= 2;

System.out.println(n);

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8 | 9 & 10 ^ 11

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boolean b1=b2=b3=b4=true

b1 | b2 & b3 ^ b4

----------------------------------------

boolean b1=b2=b3=b4=true

b4 = b4 | b1 & b2;

b3 = b3 & b1 | b2;

What is the eveluating order and results for these questions?

Pls expain it for me. Thanks.

----------------------------------------

int n = 7;

n<<=3;<br /> n = n & n + 1 | n + 2 ^ n + 3;<br /> n >>= 2;

System.out.println(n);

----------------------------------------

8 | 9 & 10 ^ 11

----------------------------------------

boolean b1=b2=b3=b4=true

b1 | b2 & b3 ^ b4

----------------------------------------

boolean b1=b2=b3=b4=true

b4 = b4 | b1 & b2;

b3 = b3 & b1 | b2;

Amit, Jhalani

Greenhorn

Posts: 21

posted 15 years ago

- 0

int n = 7;

n<<=3;<br /> n = n & n + 1 | n + 2 ^ n + 3;<br /> n >>= 2;

System.out.println(n);

The order of execution for the bitwise and or and Xor operator is as follows first it evaluates the bitwise and(&) then it evaluates the bitwise Xor(^) ann then it evaluates the bitwise or(|)

but in the expression(n = n & n + 1 | n + 2 ^ n + 3 )) all the airthmetic operators will be evaluated first then the bitwise operators and finally the assignment took place.

so the result is 56.

8 | 9 & 10 ^ 11

result will be 14.

boolean b1=b2=b3=b4=true

b1 | b2 & b3 ^ b4

you can not do assignment and initialization at the same time.

rather you can say

boolean b1,b2,b3,b4;

b1=b2=b3=b4=true;

b1 | b2 & b3 ^ b4;

then result would be true.

& is first to be evaluated then ^ and then |

same resion for the following.

b4 = b4 | b1 & b2;

b3 = b3 & b1 | b2;

correct me if i am wrong.

Thanks.

[This message has been edited by Amit, Jhalani (edited January 12, 2001).]

[This message has been edited by Amit, Jhalani (edited January 12, 2001).]

n<<=3;<br /> n = n & n + 1 | n + 2 ^ n + 3;<br /> n >>= 2;

System.out.println(n);

The order of execution for the bitwise and or and Xor operator is as follows first it evaluates the bitwise and(&) then it evaluates the bitwise Xor(^) ann then it evaluates the bitwise or(|)

but in the expression(n = n & n + 1 | n + 2 ^ n + 3 )) all the airthmetic operators will be evaluated first then the bitwise operators and finally the assignment took place.

so the result is 56.

8 | 9 & 10 ^ 11

result will be 14.

boolean b1=b2=b3=b4=true

b1 | b2 & b3 ^ b4

you can not do assignment and initialization at the same time.

rather you can say

boolean b1,b2,b3,b4;

b1=b2=b3=b4=true;

b1 | b2 & b3 ^ b4;

then result would be true.

& is first to be evaluated then ^ and then |

same resion for the following.

b4 = b4 | b1 & b2;

b3 = b3 & b1 | b2;

correct me if i am wrong.

Thanks.

[This message has been edited by Amit, Jhalani (edited January 12, 2001).]

[This message has been edited by Amit, Jhalani (edited January 12, 2001).]