Precendence Examples

4 / 3 * 2 ??? I thought this would be 4 / (3 * 2)? But it evaluates to (4 / 3) * 2


-brian

     printf("%d\n",16 % 41 / 6 * 62 + 45 / 22 % 5 );
      printf("%d\n",(((16 % 41) / 6) * 62) + ((45 / 22) % 5) );


those produce the same result, but I thought that division had higher precedence than mod, and multiplication had higher precedence than division.


-Brian

And I thought that the Logical Not was only once step higher than the logical and, why is the logical not at a higher precedence than Multiplication???


-brian

Hi BrianGEFF719,

The C grammar precedence is laid out here:

http://www.difranco.net/cop2220/op-prec.htm



Good Luck!
Kent

Hi BrianGEFF719,

> 4 / 3 * 2 ??? I thought this would be 4 / (3 * 2)? But it evaluates to  (4 / 3) * 2

/ and * have equal weight.  They are evaluated left-to-right.


Kent

Hi BrianGEFF719,

> those produce the same result, but I thought that division had higher
> precedence than mod, and multiplication had higher precedence than division.

Again, check the table mentioned above.  Multiplication, division, and modulo are equally weighted.  They're evaluated left to right.


Kent

are these correct?

i || j || k -> ( i || j ) || k
i && j || k -> (i && j) || k
i || j && k -> i || (j && k)
i || ++j  ->  i || (j = j + 1)

Hi BrianGEFF719,
> i || j || k -> ( i || j ) || k         // logical or evaluates left-to-right.  These are equivalent
> i && j || k -> (i && j) || k       //  logical and is higher precedent than logical or.  These are equivalent
> i || j && k -> i || (j && k)       //  logical and is higher precedent than logical or.  These are equivalent
> i || ++j  ->  i || (j = j + 1)     //  unary preincrement is higher precentent that logical or.  These are equivalent.

You get an 'A'.  (Assuming of course that you were using -> as a separator, not an embedded dereference.  :)  )

Kent

>>> i || ++j  ->  i || (j = j + 1)     //  unary preincrement is higher precentent that logical or.  These are equivalent.

but if it was writen with out parenthesis:

i || j = j + 1 that would be

(i || j) = j + 1, which is invalid, is this correct?


-brian

Hi BrianGEFF719,


A bit depends on the compiler.  You certainly can't store (j+1) into (i || j).  The compiler may (heavily on the may) recognize that that evaluation is illegal and evaluate it in a legal fashion, but that would certainly be an extension that I wouldn't rely upon.


Kent

but is the order correct as far as how it would be processed?

Hi BrianGEFF719,

Yes.  There are hardware reasons behind some of this.

1)  On machines so equipped, ++j uses the INCR or increment instruction.  Basically, the opcode instructs the CPU to add 1 to the value at the specified address.

2)  j = j+1 is handled by the expression evaluator.  J+1 is evaluated, and the result stored back at J.




Kent

i know on x86 its

inc register to add one to the value in the register or
dec register to subtract one from the value in the register.

but should any optimizing compiler do that anyway???


-brian

in your opinion code that is writtin gas such

i || (j = j + 1) should be written as i || ++j, correct?


-brian