Is there some one who can help me with some complex math Expressions in 'C' programming.(details given below)

please explain how the compiler calculates the following expressions and which variable get what value





1...assume x=1 i=2 j=3 k=4


x-=k*=j/=i%=5; (please explain how this works)





2... x=5 i=4 j=3 k=2


x*=1 + (i%= 1+(j/=-1+ ++k)





3....i=j=10;


i%=j=(j=5)%(i=3);





thanks

Is there some one who can help me with some complex math Expressions in 'C' programming.(details given below)
in C the evaluation for an expression goes like this:


Evaluate the right part of expression and then the left one


and i=5 returns 5


so:


1. x-=k*=j/=i%=5;


- i%=5 -%26gt; 2 and i=2


- j/=i%=5 -%26gt; j/=2 -%26gt; 1 and j = 1


- k*=j/=i%=5 -%26gt; k*= 1 -%26gt; 4


- x-=k*=j/=i%=5 -%26gt; x-=4 -%26gt; -3 and x= -3


2.x*=1 + (i%= 1+(j/=-1+ ++k)


- ++k - %26gt; 3 and k=3


- j/=-1+ ++k -%26gt; j/=-1 + 3 -%26gt; j/= 2 -%26gt; 1


-... i%= 1+(j/=-1+ ++k) -%26gt; 0


- x*=1 + (i%= 1+(j/=-1+ ++k)) -%26gt; x*= 1+0


- x = 5


3.i%=j=(j=5)%(i=3)


- i=3 -%26gt; 3 and i become 3


- j=5 -%26gt; same


- (j=5)%(i=3) 0 -%26gt; 2


- j=(j=5)%(i=3) -%26gt; j=2 -%26gt; 2 and j become 2


- i%=j=(j=5)%(i=3) -%26gt; i%= 2 -%26gt; i = 3 % 2 = 1
Reply:1...assume x=1 i=2 j=3 k=4


x-=k*=j/=i%=5;





Let's split this up, it's a simple set of operators:





"X-=" x minus equals: This means subtract a value from X. Example is X-=5: This means subtract 5 from whatever the value of X is.





"K*=" K multiplied equals - as above, only with multiplication.





"J/=" J F-Slash equals - As above only with division





"i%=" I Percent Equals - Modulus. Show the remainder after a division. Example 10%3=1 (keep removing a value until a lesser value is left)





2


You can use the same as previous - there's a new symbol here though:





"++" - Increment by 1. If the "++" is before a variable, we PREINCREMENT, otherwise we POSTINCREMENT, this determines the time to increment based on other operations.





3


There's nothing new here.





Take all of these apart using this explanation, and you should have no problems. There's also a website:





http://www.cplusplus.com





for help.





Run it through a compiler yourself, and use printf or cout statements to display variable values, it's an important part of programming to do this "lazy" debugging.








2... x=5 i=4 j=3 k=2


x*=1 + (i%= 1+(j/=-1+ ++k)





3....i=j=10;


i%=j=(j=5)%(i=3);