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what is the execution sequence of this code?
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Mr, Ancient Dragon
take it easy, I don't mean that, I just wanted to know how can I make it with myself.
take it easy, I don't mean that, I just wanted to know how can I make it with myself.
I'd give my right arm to Improve my english
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Join Date: Aug 2005
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Like I said before, do whats in the inner-most parentheses first, then work yourself out. Once you have removed all parentheses, work the statement from left to right.
For example:
f = b * c;
now the equation will look like this:
g = f / d; ===>
For example:
a - ((b * c) / d) + e; First calculate b*c, then divide that result by d, finally subtract that result from the value of a and add e.f = b * c;
now the equation will look like this:
a - (f / d) + e g = f / d; ===>
a - g + e Last edited by Ancient Dragon; Dec 4th, 2008 at 9:00 am.
Don't PM me with questions -- you might get a nasty PM in response. If you have a question then post it in one of the forums.
Anciant Dragon has given the answer of question you've asked, you asked for What is the Execution Sequence you didn't ask How to find out the execution sequence.
Let me teach you how to find it out.
when the above funciton is called, Compilers will push the PTR Puffer on stack then push the Local variables i and J on the stack.
now the interesting part.
assignement operator are evaluatted from right to left.
so the above code can be broken as follows
you will get some character in that case aren't you?
assign that character to lvalue which is
lets say right value calculated as 'C' you assign it to left handside so the right handside become 'C' as well which is not equal to '\0'
isn't it
always remember one thing there is still a world ahead
a more better explanation can be provided to you but the thing is that you don't know the assembly and compilers. Sometimes experts like Anciant Dragon doesn't explain the very minor details because either they are irrelevant to you or more advance so that you can't grasp them. be patient just take a hint and dig it out yourself one day you'll learn why Anciant doesn't tell you the complete story.
Let me teach you how to find it out.
cpp Syntax (Toggle Plain Text)
when the above funciton is called, Compilers will push the PTR Puffer on stack then push the Local variables i and J on the stack.
now the interesting part.
while((*(pbuffer+i) = *(pbuffer + j++)) != 0) assignement operator are evaluatted from right to left.
so the above code can be broken as follows
*(pbuffer + j++) the above expression is evaluated as follows C++ Syntax (Toggle Plain Text)
you will get some character in that case aren't you?
assign that character to lvalue which is
*(pbuffer + i) you can break the above statement now, lets say right value calculated as 'C' you assign it to left handside so the right handside become 'C' as well which is not equal to '\0'
isn't it
always remember one thing there is still a world ahead
a more better explanation can be provided to you but the thing is that you don't know the assembly and compilers. Sometimes experts like Anciant Dragon doesn't explain the very minor details because either they are irrelevant to you or more advance so that you can't grasp them. be patient just take a hint and dig it out yourself one day you'll learn why Anciant doesn't tell you the complete story.
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