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C/weird syntax a+++b
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Question
why if I write the following code:
int a=3, b=5, i;
i=a+++b;
printf ("%d",i); // 8 is printed
but if I write:
int a=3, b=5;
a=a+++b;
printf ("%d",a_; // 9 is printed
?
thanks,
nir.
It depends on the way computer interprets the code.
It can interpret it as (a++ + b) or (a + ++b).
So, in the first case it will result in 8 and in second case it will be 9.
Look at the assembly listing and then try to analyze.
The same code may give different result on another compiler!
-Narendra
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| Comment | thank you alot for replying fast, i will check it out,thanks, | ||
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Narendra
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I can answer questions in C related to programming, data structures, pointers and file manipulation. I use Solaris for doing C code and if you have questions related to C programming on Solaris, I will be able to help better.
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