Hierarchy of arithmetic operations in C.

Question

When I type my code like given below :

int a=10,b,c,d,e;
c= a++;
d = ++a;
e = ++a;
printf("value of c=%d, d =%d, e=%d",c,d,e);

it gives me an output like c =10 , d= 12, e=13 and when we add these values,i.e 10+12+13 becomes 35, but when I code it like :

b = a++ + ++a + ++a;
printf("value of b=%d" ,b);

it gives me output 36.

Can somebody describe what's the process behind this code and why the output of codes are different? Thank you!

http://stackoverflow.com/questions/949433/could-anyone-explain-these-undefined-behaviors-i-i-i-i-i-etc — haccks, Jun 29 '13 at 10:49

haccks · Answer 1 · 2013-06-29T11:05:44.253

0

int a=10,b,c,d,e;
c= a++;
d = ++a;
e = ++a;
printf("value of c=%d, d =%d, e=%d",c,d,e);

In statement c = a++ , value of a is used first(which is 10) and then incremented to11.
Statement d = ++a first increment a (which is 12 now) and then use its value to print in printf() statement.
Same for e = ++a.

Your second snippet

b = a++ + ++a + ++a;
printf("value of b=%d" ,b);

results in Undefined Behavior(http://en.wikipedia.org/wiki/Undefined_behavior)

edited Jun 29 '13 at 11:05

answered Jun 29 '13 at 10:51

haccks

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thanks for clarify all the things .....your answer is correct too. – Android developer Jun 29 '13 at 11:05
I am glad that it helped you. See FAQ http://stackoverflow.com/about – haccks Jun 29 '13 at 11:08

score 0 · Accepted Answer · answered Jun 29 '13 at 10:52

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Delicious Undefined Behaviour, the order of evaluation of the operands of + (and many others) is left to the implementation. So it's not even always 36 for the second case.

answered Jun 29 '13 at 10:52

Kninnug

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score -1 · Answer 3 · answered Jun 29 '13 at 10:51

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Difference between them is second expression has not ended with a++ and yet to add some other values thats why a++ is 11, not 10

answered Jun 29 '13 at 10:51

Kwag.degi

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Hierarchy of arithmetic operations in C.

3 Answers3