value of ++a+++a when a=8,how comuter calculate this?

[SMGHOSH]

int a=8,b;
b=++a+++a;

please tell me how the value of b becomes 20 ?

[weaknessforcats]

Who knows.

The rules allow that a variable can be changed only once in a statement.

Because of that rule, the compiler can evaluate the statement in any order. Right to left, or left to right and get the correct answer. Violate the rule, as this code does, and you get garbage (er, indeterminate results).

[horace1]

++ has a higher precedence than + therefor the expressions is

b=++a + ++a;

a is incremented twice (a then = 10) then a+a calculated and assigned to b which gives b=20

what about?

b=++a + ++a + ++a;

[weaknessforcats]

Did you read my post #2?

[donbock]

Some past warnings regarding undefined behavior...
Undefined behavior means anything can happen.
Don't try to figure out what the undefined behavior happens to be for your compiler.
Seriously, don't try to figure out what will happen.

[horace1]

The critical thing to remember is do not use a variable more than once in an expression if one (or more) of the references has one of ++ or -- operators attached to it. The standard does not specify the order in which the operands of an operator are evaluated and there is no guarantee when an affected variable will change its value.
in the case of

b=++a + ++a;

it is not critical, the ++a operands will be evaluated before the + operator so the result will be a=10 and b=20

consider

b=++a + ++a + ++a;

the + operator associates from left to right so it evaluated

b=(++a + ++a) + ++a;

however, when are the ++a operands evaluated?
1. the ++a operands of (++a + ++a) evaluated and the result added then the final ++a evaluated and added? the result is a=11 and b=31
2. all the ++a operands evaluated and the then the + operations executed giving a=11 and b= 33