问题
Following is the test code:
int main()
{
int a = 3;
int b = 4;
a = a + b - (b = a);
cout << \"a :\" << a << \" \" << \"b :\" << b << \"\\n\";
return 0;
}
Compiling this gives the following warning:
> $ g++ -Wall -o test test.cpp test.cpp: In function ‘int main()’:
> test.cpp:11:21: warning: operation on ‘b’ may be undefined
> [-Wsequence-point]
Why can the operation be undefined?
According to my understanding, first the subexpression (b = a) should be evaluated because of higher precedence of (), thus setting b = a. Then, since \'+\' and \'-\' have same precedence, the expression would be evaluated left-associatively. Thus, a + b should be evaluated next, and finally the result of (b = a) should be subtracted from a + b. I can\'t see any sequence-point rule being violated here.
回答1:
There is a difference between an expression being evaluated and completing its side effects.
The b = a assignment expression will be evaluated ahead of subtraction due to higher precedence of the parentheses. It will provide the value of a as the result of the evaluation. The writing of that value into b, however, may not complete until the next sequence point, which in this case is the end of the full expression. The end result of the overall expression is therefore undefined, because the subtraction may take the value of b before or after the assignment.
回答2:
In C++, subexpressions in arithmetic expressions do not have temporal ordering.
a = x + y;
Is x evaluated first, or y? The compiler can choose either, or it can choose something completely different. The order of evaluation is not the same thing as operator precedence: operator precedence is strictly defined, and order of evaluation is only defined to the granularity that your program has sequence points.
In fact, on some architectures it is possible to emit code that evaluates both x and y at the same time -- for example, VLIW architectures.
回答3:
To solve it separate them in two different statements.
PS: Don't forget that humans may make mistakes performing arithmetic operations. Therefore is better to make the operations clearer by separating them in different statements. I hope I helped.
int main()
{
int a = 3;
int b = 4;
/* Two different Statements*/
b = a;
/* or a = a + b - a */
a = a + b - b;
cout<<"a :"<<a<<" "<<"b :"<<b<<"\n";
return 0;
}
回答4:
a = b + a - a;
is just written as
a = b + a - (b = a)------>> (exp 1)
The following three results same as (exp 1)
a = (b + a - (b = a));
a = ((b + a) - (b = a));
a = (b + a) - (b = a);
Observations +, - operators has got same precedence and also left to right associativity Hence 'b+a' gets executed first and then 'a' value gets assigned to 'b' before subtraction
Now observe the following When a = 10 and b = 20;
a = (b = a) - b + a; =======> a = 10; b = 10
a = ((b = a) - b + a); =======> a = 10; b = 10
a = ((b = a) - (b + a)); =======> a = -10; b = 10
From the above expressions its clear that even if innermost parenthesis gets executed first the associativity is followed first and then the precedence
Note:
To avoid confusion between the precedence of outer and inner parenthesis
Consider the following expression
a = (b + a - (b = a)) =====> Actual Result => a = 20, b = 10;
would have been a = 10, b = 10; (if precedence is primary when compared to associativity)
Thus by the above example we can say that associativity is primary when compared to precedence
来源：https://stackoverflow.com/questions/13317684/why-does-the-expression-a-a-b-b-a-give-a-sequence-point-warning-in-c