C中的序列点和副作用

Question

In this C-FAQ it is give about sequence point ;

The Standard states that;
Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored.

In the examples

i = i++;
a[i] = i++;

it is clear from first sentence of the statement that these examples are results in undefined behavior .
In explaining second sentence of the statement it is said that;

second sentence says: if an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written. This rule effectively constrains legal expressions to those in which the accesses demonstrably precede the modification. For example, the old standby

 i = i + 1 

is allowed, because the access of i is used to determine i's final value. The example

a[i] = i++

is disallowed because one of the accesses of i (the one in a[i]) has nothing to do with the value which ends up being stored in i (which happens over in i++), and so there's no good way to define.

My questions are;
1.What does it mean by, if an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written. ?

2.what does it mean by, The example a[i] = i++ is disallowed because one of the accesses of i (the one in a[i]) has nothing to do with the value which ends up being stored in i (which happens over in i++)
Could someone explain it in some easy way?

Answer 1

My question are; 1.What does it mean by, if an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written.?

With a sub-expression like i++ , i is written to. Moreover, assignment is an expression, so in i = 2 , i is written to. It may not be immediately obvious that a = b is an expression, but it is. This is why you can do things like a = b = c , which is good, and if (a = b) which is less good.

So what it is saying is that if you write to i , with = , or pre- or post- increment, then any access to i must be as part of the calculation of the new value of i. However, and this is important, the only thing involved in the calculation of pre and post increment is the value of i at the start of the statement .

2.what does it mean by, The example a[i] = i++ is disallowed because one of the accesses of i (the one in a[i]) has nothing to do with the value which ends up being stored in i (which happens over in i++)

Precisely what it says. When you access i in a[i] it is not part of the calculation of the new value of i that results from i++ .

Could someone explain it in some easy way?

Easy way: Don't use pre or post increment in an expression. always use them in statements by themselves. If you really really must, do NOT use the same variable anywhere else in the entire statement .

Answer 2

This explanation is weird and I can't make sense of it.

The true explanation is that the expression i++ has a side-effect (incrementing i ) that can be applied anytime after this particular i has been evaluated.

Since outside of sequence points , the C language doesn't guarantee the evaluation order or the time when post-incrementations are applied (likely for performance reasons), three things can happen in the second example (let's assume i = 5 before the line) :

• The leftmost i (the one in a[i] ) is evaluated first, to calculate the address of a[i] for storage. Then the rightmost i is evaluated, then the post-incrementation is applied: The line is performs a[5] = 5; i=6; a[5] = 5; i=6;
• The rightmost i is evaluated first, then the leftmost one, and then i the post-incrementation is applied when all is said and done: In this case the effects are identical to the above case.
• The rightmost i is evaluated first, the post-incrementation applied immediately, and then the leftmost i is evaluated to calculate the address of a[i] for storage. This time the effect is a[6] = 5; i=6 a[6] = 5; i=6 .

The choice of outcome depends not only on the choice of compiler, but on the compiler's setting: Microsoft Visual C++ can give different outcomes depending on whether you compile in Debug or Release.

Answer 3

Finally I got an explanation on SO about this point . After reading it and FAQ I concluded that;

1.The last sentence

Furthermore, the prior value shall be accessed only to determine the value to be stored

would be like this;

Furthermore, the prior value of an object shall be accessed only to determine the modified/new value( of same object ) to be stored.

As it is clear by the example

 int i = 1, j, a[5];    
 i = i + 1;
 j = i + 1;
 a[i] = i; 

in case of expression i = i + 1 prior value (which is 1 here) of i (in RHS) is accessed to determine the value of i to be stored and this is what the statement

if an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written .

says.
While in case of j = i + 1 and a[i] = i , the accessed value of i is just value not prior value as no where i is modified in these statements.

2.The second question can be explained as;
In case of expression a[i] = i++ or a[i++] = i , first sentence of above statement

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression.

get failed as i is modified only once between two consecutive sequence point . And that's why we need second sentence.
Both of these examples are disallowed in C because the prior value of i accessed two times ie, i++ itself access prior value of i in the expression to modify it and hence other access of prior value / value of i is needless as it is not accessed to determine the modified value to be stored.