Thread: quick syntax question

In effect, yes. Alternatively:

Code:

*(s->top) = c;
(s->top)++;

Originally Posted by laserlight

In effect, yes. Alternatively:

Code:

*(s->top) = c;
(s->top)++;

Ixnay. The post-increment will occur before the assignment. If the point is to add one to the value of s->top, the ++ does nothing in the end.

If the point is to make s->top point to what was the equivalent of s->top[1], then that space better be part of s, because it will be set to the value of c.

Originally Posted by MK27

Ixnay. The post-increment will occur before the assignment.

The post-increment does not happen until after s->top has been fully evaluated. Whether it occurs before or after the assignment does not change the result.

The line of code stores 'c' into the location pointed to by s->top, then increments s->top.

Originally Posted by brewbuck

The post-increment does not happen until after s->top has been fully evaluated. Whether it occurs before or after the assignment does not change the result.

The line of code stores 'c' into the location pointed to by s->top, then increments s->top.

Sorry -- I was looking at this order of precedence table:

C Operator Precedence Table

However, it's obviously wrong as even the simple:

Code:

int a = 5, b = a++;

demonstrates (which if it were my code, I'd have been writing it and noticed )

Originally Posted by MK27

Sorry -- I was looking at this order of precedence table:

C Operator Precedence Table

However, it's obviously wrong as even the simple:

Code:

int a = 5, b = a++;

demonstrates (which if it were my code, I'd have been writing it and noticed )

Precedence controls order of binding, not order of operations. It specifies which operations will occur to which expressions, not when they will occur. The order of operations is controlled by associativity and sequence points. Operators in different precedence classes are treated independently when determining associativity.

On top of all that, the compiler can evaluate the expression any way it likes, so long as the result is the same as if it had done it in the order defined by the language.

The code in question looks like it's pushing something onto a stack.

Well that is very deceptive then isn't it

I guess the best means of resolving uncertainty is still actual experimentation.

Originally Posted by MK27

Well that is very deceptive then isn't it

In most cases, binding order completely defines order of operations. It's just that it doesn't cover every case. Even expressions that don't involve increment/decrement operators can have ambiguous evaluation order:

Code:

x = a * b + c * d;

You can't tell whether a * b will evaluate before, or after, c * d. Left-to-right associativity doesn't factor in, because of the intervening addition. Even parenthesizing the expression doesn't remove the ambiguity. The only way to force a particular order is to insert a sequence point by splitting across more than one statement.