prefix increment operator and side-effects

[subramanian100in]

Stroustrup in his book "The C++ Programming Language - Third Edition
(NOT the special third edition)", mentions the following in page 125
in section "6.2.5 Increment and Decrement".
"By definition, ++lvalue means lvalue += 1, which again means lvalue =
lvalue + 1 provided lvalue has no side-effects".

Here I am unable to understand why Stroustrup has mentioned "provided
lvalue has no side-effects". Isn't the expression ++lvalue the same as
lvalue = lvalue + 1 always ? Kindly clarify with an example.

Thanks
V.Subramanian

 

[Niels Dekker - no reply address]

Stroustrup in his book "The C++ Programming Language - Third Edition
(NOT the special third edition)", mentions the following in page 125
in section "6.2.5 Increment and Decrement".
"By definition, ++lvalue means lvalue += 1, which again means lvalue =
lvalue + 1 provided lvalue has no side-effects".

Here I am unable to understand why Stroustrup has mentioned "provided
lvalue has no side-effects".

For example, suppose you have an object v, of a type similar to
std::vector<int>. You would expect ++v[0] to be equivalent to v[0] = v[0] +
1. But clearly that assumes that doing v[0] doesn't have some side-effect,
like doing std::cout << "Hey, operator[] is called".

Does that answer your question?

 

[James Kanze]

Stroustrup in his book "The C++ Programming Language - Third
Edition (NOT the special third edition)", mentions the
following in page 125 in section "6.2.5 Increment and
Decrement". "By definition, ++lvalue means lvalue += 1, which
again means lvalue = lvalue + 1 provided lvalue has no
side-effects".

Here I am unable to understand why Stroustrup has mentioned
"provided lvalue has no side-effects". Isn't the expression
++lvalue the same as lvalue = lvalue + 1 always ? Kindly
clarify with an example.

int* p;
++(*p ++); // which is NOT the same as:
*p ++ = *p ++ + 1; // undefined behavior...