Order of evaluation -- clash w/sequentiel expression operator

[Jan Engelhardt]

Hi,
I was told that order of evaluation is unspecified for functions, i.e.

int f = 0;
print_results(modify(&f), modify(&f), modify(&f));

where i.e. modify() increases f by one. In my case w/gcc, it was evaluated from
right-to-left (gcc does a nice stack optimization). Not what I expected though.
Anyway:

Taking a look at the C operator precedence table, I find

, sequential expression left-to-right

at the bottom. So what's wrong here?

--
- Jan Engelhardt

[Ian Woods]

Jan Engelhardt <je*****@linux01.gwdg.de> wrote in
news:Pi*******************************@yvahk01.tjq t.qr:

Hi,
I was told that order of evaluation is unspecified for functions, i.e.

int f = 0;
print_results(modify(&f), modify(&f), modify(&f));

where i.e. modify() increases f by one. In my case w/gcc, it was
evaluated from right-to-left (gcc does a nice stack optimization). Not
what I expected though. Anyway:

Taking a look at the C operator precedence table, I find

, sequential expression left-to-right

at the bottom. So what's wrong here?

From memory, the comma as an operator and a comma in a function invokation
are different things entirely. The order of evaluation of function
arguments is undefined.

Ian Woods

[Christian Bau]

In article <Pi*******************************@yvahk01.tjqt.qr >,
Jan Engelhardt <je*****@linux01.gwdg.de> wrote:

Hi,
I was told that order of evaluation is unspecified for functions, i.e.

int f = 0;
print_results(modify(&f), modify(&f), modify(&f));

where i.e. modify() increases f by one. In my case w/gcc, it was evaluated
from
right-to-left (gcc does a nice stack optimization). Not what I expected
though.
Anyway:

Taking a look at the C operator precedence table, I find

, sequential expression left-to-right

at the bottom. So what's wrong here?

You are confusing "comma expression" and function arguments. If you look
at the syntax of a function call you will find that the arguments are
not "expression"s but "assignment-expression"s. For example, the
comma-exppression x,y,z is _not_ an assignment expression, so f(x,y,z)
has three arguments, and not one.