Re: Function call evaluation order


"Cheng" <chengwuchew@xxxxxxxxxxx> escreveu na mensagem
news:u8X%23alhTGHA.4340@xxxxxxxxxxxxxxxxxxxxxxx

Hi folks,

I need help to explain the behavior with the code below. I would like to
chain up function call into a single line, while the order of the function
call is correct, I cannot be certain with order the expression within the
function call is evaluated.

Woud it be better, then, to write all the function call into seperate
lines?

With regards,
Cheng Wu


Code:

#include <stdio.h>
#include <string>

using namespace std;

void main()
{
char* data[] = {" zero ", " one ", " two ", " three "};
string s;
int i;

i = 0;
s.assign(data[i++]).append(data[i++]).append(data[i++]);
printf("%s\n", s.c_str());

i = 0;
s.assign(data[++i]).append(data[++i]).append(data[++i]);
printf("%s\n", s.c_str());

i = 0;
s.assign(data[i++]);
s.append(data[i++]);
s.append(data[i++]);
printf("%s\n", s.c_str());
}


Output:

zero zero zero
three three three
zero one two


It seems obvious to me that assign() will be performed before the first
append(), and this one before the second one. Obvious because member
selecion operator "." has the higher precedence and it's evaluated left to
right.

At the same time, in the first s.assign()... the routine uses post-increment
operators, so "i" will be incremented AFTER the expression is evaluated. The
same happens in the second s.assing().... but with pre-increment.

The last sentences are evaluated separately, so "s.assing(data[i++])" is the
same as "s.assign(data[i]); i++;".

Take a look at the precedence table in any good C book (or MSDN)... you'll
notice that . has the higher precedence... followed by ->, followed by [],
followed by funcion call "()"...

Hope it helped!

[]s
Fred


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