Increment/decrement question

[Kevin Walzer]

This code:

#include <stdio.h>

int main(void) {

int n1, n2; //two integers

n1 = 1;
n2 = 1;

printf("At first, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1++;
printf("After n2 = n1++, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1--;

printf("After n2 = n1--, n1 is %d, n2 is %d.\n\n", n1, n2);

return 0;
}

yields this output:

At first, n1 is 1, n2 is 1.
After n2 = n1++, n1 is 2, n2 is 1.
After n2 = n1--, n1 is 1, n2 is 2.
Question:

I would expect the statement:

n2 = n1++

to be equivalent to

n2=(n1 +1),

which, given that n1 = 1, should return a value of 2.

So why is n1's value 2 but n2's value is 1?

[cipher]

On 15 Okt., 16:44, Kevin Walzer <[e-mail]:

This code:

#include <stdio.h>

int main(void) {

int n1, n2; //two integers

n1 = 1;
n2 = 1;

printf("At first, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1++;

printf("After n2 = n1++, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1--;

printf("After n2 = n1--, n1 is %d, n2 is %d.\n\n", n1, n2);

return 0;
}

yields this output:

At first, n1 is 1, n2 is 1.
After n2 = n1++, n1 is 2, n2 is 1.
After n2 = n1--, n1 is 1, n2 is 2.

Question:

I would expect the statement:

n2 = n1++

to be equivalent to

n2=(n1 +1),

which, given that n1 = 1, should return a value of 2.

So why is n1's value 2 but n2's value is 1?

Hi Kevin,

n1++ and n1-- are called post-increment / post-decrement operators. If
you write
n2 = n1++, then value of n1 gets incremented / decremented *after*
assigning it's value to n2.
++n1 and --n1 is also possible, but with a little difference:
n2 = ++n1
first increments n1 and then assigns the new value of n1 to n2 (pre-
increment / pre-decrement operators).

Greetings,

Markus

[jameskuyper]

Kevin Walzer wrote:
....

I would expect the statement:

n2 = n1++

to be equivalent to

n2=(n1 +1),

Re-read your textbook, and try to understand the difference between n1+
+ and ++n1. Once you understand that difference, the behavior of your
program will be much clearer.

[Kevin Walzer]

Kevin Walzer wrote:

This code:

<snip>

Got it--thanks to all for the quick correction.

[Richard Heathfield]

husterk said:

<snip>

Try putting parentheses around the increment and decrement operations
and that should fix your code.

You might want to try compiling that before asserting it. :-)

example:
n2 = n1++; results in n2 = n1 then n1 is incremented.
n2 = ++n1; results in n1 is incremented then n2 = n1.
n2 = (n1++); results in n1 is incremented then n2 = n1.

No, n2 = (n1++); is equivalent to n2 = n1++;

Run this program:

#include <stdio.h>

int main(void)
{
int n1 = 1;
int n2 = 1;
n2 = (n1++);
printf("n1 = %d, n2 = %d\n", n1, n2);
return 0;
}

If you are right, this will print n1 = 2, n2 = 2

But it doesn't.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Willem]

Kenneth wrote:
) For example, it's perfectly reasonable for the compiler to generate
) something like this pseudo-code:
)
) LOAD R1,n1 ; Get the current n1 value
) INCR n1 ; Increment n1
) STOR R1,n2 ; Set n2 to the pre-incremented value

Compilers can do a lot worse. I've seen an optimizer effectively translate
n2 = n1++;
to
n2 = (++n1)-1;

Or in pseudo-assembly:

INCR n1
MOVE n2, (n1 -1)

The second instruction is one that assigns a register to another register,
while at the same time adding something to it. 'LEAL' on x86, iirc.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

[Barry Schwarz]

On Mon, 15 Oct 2007 10:44:31 -0400, Kevin Walzer <[e-mail]>
wrote:

>This code:

#include <stdio.h>

int main(void) {

int n1, n2; //two integers

n1 = 1;
n2 = 1;

printf("At first, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1++;
printf("After n2 = n1++, n1 is %d, n2 is %d.\n", n1, n2);

n2 = n1--;

printf("After n2 = n1--, n1 is %d, n2 is %d.\n\n", n1, n2);

return 0;
}

yields this output:

At first, n1 is 1, n2 is 1.
After n2 = n1++, n1 is 2, n2 is 1.
After n2 = n1--, n1 is 1, n2 is 2.
Question:

I would expect the statement:

n2 = n1++

to be equivalent to

n2=(n1 +1),

It is actually closer to
n2 = n1;
n1 += 1;

To be more precise, n1++ is an expression that evaluates to the
original (unincremented) value of n1. The ++ post-increment operator
also has a side effect of incrementing n1. There is no specified
temporal relationship between the two; either can occur first.

>
which, given that n1 = 1, should return a value of 2.

So why is n1's value 2 but n2's value is 1?

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