what is the behavior of a statement logically operated with 0?

shaktikanta

ex- if( 0 < -1 ) I remember "C" compiler takes bits data of -1 to signed format and returns true for the above statement. Whats behavior in C++. Is the behavior same/different.Any updated standard(C/C++) brought changes to this behavior.Please answer.

CPallini

In C a zero value is assigned to (0 < -1) expression, hence, in the following code

if ( 0 < -1)
{
k = 5;
}
else
{
k = 10;
}

the statement k = 10; is executed. In C++ the esame expression is evaluated as false and (like happened in the C program) the statement k = 10; is executed.

Veni, vidi, vici.

shaktikanta

I tried on gcc compiler, the statement 0 < -1 always false.please help.

CPallini

Of course. I made a blunder. :( I am very sorry. Please see my (hopefully) fixed answer.

Veni, vidi, vici.

shaktikanta

Oh.. fine..But I read somewhere that when a signed value is compared with unsigned, signed converted to unsigned first. 0 is by default unsigned, hence signed version of -1 is greater than 0. May be old compiler showing this behavior.Do u have any idea?

CPallini

Oh, now I see. You are looking for an example of 'idiosyncrasy of integer promotions' (see, for instance, , however

shaktikanta wrote:

0 is by default unsigned

is a wrong assumption. It is signed, by default. You have to write:

#include <stdio.h>
int main()
{
int k;
if ( 0u < -1)
{
k=5;
}
else
{
k=10;
}
printf("%d\n", k);
return 0;
}

Veni, vidi, vici.