Hi All, I find some difference in auto increment operator precedence between C and Perl. As almost say that for Perl, the operators derived from C has the same precedence, this is somewhat different. Here is the sample code: Perl: 
#!/usr/local/bin/perl -W
$i = 2;
$j = 10;
$k = $i++ - $i++;
print $k,"\n";
$i = 2;
$j = 10;
$k = ++$i - ++$i;
print $k,"\n";
$i = 2;
$j = 10;
$k = $i++ + $i++;
print $k,"\n";
$i = 2;
$j = 10;
$k = ++$i + ++$i;
print $k,"\n";

which outputs:
-1
0
5
8

[download]
and the equivalent C program: 
#include <stdio.h>
main()
{
int i,j,k;
i = 2;
j = 10;
k = i++ - i++;
printf("i=%d, j = %d, k = %d\n",i,j,k);

i = 2;
j = 10;
k = ++i - ++i;
printf("i=%d, j = %d, k = %d\n",i,j,k);

i = 2;
j = 10;
k = i++ + i++;
printf("i=%d, j = %d, k = %d\n",i,j,k);

i = 2;
j = 10;
k = ++i + ++i;
printf("i=%d, j = %d, k = %d\n",i,j,k);
}

which outputs:
k = 0
k = 0
k = 4
k = 8

[download]
as you can observe the difference in outputs. Could anyone help me explaining whats happening here . why is the output different different especially with post-increment operator? Thanks,

In reply to Autoincrement operator precedence difference between C and Perl by Mbk

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