$a places scalar $a on the stack (the variable $a, not its value 10).

$a-- places a scalar containing the original value of $a on the stack, so $a == $a-- compares $a with the original value of $a. They will never be equal.

--$a places scalar $a on the stack, so $a == --$a compares $a and $a. They will always be equal.

C behaves differently since it places values on the stack or in the registers. You can replicate the behaviour using pointers in C or references in C++. 

#include <stdio.h>
#include <stdlib.h>

int& postdec(int& a) {
   int* rv_ptr = (int*)malloc(sizeof(int));
   *rv_ptr = a;
   --a;
   return *rv_ptr;
}

int& predec(int& a) {
   --a;
   return a;
}

bool& eq(int& a, int& b) {
   bool* rv_ptr = (bool*)malloc(sizeof(bool));
   *rv_ptr = a == b;
   return *rv_ptr;
}

int main() {
   int a = 10;
   if (eq(a, postdec(a))) {
      printf("T1\n");
   }

   a = 10;
   if (eq(a, predec(a))) {
      printf("T2\n");
   }
}

[download]

Update: Removed overspecification identified by BrowserUk.

In reply to Re: Help me to understand increment / decrement operator behavior in perl by ikegami
in thread Help me to understand increment / decrement operator behavior in perl by sam_bakki

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