This is correct, reference this in C if you don't believe me. Technically it
is because the operator is unary right associative and is 'pre' increment.
Hence, the instruction set generated is (roughly):
INC CNT in place ; CNT is now 4
INC CNT in place ; CNT is now 5
LD CNT ; 5
ADD CNT ; Add 5
And the answer is 10. I presume that you can derive the sequence for your
second example from this.
Now contrast that with 'post' increment.
Jim
#include <stdio.h>
int
main(int cnt, char * argc[])
{
int c, c1;
c = 3;
printf("Expr is %d\n, c is %d\n\n", ++c + ++c, c);
c = 3;
printf("Expr is %d\n, c is %d\n\n", c++ + c++, c);
}
From: [email protected] [mailto:[email protected]] On Behalf Of [
Xze ]
Sent: Wednesday, September 01, 2010 5:35 AM
To: [email protected]
Subject: Pre-increment
Hi all!
My question pertains to the way that jbase treats pre-increments
Consider the following routine:
0004 PROGRAM MY.MATH
0005
0006 CNT = 3
0007 RES = ++CNT + ++CNT
0008
0009 PRINT 'RES = ':RES
0010 RETURN
0011 END
The output result is RES = 10
if i change line 007 to RES = ++CNT + ++CNT + ++CNT,
then the output result is RES = 16
Can anybody explain this behaviour? (I expected the first case to output RES
= 9 and the second RES = 15)
My system configuration:
OS : AIX 5.3
jbase : Major 5.0 , Minor 20 , Patch 0364 (Change 85159)
Regards
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