i=i++ + i--; 怎么算

wangzhonnew · 发表于 2009-1-25 03:20

(taoewang@panipuri) /home/taoewang/test $ more test.c
#include <stdio.h>
int main()
{
            int i=5;
            i=i++ + --i;
            printf("%d\n", i);
            return 0;
}
(taoewang@panipuri) /home/taoewang/test $ xlc test.c
(taoewang@panipuri) /home/taoewang/test $ ./a.out
10

still IBM xlc v9.0

.main:                               # 0x00000000 (H.10.NO_SYMBOL)
      mfspr r0,LR
      st    r31,-4(SP)
      st    r0,8(SP)
      stu    SP,-80(SP)
      l    r31,T.18.__STATIC(RTOC)
      cal    r4,5(r0)
      st    r4,64(SP)
      cal    r3,-1(r4)
      st    r4,68(SP)
      st    r3,64(SP)
      l    r4,68(SP)
      cax    r3,r3,r4
      st    r3,64(SP)
      cal    r4,1(r3)
      st    r4,64(SP)
      oril r3,r31,0x0000
      bl    .printf{PR}
      oril r0,r0,0x0000
      cal    r3,0(r0)
      l    r0,88(SP)
      mtspr LR,r0
      cal    SP,80(SP)
      l    r31,-4(SP)
      bcr    BO_ALWAYS,CR0_LT
      .long 0x00000000

so it's
itemp=i-1
i=i+itemp
i=i+1

wangzhonnew · 发表于 2009-1-25 03:27

by using gcc, we have:
(taoewang@panipuri) /home/taoewang/test $ more test.c
#include <stdio.h>
int main()
{
            int i=5;
            i=i++ + --i;
            printf("%d\n", i);
            return 0;
}
(taoewang@panipuri) /home/taoewang/test $ gcc -S test.c
(taoewang@panipuri) /home/taoewang/test $ ./a.out
9
(taoewang@panipuri) /home/taoewang/test $ gcc -v
Reading specs from /afs/torolab.ibm.com/common/progs/gcc-3.4.3/aix51/usr/local/bin/../lib/gcc-lib/powerpc-ibm-aix5.1.0.0/3.3.4/specs
Configured with: ../gcc-3.3.4/configure --disable-nls
Thread model: aix
gcc version 3.3.4

.main:
      mflr 0
      stw 31,-4(1)
      stw 0,8(1)
      stwu 1,-80(1)
      mr 31,1
      li 0,5
      stw 0,56(31)
      lwz 11,56(31)
      lwz 9,56(31)
      addi 0,9,-1
      mr 9,0
      stw 9,56(31)
      mr 0,11
      add 0,0,9
      addi 11,11,1
      stw 11,56(31)
      stw 0,56(31)
      lwz 3,LC..1(2)
      lwz 4,56(31)
      bl .printf
      nop
      li 0,0
      mr 3,0
      lwz 1,0(1)
      lwz 0,8(1)
      mtlr 0
      lwz 31,-4(1)
      blr

looks like
i=i-1
i=i+i
i++

wangzhonnew · 发表于 2009-1-25 03:34

for java:
(taoewang@panipuri) /home/taoewang/test $ more test.java
import java.io.*;
class test
{
      public static void main( String args[] )
      {
            int i=5;
            i=i++ + i--;
            System.out.println(i);
      }
}
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/bin/javac test.java
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/jre/bin/java test
11
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/jre/bin/java -version
java version "1.5.0"
Java(TM) 2 Runtime Environment, Standard Edition (build pap64devifx-20060310a (SR1))
IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 AIX ppc64-64 j9vmap6423ifx-20060310a (JIT enabled)
J9VM - 20060220_05389_BHdSMr
JIT  - 20060220_2133_r8
GC - 20060214_AA)
JCL  - 20060222a

it's
i=i+1
i=i+i
i=i-1

================
(taoewang@panipuri) /home/taoewang/test $ more test.java
import java.io.*;
class test
{
      public static void main( String args[] )
      {
            int i=5;
            i=i++ + --i;
            System.out.println(i);
      }
}
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/bin/javac test.java
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/jre/bin/java test
10
(taoewang@panipuri) /home/taoewang/test $ /home/taoewang/sqllib/java/jdk64/jre/bin/java -version
java version "1.5.0"
Java(TM) 2 Runtime Environment, Standard Edition (build pap64devifx-20060310a (SR1))
IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 AIX ppc64-64 j9vmap6423ifx-20060310a (JIT enabled)
J9VM - 20060220_05389_BHdSMr
JIT  - 20060220_2133_r8
GC - 20060214_AA)
JCL  - 20060222a

justforregister · 发表于 2009-1-25 06:46

牛年现牛人

erwinchn · 发表于 2009-1-25 12:31

My test
=======================

C:\J\test\basic>type Oper.java
public class Oper{
      public static void main(String[] args){
            int i=5;
            i=i++ + i--;
            System.out.println(i);
            i=5;
            i=i++ - i--;
            System.out.println(i);
            i=5;
            i=i++ * i--;
            System.out.println(i);
      }
}
C:\J\test\basic>javac Oper.java
C:\J\test\basic>java Oper
11
-1
30
C:\J\test\basic>

My understanding:
=========================
i++ is evaluated as 5
i-- is evaluated as 6(because i++ take into effect)
so the test I did are evaluated as:
5 + 6 = 11
5 - 6 = -1
5 * 6 = 30

so statement i=i++ + i--
=====================
there are 4 operators in this statement:
1) =
2) ++
3) +
4) --
operator precedence is:
2)++
4)--
3)=
1)=

so the statement is evaluated in the following order:
i++: i become 6, but 5 participate in addition action
i--: i become 5 again, but 6 participate in addition action
+: 5 + 6 = 11, but before = is evaluated, i = 5(because of i-- in last step)
=: i = 11 (assign 5 + 6 to i)

wanglei2924 · 发表于 2009-1-26 12:27

wangzhonnew · 发表于 2009-1-27 01:08

有没有人能知道怎么把java里面的byte code给翻译出来吗？如果能像看汇编一样看byte code，那这个逻辑不就简单了么

erwinchn · 发表于 2009-1-27 06:09

Source code:

public class Oper{
      public static void main(String[] args){
            int i=5;
            i=i++ + i--;
            System.out.println(i);
      }
}

================
from Bytecode viewer:

0 iconst_5
1 istore_1
2 iload_1
3 iinc 1 by 1
6 iload_1
7 iinc 1 by 255
10 iadd
11 istore_1
12 getstatic #2 <java/lang/System.out>
15 iload_1
16 invokevirtual #3 <java/io/PrintStream.println>
19 return

step 2 loads local variable i (=5) pushes 5 on the operand stack,
step 3 changes local variable i from 5 to 6,
step 6 loads local variable i (=6) pushes it on the operand stack,
step 7 changes local variable i from 6 to 5,
step 10 pops add operand and the next 2 stacks(5 and 6) and add them up to 11, pushes 11 back to stack
step 11 pop top stack up and stores it to local variable i (i was changed from 5 to 11)

so i = 11 in the end

freedom2k · 发表于 2009-1-27 21:35

有啥麻烦得，这东西就看符号在前还是在后
i++就是先使用i得值再把i得值加1，比如
i=i++
那就是先i=i然后再i=i+1，这样得到得
++i就是先i=i+1，然后再i=i

int i=5;
i=i++ + i--;
i++;
说这一个例子吧
i=i（此时i=5）
i=5+（i+1）（此时i=6）
i=5+6=11
然后i--也就是i=10
最后i++得到11
没去写程序验证，应该是上面这样得过程，符号放在后面就是先使用变量得值，然后再+1（或者-1），符号在后就是先+1（或-1）再使用变量得值

erwinchn · 发表于 2009-1-27 22:53

19楼，你的例子答案应是１２，不是１１