Procedure to combine the 0x89ABCDEF and 0x76543210 as 0x765432EF

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

/*
Procedure to combine the
0x89ABCDEF and 0x76543210
as 0x765432EF
*/

int main(int argc, char* argv[])
{
int x = 0x89ABCDEF;
int y = 0x76543210;
int z = (x & 0xFF) + (y & ~0xFF);
printf("%X \n",z);
system("PAUSE");
return 0;
}

--
Happy day, happy life!

Procedure to identify the machine is little endian or big endian

#include <stdio.h>
#include <stdlib.h>
/*
 Procedure to identify the machine is
 little endian or big endian
*/

// Using
// Compile it and run it on the target machine which you want to check.

typedef unsigned char* byte_pointer;

int is_little_endian()
{
    int test = 0x0000FFFF;
    int i;
    byte_pointer p = (byte_pointer)&test;
    return p[0] == 0xFF;
}

int main(int argc, char* argv[])
{
    printf("Current machine is%slittle endian \n",is_little_endian()? " ":"not");
    system("PAUSE");
    return 0;
}

--
Happy day, happy life!

msup 微软访谈

ref: http://www.msup.com.cn/msdonwload.aspx

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--
Happy day, happy life!

Think it over……好好想想……

Today we have higher buildings and wider highways,but shorter temperaments and narrower points of view;
今天我们拥有了更高层的楼宇以及更宽阔的公路，但是我们的性情却更为急躁，眼光也更加狭隘；

We spend more,but enjoy less;
我们消耗的更多，享受到的却更少；

We have bigger houses,but smaller famillies;
我们的住房更大了，但我们的家庭却更小了；

We have more compromises,but less time;
我们妥协更多，时间更少；

We have more knowledge,but less judgment;
我们拥有了更多的知识，可判断力却更差了；

We have more medicines,but less health;
我们有了更多的药品，但健康状况却更不如意；

We have multiplied out possessions,but reduced out values;
我们拥有的财富倍增，但其价值却减少了；

We talk much,we love only a little,and we hate too much;
我们说的多了，爱的却少了，我们的仇恨也更多了；

We reached the Moon and came back,but we find it troublesome to cross our own street and meet our neighbors;
我们可以往返月球，但却难以迈出一步去亲近我们的左邻右舍；

We have conquered the uter space,but not our inner space;
我们可以征服外太空，却征服不了我们的内心；

We have highter income,but less morals;
我们的收入增加了，但我们的道德却少了；

These are times with more liberty,but less joy;
我们的时代更加自由了，但我们拥有的快乐时光却越来越少；

We have much more food,but less nutrition;
我们有了更多的食物，但所能得到的营养却越来越少了；

These are the days in which it takes two salaries for each home,but divorces increase;
现在每个家庭都可以有双份收入，但离婚的现象越来越多了；

These are times of finer houses,but more broken homes;
现在的住房越来越精致，但我们也有了更多破碎的家庭；

That's why I propose,that as of today;
这就是我为什么要说，让我们从今天开始；

You do not keep anything for a special occasion.because every day that you live is a SPECIAL OCCASION.
不要将你的东西为了某一个特别的时刻而预留着，因为你生活的每一天都是那么特别；

Search for knowledge,read more ,sit on your porch and admire the view without paying attention to your needs;
寻找更我的知识，多读一些书，坐在你家的前廊里，以赞美的眼光去享受眼前的风景，不要带上任何功利的想法；

Spend more time with your family and friends,eat your favorite foods,visit the places you love;
花多点时间和朋友与家人在一起，吃你爱吃的食物，去你想去的地方；

Life is a chain of moments of enjoyment;not only about survival;
生活是一串串的快乐时光；我们不仅仅是为了生存而生存；

Use your crystal goblets.Do not save your best perfume,and use it every time you feel you want it.
举起你的水晶酒杯吧。不要吝啬洒上你最好的香水，你想用的时候就享用吧！

Remove from your vocabulary phrases like"one of these days"or "someday";
从你的词汇库中移去所谓的" 有那么一天"或者 "某一天"；

Let's write that letter we thought of writing "one of these days"!
曾打算"有那么一天 "去写的信，就在今天吧！

Let's tell our families and friends how much we love them;
告诉家人和朋友，我们是多么地爱他们；

Do not delay anything that adds laughter and joy to your life;
不要延迟任何可以给你的生活带来欢笑与快乐的事情；

Every day,every hour,and every minute is special;
每一天、每一小时、每一分钟都是那么特别；

And you don't know if it will be your last.
你无从知道这是否最后刻。

--
Happy day, happy life!

Intel IA32 Floating-Point Arithmetic problems with GCC

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

double recip( int denom)
{
    return 1.0 / (double)denom;
}

void do_nothing(){}

void test1( int denom)
{
    double r1,r2;
    int t1,t2;

    r1 = recip(denom);
    r2 = recip(denom);
    t1 = r1 == r2;
    printf("test1 t1: r1 %f %c= r2 %f\n",r1,t1 ? '=' : '!',r2);
    do_nothing();
    t2 = r1 == r2;
    printf("test1 t2: r1 %f %c= r2 %f\n",r1,t2 ? '=' : '!',r2);
}

void test2(int denom)
{
    double r1;
    int t1;
    r1 = recip(denom);
    t1 = r1 == 1.0 /(double)denom;
    printf("test2 t1: r1 %f %c= 1.0/10.0 \n",r1,t1 ? '=' : '!');
}

long double recip_l(int denom)
{
    return 1.0/(long double)denom;
}

void test3(int denom)
{
    long double r1,r2;
    int t1, t2, t3;

    r1 = recip(denom);
    r2 = recip(denom);
    t1 = r1 == r2;
    printf("test3 t1: r1 %f %c= r2 %f\n",(double)r1,t1 ? '=' : '!',(double)r2);

    do_nothing();
    t2 = r1 == r2;
    printf("test3 t2: r1 %f %c= r2 %f\n",(double)r1,t2 ? '=' : '!',(double)r2);

    t3 = r1 == 1.0 /(long double)denom;
    printf("test3 t3: r1 %f %c= 1.0/10.0 \n",(double)r1,t3 ? '=' : '!');
}

int main(int argc, char* argv[])
{
    test1(10);
    test2(10);
    test3(10);
    system("PAUSE");
    return 0;
}

// Using
// Compile the code with GCC with compile parameters repectively
// a. -O2
// b. -ffloat-store

//Result running on redhat linux
//[root@localhost download]# gcc -O2 2.4.6.c
//[root@localhost download]# ./a.out
//test1 t1: r1 0.100000 != r2 0.100000
//test1 t2: r1 0.100000 == r2 0.100000
//test2 t1: r1 0.100000 == 1.0/10.0
//test3 t1: r1 0.100000 == r2 0.100000
//test3 t2: r1 0.100000 == r2 0.100000
//test3 t3: r1 0.100000 == 1.0/10.0
//sh: line 1: PAUSE: command not found
//
//
//[root@localhost download]# gcc -ffloat-store 2.4.6.c
//[ root@localhost download]# ./a.out
//test1 t1: r1 0.100000 == r2 0.100000
//test1 t2: r1 0.100000 == r2 0.100000
//test2 t1: r1 0.100000 != 1.0/10.0
//test3 t1: r1 0.100000 == r2 0.100000
//test3 t2: r1 0.100000 == r2 0.100000
//test3 t3: r1 0.100000 == 1.0/10.0
//sh: line 1: PAUSE: command not found



--
Happy day, happy life!

How to get the decimal value from the bit representation of a float value?(3)

For example: Float value is 3490593.0

Bit representation on little endian machine: 840C554A

Actual Bit representation is : 4A550C84

Bit representation: 01001010010101010000110010000100

Float interpretation: sign 1 bit, exponent 8 bits, fraction 23 bits.

         Sign    Exponent          fraction

           0     10010100     10101010000110010000100

Decimal

value              148         5573764/2^23

 

Use normal interpretation:

E = 148-(2^(8-1)-1)

(1 + 5573764/2^23)*(2^(148-(2^(8-1)-1)))

2^23 = 8388608;

148-(2^(8-1)-1) = 21;

(1 + 5573764/2^23) = 1. 664444446563720703125

(2^(148-(2^(8-1)-1))) = 2097152

 

(1 + 5573764/2^23)*(2^(148-(2^(8-1)-1))) = 1. 664444446563720703125* 2097152= 3490593

--
Happy day, happy life!

How to get the decimal value from the bit representation of a float value?(2)

For example: Float value is 54321.1

Bit representation on little endian machine: 1A315447

Actual Bit representation is : 4754311A

Bit representation: 01000111010101000011000100011010

Float interpretation: sign 1 bit, exponent 8 bits, fraction 23 bits.

         Sign    Exponent          fraction

           0     10001110     10101000011000100011010

Decimal

value              142         5517594/2^23

 

Use normal interpretation:

E = 142-(2^(8-1)-1)

(1 + 5517594/2^23)*(2^(142-(2^(8-1)-1)))

2^23 = 8388608;

142-(2^(8-1)-1) = 15;

(1 + 5517594/2^23) = 1. 6577484607696533203125

(2^(142-(2^(8-1)-1))) = 32768

 

(1 + 5517594/2^23)*(2^(142-(2^(8-1)-1))) = 1. 6577484607696533203125 * 32768= 54321.1015625

~54321.1

--
Happy day, happy life!

How to get the decimal value from the bit representation of a float value?

For example: Float value is 12345.0

Bit representation on little endian machine: 00E44046

Actual Bit representation is : 4640E400

Bit representation: 01000110010000001110010000000000

Float interpretation: sign 1 bit, exponent 8 bits, fraction 23 bits.

         Sign    Exponent          Fraction

           0      10001100    10000001110010000000000

Decimal

value              140         4252672/2^23

 

Use normal interpretation:

E = 140-(2^(8-1)-1)

(1 + 4252672/2^23)*(2^(140-(2^(8-1)-1)))

2^23 = 8388608;

140-(2^(8-1)-1) = 13;

(1 + 4252672/2^23) = 1.5069580078125

(2^(140-(2^(8-1)-1))) = 8192

 

(1 + 4252672/2^23)*(2^(140-(2^(8-1)-1))) = 1.5069580078125 * 8192 = 12345

 

--
Happy day, happy life!