萌新初来乍到，请多指教 - 编程论坛

#2

自由而无用2021-10-12 13:46

//online parser: https://www.bccn.net/run/

程序代码：

#include <stdio.h>

int main(int argc, char *argv[])
{
    int a, b;
    //add volatile to mask gcc overdoing optimization
    volatile double x, y;
    double formula;

#define critical_section_in
    ((int *)&a)[0] = 10;
    ((int *)&a)[1] = 20;
    *(double *)&((int *)&a)[2] = 1.25;
    *(double *)&((int *)&a)[4] = 12.17;
#define critical_section_out

//attention: while closed print_on gcc optimized the statment (7.0 * x * y)
#define PRINT_ON
#ifdef PRINT_ON
    printf("a = %p\n", &a);
    printf("a = %d\n", a);
    printf("b = %d\n", b);
    printf("(a->x) = %p-----", &((int *)&a)[2]);
    printf("x = %p\n", &x);
    printf("x = %lf\n", x);
    printf("y = %lf\n", y);
#endif

    formula =  6 * a * b;
    formula = formula / (7.0 * x * y);
    printf("f = %.2lf\n", formula);

    return 0;
}

output sample:
a = 0x7ffee15347f8
a = 10
b = 20
(a->x) = 0x7ffee1534800-----x = 0x7ffee1534800
x = 1.250000
y = 12.170000
f = 11.27

[此贴子已经被作者于2021-10-12 13:53编辑过]

#3

rjsp2021-10-13 07:54

交代一下必要的上下文信息吧，没头没脑的你让人怎么回答？
比如我问你“俺隔壁村子的马二狗为什么半夜起床？”，你怎么答？撒尿、拉屎、梦游、打蚊子、去做贼、闻鸡起舞、……都有可能，而可能的种类无限多。

可能是作者处理的数据中，a*b 过大会溢出，而作者又不知道可以写成 6.0*a*b/(7*x*y)；
也可能是作者处理的数据中，x更接近6，先计算 6/x 可以提高精度；
也可能是作者处理的数据中，6/x*a更接近1，先计算 6/x*a 可以提高精度；
也可能是……
也可能是其它人就是这么写的，因为浮点数不满足结合律，所以算式不同会导致结果不同。
也可能是……

#4

自由而无用2021-10-13 09:10

//online parser: https://www.bccn.net/run/?lang=c

程序代码：

#include <stdio.h>

int main(int argc, char *argv[])
{
    int a, b;
//#define MASK_GCC_OVERDOING
#ifdef MASK_GCC_OVERDOING
    //add volatile to mask gcc CT(compile-time) overdoing optimization
    /* runtime check */
    volatile double x, y;
#else
    /* compile-time check */
    double x, y;
#endif
    double formula;

#define runtime_block__IN
/* as gcc unable to check runtime code */
    ((int *)&a)[0] = 10;
    ((int *)&a)[1] = 20;
    *(double *)&((int *)&a)[2] = 1.25;
    *(double *)&((int *)&a)[4] = 12.17;
#define runtime_block__OUT

#ifdef MASK_GCC_OVERDOING
//attention: while closed print_on gcc optimized the expression (7.0 * x * y)
#define PRINT_ON
#ifdef PRINT_ON
    printf("a = %p\n", &a);
    printf("a = %d\n", a);
    printf("b = %d\n", b);
    printf("(a->x) = %p-----", &((int *)&a)[2]);
    /*while CT checking, until here gcc finally realized the exsitence of x, y*/
    printf("x = %p\n", &x);
    printf("x = %lf\n", x);
    printf("y = %lf\n", y);
#endif
#endif /*MASK_GCC_OVERDOING*/

    formula =  6 * a * b;
    /* CT: 7.0 * x * y = 0 so got the f = inf[divide zero] */
    formula = formula / (7.0 * x * y);
    printf("f = %.2lf\n", formula);

    return 0;
}

output sample:
f = inf

#5

diycai2021-10-13 11:21

回复 3楼 rjsp

让我想起了20年前的一个很火的网游MU，角色死亡惩罚是身上携带的金钱扣除30%。
程序员是这么处理的：
int money = xxx;//角色持有金钱数量
money = money - (moeny*3/10);
然后就被玩家发现了bug，大家争先恐后带着巨额金钱出城自杀。

#6

jianghushife2021-10-14 16:10