【C语言】练习2-1
2024-08-28 07:21:32
- 题目来源:《The C programming language》中的习题P27
- 练习2-1: 编写一个程序以确定分别由signed及unsigned限定的char、short、int与long类型变量的取值范围。采用打印标准头文件中的相应值以及直接计算两种方式实现。后一种方法的实现较困难一些,因为要确定各种浮点类型的取值范围。
采用打印标准头文件中的相应值:
#include <stdio.h>
#include <limits.h> int main()
{
printf("采用打印标准头文件limits.h中的相应值:\n");
//signed types
printf("signed char min = %d\n", SCHAR_MIN);
printf("signed char max = %d\n", SCHAR_MAX);
printf("signed short min = %d\n", SHRT_MIN);
printf("signed short max = %d\n", SHRT_MAX);
printf("signed int min = %d\n", INT_MIN);
printf("signed int max = %d\n", INT_MAX);
printf("signed long min = %ld\n", LONG_MIN);
printf("signed long max = %ld\n", LONG_MAX);
//unsigned types
printf("unsigned char max = %u\n", UCHAR_MAX);
printf("unsigned short max = %u\n", USHRT_MAX);
printf("unsigned int max = %u\n", UINT_MAX);
printf("unsigned long max = %lu\n", ULONG_MAX);
return ;
}
执行结果:
直接计算的方式:
#include <stdio.h>
//determine ranges of types
int main()
{
printf("采用直接计算的方式:\n");
//signed types
printf("signed char min = %d\n", -(char)((unsigned char)~>>)-);
printf("signed char max = %d\n", (char)((unsigned char)~>>));
printf("signed short min = %d\n", -(short)((unsigned short)~>>)-);
printf("signed short max = %d\n", (short)((unsigned short)~>>));
printf("signed int min = %d\n", -(int)((unsigned int)~>>)-);
printf("signed int max = %d\n", (int)((unsigned int)~>>));
printf("signed long min = %ld\n", -(long)((unsigned long)~>>)-);
printf("signed long max = %ld\n", (long)((unsigned long)~>>));
//unsigned types
printf("unsigned char max = %u\n", (unsigned char)~);
printf("unsigned short max = %u\n", (unsigned short)~);
printf("unsigned int max = %u\n", (unsigned int)~);
printf("unsigned long max = %lu\n", (unsigned long)~);
return ;
}
执行结果:
备注:
- <limits.h>包含了定义char int short long类型取值的最大值和最小值用常量表示(例如SCHAR_MIN = -128), 所以直接打印即可。
- 如果直接计算的话,则需要用到按位运算符~和右移运算符>>,对于-(char)((unsigned char)~0 >> 1)解释:
a: (unsigned char)~0用二进制表示为:11111111
b: (unsigned char)~0 >> 1,将(unsigned char)~0逻辑右移1位,结果为:01111111
c: (char)((unsigned char)~0 >> 1),将((unsigned char)~0 >> 1)由无符号类型强制转换为有符号类型
d: -(char)((unsigned char)~0 >> 1),得到的二进制结果为11111111,最高位为符号为,对应的十进制数为:-127
- 因为绝大部分机器采用补码方式表示有符号整数,所以需要在-(char)((unsigned char)~0 >> 1)的基础上再减去1,才是真正的最小值。
参考资料:
[1] http://www.cnblogs.com/wangzhiyu811/archive/2011/07/12/2104377.html
[2] http://blog.csdn.net/roma823/article/details/6367142
最新文章
- Oracle学习笔记十 使用PL/SQL
- Android在layout xml中使用include
- 利用AdaBoost元算法提高分类性能
- javascript 事件传播与事件冒泡,W3C事件模型
- MONO 架构
- Javascript 中 null、NaN和undefined的区别
- C99新特性
- Qt浅译:JSON Support in Qt(JSON只有六种数据类型)
- POJ输出状态的逻辑。
- Android SDCard Mount 流程分析
- crm创建和编辑全局选项集
- mysql 中 SQL_CALC_FOUND_ROWS 功能
- ecshop 去版权(前台)
- 使用xUnit为.net core程序进行单元测试(上)
- 第I篇PCI体系结构概述
- BZOJ 1510: Kra-The Disks
- Guava新增集合类型-Bimap
- 41.找出所有和为S的连续正数序列
- 20172310 2017-2018-2 《程序设计与数据结构》实验三报告(敏捷开发与XP实践)
- easyui datagrid 禁止选中行