面试笔试问题总结（一）—位制、位操作、位图

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作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/38925425

时间：2014-8-29

前言

正文

1. 关于位的考法

1.1 64位机 VS 32位机

#include<stdio.h>#include<stdlib.h>int main(){    unsigned char uint8 = 0;    signed char int8 = 0;    unsigned short uint16 = 0;    signed short int16 = 0;    unsigned int uint32 = 0;    signed int int32 = 0;    float fp32 = 0;    double fp64 = 0;    unsigned long ulong = 0;    printf("unsigned char is %d bit\n\r", sizeof(uint8)*8);    printf("signed char is %d bit\n\r", sizeof(int8)*8);    printf("unsigned short is %d bit\n\r", sizeof(uint16)*8);    printf("signed short is %d bit\n\r", sizeof(int16)*8);    printf("unsigned int is %d bit\n\r", sizeof(uint32)*8);    printf("signed int is %d bit\n\r", sizeof(int32)*8);    printf("float fp32 is %d bit\n\r", sizeof(fp32)*8);    printf("double fp64 is %d bit\n\r", sizeof(fp64)*8);    printf("unsigned long is %d bit\n\r", sizeof(ulong)*8);    system("pause");    return 0;}

64位结果是

unsigned char is 8 bit
signed char is 8 bit
unsigned short is 16 bit
signed short is 16 bit
unsigned int is 32 bit
signed int is 32 bit
float fp32 is 32 bit
double fp64 is 64 bit
unsigned long is 32 bit

32位运行结果是：

32位编译器：

      char ：1个字节
      char*（即指针变量）: 4个字节（32位的寻址空间是2^32, 即32个bit，也就是4个字节。同理64位编译器）
      short int : 2个字节
      int：  4个字节
      unsigned int : 4个字节
      float:  4个字节
      double:   8个字节
      long:   4个字节
      long long:  8个字节
      unsigned long:  4个字节

64位编译器：

      char ：1个字节
      char*(即指针变量): 8个字节
      short int : 2个字节
      int：  4个字节
      unsigned int : 4个字节
      float:  4个字节
      double:   8个字节
      long:   8个字节
      long long:  8个字节
      unsigned long:  8个字节

某页式存储管理系统中，地址寄存器长度为24位，其中页号占14位，则主存的分块大小是（）字节。

A、10 B、2^10 C、2^14 D、2^24

Answer：24-14是寻址位数 B

1.2 位操作

位操作的面试题以后再上，先总结一下笔试题。

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