文件中的字节数“0”。和“一个” [英] Number of bytes in file "zero" and "one"

问题描述

嘿朋友

我使用visual studio 2010，我有一个文件，如文件包含的输入流，例如（a，a，b，a，c，d，e）

-首先我想要计算文件的大小，（可能是文件是图像灰度）

- 更改二进制的ASCII字符？



-例如：

a in ASCII = 61

a in binary = 00111101

b in ASCII = 62
b in binary = 00111110

等...直到EOF

所以我需要帮助我可以计算：

零（a）= 3

一个数字（a）= 5

数字pf零（b）= 3

一个数字（ b）= 5

....

thnx为你提供帮助





[更新]

最近尝试了什么：

hey friends
I work with visual studio 2010 , I have a file like as input stream the file contains for example (a,a,b,a,c,d,e)
-firstly I want just calculate the size of the file , (maybe file was a image greyscale)
-change ASCII character in binary?

-for example :
a in ASCII =61
a in binary = 00111101
b in ASCII =62
b in binary =00111110
etc ... until EOF
so I need help for I can calculate:
number of zero (a)=3
number of one (a)=5
number pf zero (b)=3
NUMER of one (b)=5
....
thnx for ur help


[update]
What have tried more recently:

// Make_Example.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

#include <stdio.h>
#include <conio.h>

#include<stdlib.h>

 

void printCharAsBinary(char c) {
 
	int i;
 
	for(i = 0; i < 8; i++)
	{
 
	printf("%d", (c >> i) & 0x1);
 
	}
 
}
 

void printStringAsBinary(char* s){
	int j;
 
	for(j=0; j<4; j++)
	
	{
 
		printCharAsBinary(*s);
		printf("\n");	
 
	}
	printf("\n");
}
 
int _tmain(int argc, _TCHAR* argv[])
 
{
	char s1[] = "aabc";
 
	printStringAsBinary(s1);
	getch();
	
}

[/ update]



我有什么尝试过：



我没试过！我不知道

[/update]

What I have tried:

I nothing tried ! I haven't idea

推荐答案

如果你用google搜索过，你应该找到一些像这样的教程文件i / o的介绍。对于初学者来说，更容易在Youtube上播放一些视频。



您还必须了解在文件api中您编写的字节 - 并且每种数据类型都有一些字节大小。 />


提示：想象一个文件自己作为字节数组，需要由你的代码解释。

If you had googled you should had found some tutorials like this introduction to the file i/o. Easier can be some videos on Youtube for beginners.

You also must understand that in the file api you write bytes - and every data type has some byte size.

tip: Imagine a file yourself as byte array which needs to be interpreted by your code.

提示：为了转换a在二进制格式的字节中，您可以迭代地测试第7位然后向左移位数字，例如迭代以下代码

Hint: in order to convert a byte in binary format, you may, iteratively, test bit 7 and then shift left the number, e.g. iterate the following code

char binary[9];
binary[8] ='\0'; // NULL terminate the string
unsigned char a = 'a'; a is 0x61
binary[0] = (a & 0x80) ? '1' : '0';
a <<= 1;
binary[1] = (a & 0x80) ? '1' : '0';
a <<= 1;
//...
// put this in a loop

'转换过程'代码也可以计算收集的零（或一个）的总数。



[update]

您的 printStringAsBinary 不会迭代字符。尝试：

the 'conversion process' code could also count the total number of collected zeroes (or ones).

[update]
Your printStringAsBinary doesn't iterate over characters. Try:

void printStringAsBinary(char* s)
{
  while ( *s )
  {
    printCharAsBinary(*s);
    printf("\n");
    ++s;
  }
}

[/ update]





[另一个更新]

printCharAsBinary的修改允许计算字节中有多少个：

[/update]


[another update]
A modification of printCharAsBinary allows to count how many ones you have in the byte:

int printCharAsBinary(char c) {

  int ones = 0;
  int i;

  for(i = 0; i < 8; i++)
  {
    char bit = (c & 0x80) ? 1 : 0;
    printf("%c", bit+'0');
    ones += bit;
    c <<= 1;
  }
  printf("\n");
  return ones;
}

[/ another update]

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