将unsigned char数组的内容分成两半 [英] Divide contents of a unsigned char array into 2 halves

问题描述

我想知道是否有人可以帮助我我只学习c，我试图将无符号字符数组的内容分成两半，结果可以存储在两个无符号整型数组中，

举例来说，我有一些代码将一个十六进制值添加到BYTE数组中，所以如何将val []的内容拆分为两个，但保持相同的顺序

  #include< stdio.h> 
 typedef unsigned char BYTE; 
 int main（）
 {
 //将十六进制值放入val的示例用途[8] 
 int i，j; 
 long long hex = 0x78661EB54FE76763; 
 BYTE val [8]; （j = 0，i = 7; i> = 0; i  - ，j ++）{
 val [j] =（hex>（i * 8））的
 
 &安培; 0xff的; 
} 
 
 //如何分割val [8]的内容（现在包含十六进制）
 
 return 0; 
} 
  

我试图将十六进制值拆分为78661EB5,4FE76763，里面有一个unsigned int，它存储在我的例子中的val [8]里面

解决方案

创建一个帮助函数， 无符号字符的数组，并调用它两次。

给OP一些事情发现。

$ unsigned long form_big_endian4（const unsigned char * x）{
unsigned long y = TBB; //初始值应该是多少？
for（int i = 0; i y * =待定; //提示：8位的最大值+ 1
y + = x [i];
}
return y;
}

#include< stdio.h>
int main（void）{
const unsigned char val [8] =
{0x78，0x66，0x1E，0xB5，0x4F，0xE7，0x67，0x63};
unsigned long half = form_big_endian4（val）; //为什么长：提示可以有多小的无符号数？
printf（％08lX \\\
，一半）; //为什么在格式中为0？为什么`lX`？
half = form_big_endian4（val + TBD）; //偏移量有多远？
printf（％08lX \\\
，一半）;
返回0;
}

I am wondering if anyone can help me I am only learning c, I am trying to Divide contents of a unsigned char array into 2 halves, which the result can stored in two unsigned int's,

For example purposes, I have some code below which adds a hex value into a BYTE array, so How would split the contents of val[] into two but keep the same order

#include <stdio.h>
typedef unsigned char BYTE;
int main()
{
    // Sample purposes putting hex into val[8]
    int i,j;
    long long hex=0x78661EB54FE76763;
    BYTE val[8];

    for(j=0,i=7; i>=0; i--,j++){
        val[j]= (hex>>(i*8))&0xff;
    }       

    // How to split the contents of val[8] which now holds the hex

    return 0;  
}

I am trying to split the hex value into 78661EB5, 4FE76763 and store each one inside an unsigned int which is stored inside val[8] in my example

解决方案

Create a helper function that takes a pointer to an array of unsigned char and call it twice.

Left some things out for OP to discover.

unsigned long form_big_endian4(const unsigned char * x) {
  unsigned long y = TBB;           // What should the initial value be?
  for (int i = 0; i < TBD; i++) {  // How many times to loop?
    y *= TBD;                      // Hint: max value of 8 bits + 1
    y += x[i];                     
  }
  return y;
}

#include <stdio.h>
int main(void) {
  const unsigned char val[8] =
      { 0x78, 0x66, 0x1E, 0xB5, 0x4F, 0xE7, 0x67, 0x63 };
  unsigned long half = form_big_endian4(val); // Why long: Hint how small can unsigned be?
  printf("%08lX\n", half);                    // Why 0 in format? Why `lX`?
  half = form_big_endian4(val + TBD);         // How far an offset?
  printf("%08lX\n", half);
  return 0;
}

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