将char数组转换为整数 [英] Casting a char array to an integer

问题描述

基本上我正在读一个二进制格式，其中4个字节指定字符串的大小。所以我想把4个字符从一个缓冲区读到一个整数。

这是我有的。

  int FileReader :: getObjectSizeForMarker（int cursor，int eof，char * buffer）{
 //跳过标记并读下4个字符
 int cursor = cursor + 4; // skip marker and read 4 
 char tmpbuffer [4] = {buffer [cursor]，buffer [cursor + 1]，buffer [cursor + 2]，buffer [cursor + 3]}; 
 int32_t objSize = tmpbuffer; 
 return objSize; 
 
} 
  

想法？

  unsigned char * ptr =（unsigned char *）（buffer + cursor）; 
 // unpack big-endian order 
 int32_t objSize =（ptr [0]<< 24）| （ptr [1]<< 16）| （ptr [2] << 8）| ptr [3]; 
  

Basically I am reading a binary format where 4 bytes specify the size of the string to follow. So i want to cast 4 chars I am reading from a buffer to 1 integer.

Here is what I have.

int FileReader::getObjectSizeForMarker(int cursor, int eof, char * buffer) {
  //skip the marker and read next 4 byes
  int cursor = cursor + 4; //skip marker and read 4
  char tmpbuffer[4] = {buffer[cursor], buffer[cursor+1], buffer[cursor+2], buffer[cursor+3]};
  int32_t objSize = tmpbuffer;
  return objSize;

}

thoughts?

解决方案

It's pretty easy to do the unpacking manually:

unsigned char *ptr = (unsigned char *)(buffer + cursor);
// unpack big-endian order
int32_t objSize = (ptr[0] << 24) | (ptr[1] << 16) | (ptr[2] << 8) | ptr[3];

这篇关于将char数组转换为整数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！