将原始数据打印到固定长度的十六进制输出

I have a struct, well pointer to a struct, and I wish to printf the first n bytes as a long hex number, or as a string of hex bytes.

我有一个结构,指向结构的指针,我希望将前n个字节作为长十六进制数字打印,或者作为十六进制字节的字符串。

Essentially I need the printf equivalent of gdb's examine memory command, x/nxb .

基本上我需要printf等效的gdb的检查内存命令,x / nxb。

If possible I would like to still use printf as the program's logger function just variant of it. Even better if I can do so without looping through the data.

如果可能的话,我仍然希望使用printf作为程序的记录器功能,只是它的变体。如果我可以在不循环数据的情况下这样做,那就更好了。

2 个解决方案

#1

Just took Eric Postpischil's advice and cooked up the following :

刚刚接受了Eric Postpischil的建议并制作了以下内容:

struct mystruc
{
  int a;
  char b;
  float c;
};

int main(int argc, char** argv)
{
  struct mystruc structVar={5,'a',3.9};
  struct mystruc* strucPtr=&structVar;
  unsigned char* charPtr=(unsigned char*)strucPtr;
  int i;
  printf("structure size : %zu bytes\n",sizeof(struct mystruc));
  for(i=0;i<sizeof(struct mystruc);i++)
      printf("%02x ",charPtr[i]);

  return 0;
}

It will print the bytes as fas as the structure stretches.

它将字节打印为结构拉伸时的fas。

Update : Thanks for the insight Eric :) I have updated the code.

更新:感谢洞察Eric :)我已经更新了代码。

#2

Try this. Say you have pointer to struct in pstruct.

试试这个。假设您在pstruct中有指向struct的指针。

unsigned long long *aslong = (unsigned long long *)pstruct;
printf("%08x%08x%08x%08x%08x%08x%08x%08x", 
       aslong[0],
       aslong[1],
       aslong[2],
       aslong[3],
       aslong[4],
       aslong[5],
       aslong[6],
       aslong[7],
);

As Eric points out, this might print the bytes out-of-order. So it's either this, or using unsigned char * and (having a printf with 64 arguments or using a loop).

正如Eric指出的那样,这可能会无序地打印字节。所以它或者是这个,或者使用unsigned char *和(使用带有64个参数的printf或使用循环)。

#1