在 xv6 中读取 hex 文件

Question

I am trying to modify this xv6 c file into a hex viewer. I tried to utilize printf(%02X) but to no avail. I tried doing different combinations of placement for the printf(%02X) but it will either not print hex or just print "%02X" at the end of the file view. For the printf functionality, it specifies these input arguments: Print to the given fd. Only understands %d, %x, %p, %s. void printf(int fd, const char *fmt, ...) Here is the code I am working with:

#include "types.h"
#include "stat.h"
#include "user.h"

char buf[512];

void
hex(int fd)
{
  int n;

  while((n = read(fd, buf, sizeof(buf))) > 0) {
   if (write(1, buf, n) != n) {
     printf(1, "hex: write error\n");
     exit();
     }
    printf(n,"%02X");
   } 
   if(n < 0){
     printf(1, "hex: read error\n");
     exit();
   }

}

int
main(int argc, char *argv[])
{
  int fd, i;

  if(argc <= 1){
    hex(0);
    exit();
  }

  for(i = 1; i < argc; i++){
    if((fd = open(argv[i], 0)) < 0){
      printf(1, "hex: cannot open\n", argv[i]);
      exit();
    }
    hex(fd);
    close(fd);
  }
  exit();
}

Answer 1

As you found yourself, in xv6, printf is very limited and you can't use %02x modifier. You have to restrict your self to %x (or you can extend the printf function, but it's an exercise beyond what is asked to you.

For each character, you have to call printf(1, "%x ", the_character); For characters for 0x10 to 0xFF , it will print two character, for characters under 0x10 , you have to add the extra 0 .

-- To print char by char, you need a for loop on the buffer you got after the read call. Add some space and LF to improve visibility and you're done.

void hex(int fd)
{
  char buf[16];

  int n, i;

  while((n = read(fd, buf, sizeof(buf))) > 0) {
    for (i = 0; i < n; ++i) {
        if (buf[i] < 0x10) {
            printf(1, "0");
        }
        printf(1,"%x ", buf[i]);
    }
    printf(1, "\n");
   } 
}