Storing hex value in C string?

Question

I have a string `"SOMETHING /\xff"` and I want to save the hex representation of `0xff` into a `char` buffer. So I `strncpy` everything after the forward slash (`/`) into my buffer (let's call it `buff`). However, when I use the gdb command `print \x buff` to view the contents of `buff`, it doesn't show `0xff`. Any ideas as to what could be wrong? Is it possible that my forward slash is messing things up?

Accepted Answer

I think your problem has to do with the way you're printing the variable in gdb. Take this simple program, compile as debug (`-g` if using gcc), and run and break at the first `puts` statement:

```
int main(void)
{
   char *ptr = "str \xff";
   char arr[] = "str \xff";
   puts(ptr);
   puts(arr);

return 0;
}
```

If I try and print `ptr` the way you've mentioned, `p /x ptr`, it'll print the value of `ptr` (the address it's pointing to):

```
(gdb) p /x ptr
$1 = 0x4005f8
```

However if I do the same command for `arr`, I'll get:

```
(gdb) p /x arr
$2 = {0x73, 0x74, 0x72, 0x20, 0xff, 0x0}
```

That's because gdb can see that `arr` is of type `char[6]` and not `char*`. You can get the same results with the command `p /x "str \xff"`, which is useful for testing things:

```
(gdb) p /x "str \xff"
$3 = {0x73, 0x74, 0x72, 0x20, 0xff, 0x0}
```

Now if you want to be able to print a certain amount of bytes from the address pointed to by a pointer, use the examine (`x`) memory command instead of print (`p`):

```
(gdb) x/6bx ptr
0x4005f8 <__dso_handle+8>:  0x73    0x74    0x72    0x20    0xff    0x00
```

That'll print 6 bytes in hex from the address pointed to by `ptr`. Try that with your `buff` variable and see how you go.

Alternatively, another thing you can try is:

```
(gdb) p /x (char[6])*ptr
$4 = {0x73, 0x74, 0x72, 0x20, 0xff, 0x0}
```

This will treat the `char` pointed to by `ptr` as the first element in an array of 6 `char`s, and will allow you to use the print command.

Storing hex value in C string?

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