You can use a function such as this, to print the bytes:
void print_bytes(const void *object, size_t size)
{
// This is for C++; in C just drop the static_cast<>() and assign.
const unsigned char * const bytes = static_cast<const unsigned char *>(object);
size_t i;
printf("[ ");
for(i = 0; i < size; i++)
{
printf("%02x ", bytes[i]);
}
printf("]\n");
}
Usage would look like this, for instance:
int x = 37;
float y = 3.14;
print_bytes(&x, sizeof x);
print_bytes(&y, sizeof y);
This shows the bytes just as raw numerical values, in hexadecimal which is commonly used for "memory dumps" like these.
On a random (might even be virtual, for all I know) Linux machine running a "Intel(R) Xeon(R)" CPU, this prints:
[ 25 00 00 00 ] [ c3 f5 48 40 ]
This handily also demonstrates that the Intel family of CPU:s really are little endian.
@user1721803 Because you can't dereference
void *'. I think it's cleaner to have a function like this accept aconst void `, since that makes *using the function very simple and cast-less. Casts are to be avoided, much better to contain them where needed instead of inflicting them on others.@unwind why do you do that bitwise and of each byte with
0xff? Is it necessary?@JoãoAlmeida No, it shouldn't really be. Just kind of belt-and-suspenders thinking, I guess. Which is something I dislike, so I'll edit that out now. Thanks.
error: invalid conversion from ‘const void*’ to ‘const unsigned char*’ [-fpermissive] const unsigned char * const bytes = object;