memcpy将ff ff ff添加到一个字节的开头

Question

I have an array that is like this:

unsigned char array[] = {'\xc0', '\x3f', '\x0e', '\x54', '\xe5', '\x20'};
unsigned char array2[6];

When I use memcpy:

memcpy(array2, array, 6);

And print both of them:

printf("%x %x %x %x %x %x", array[0],  // ... etc
printf("%x %x %x %x %x %x", array2[0], // ... etc

one prints like:

c0 3f e 54 e5 20

but the other one prints

ffffffc0 3f e 54 ffffffe5 20

what happened?

Answer 1

I've turned your code into a complete compilable example. I also added a third array of a 'normal' char which on my environment is signed.

#include <cstring>
#include <cstdio>

using std::memcpy;
using std::printf;

int main()
{

        unsigned char array[] = {'\xc0', '\x3f', '\x0e', '\x54', '\xe5', '\x20'};
        unsigned char array2[6];
        char array3[6];

        memcpy(array2, array, 6);
        memcpy(array3, array, 6);

        printf("%x %x %x %x %x %x\n", array[0], array[1], array[2], array[3], array[4], array[5]);
        printf("%x %x %x %x %x %x\n", array2[0], array2[1], array2[2], array2[3], array2[4], array2[5]);
        printf("%x %x %x %x %x %x\n", array3[0], array3[1], array3[2], array3[3], array3[4], array3[5]);

        return 0;
}

My results were what I expected.

c0 3f e 54 e5 20
c0 3f e 54 e5 20
ffffffc0 3f e 54 ffffffe5 20

As you can see, only when the array is of a signed char type do the 'extra' ff get appended. The reason is that when memcpy populates the array of signed char , the values with a high bit set now correspond to negative char values. When passed to printf the char are promoted to int types which effectively means a sign extension.

%x prints them in hexadecimal as though they were unsigned int , but as the argument was passed as int the behaviour is technically undefined. Typically on a two's complement machine the behaviour is the same as the standard signed to unsigned conversion which uses mod 2^N arithmetic (where N is the number of value bits in an unsigned int ). As the value was only 'slightly' negative (coming from a narrow signed type), post conversion the value is close to the maximum possible unsigned int value, ie it has many leading 1 's (in binary) or leading f in hex.

Answer 2

The problem is not memcpy (unless your char type really is 32 bits, rather than 8), it looks more like integer sign extension while printing.

you may want to change your printf to explicitly use unsigned char conversion, ie.

printf("%hhx %hhx...", array2[0], array2[1],...);

As a guess, it's possible that your compiler/optimizer is handling array (whose size and contents are known at compile time) and array2 differently, pushing constant values onto the stack in the first place and erroneously pushing sign extended values in the second.

Answer 3

您应该屏蔽高位，因为在调用varargs函数时，字符将扩展为int大小：

printf("%x %x %x %x %x %x", array[0] & 0xff,  // ..

Answer 4

%x format expects integer type. Try to use casting:

printf("%x %x %x %x %x %x", (int)array2[0], ...

Edit: Since there are new comments on my post, I want to add some information. Before calling the printf function, compiler generates code which pushes on the stack variable list of parameters (...). Compiler doesn't know anything about printf format codes, and pushes parameters according to their type. printf collects parameters from the stack according to formatting string. So, array[i] is pushed as char, and handled by printf as int. Therefore, it is always good idea to make casting, if parameter type doesn't match exactly format specification, working with printf/scanf functions.