memcpy将ff ff ff添加到字节的开头 [英] memcpy adds ff ff ff to the beginning of a byte

问题描述

我有一个像这样的数组:

I have an array that is like this:

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

当我使用memcpy时:

When I use memcpy:

memcpy(array2, array, 6);

并打印两个文件:

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

一个打印像:

c0 3f e 54 e5 20

但其他人会打印

ffffffc0 3f e 54 ffffffe5 20

发生了什么事?

推荐答案

我已经将您的代码变成了完整的可编译示例.我还添加了第三列正常" char，该数组在我的环境中已签名.

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;
}

我的结果符合我的预期.

My results were what I expected.

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

如您所见，仅当数组为带符号char类型的数组时，才附加额外" ff.原因是，当memcpy填充有符号char的数组时，现在将设置高位的值与char负值相对应.当传递给printf时，char被提升为int类型，这实际上意味着符号扩展.

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将它们以十六进制格式打印，就像它们是unsigned int一样，但是随着参数作为int传递，行为在技术上是不确定的.通常，在二进制补码机上，其行为与使用mod 2 ^ N算术的标准有符号到无符号转换(其中N是unsigned int中的值位数)相同.由于该值仅是略微"负值(来自窄带正负号的类型)，转换后的值接近最大可能的unsigned int值，即它具有许多前导1(二进制)或前导f以十六进制表示.

%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, i.e. it has many leading 1's (in binary) or leading f in hex.

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