在 C 中使用 memcpy 复制无符号字符数组

Question

int n=50;
unsigned char bytes[4];
unsigned char *buffer=(unsigned char*)malloc(sizeof(int));
bytes[0]=(n>>24)&0xFF;
bytes[1]=(n>>16)&0xFF;
bytes[2]=(n>>8)&0xFF;
bytes[3]=n&0xFF;

memcpy(buffer,bytes,sizeof(bytes));

This does not copy the bytes array to buffer. Any idea why?? What can be done to copy the array to buffer??

When I try to print the sizeof(buffer) or the contents after memcpy it shows 0 and no contents. I need to copy it to buffer as I have other information to append to the buffer along with bytes.

Answer 1

 int n=50; 

Assuming that you are on a 32-bit machine, 'n' will be a 4-byte value. n = 0x00000032 = 00000000b 00000000b 00000000b 00110010b

 unsigned char bytes[4];

bytes will ve a 4-byte value :

 bytes[0]=(n>>24)&0xFF;

byte[0] = 00000000b

 bytes[1]=(n>>16)&0xFF;

byte[1] = 00000000b

 bytes[2]=(n>>8)&0xFF;

byte[2] = 00000000b

 bytes[3]=n&0xFF;

byte[3] = 00110010b

memcpy(buffer,bytes,sizeof(bytes));

Copy (sizeof(bytes)) 4 bytes from bytes to buffer.

Whether or not this does what you expect is perhaps the question. (More assumptions)

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Assuming that buffer is:

int buffer[1];

the above statement would copy as expected. However, if you test this assumption using code usch as:

printf("buffer = %d\n", buffer[0]);

The output will depend on what kind of machine you run it on; little endian, or bit endian.

On one, it will out put "buffer = 50" On the other, it will output the decimal value which is equivelant to:

0x32000000 (00110010b 00000000 00000000b 00000000b 00000000b)

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Assuming:

int buffer;

Will most likely generate a compiler warning (or error); and is probably not what you want, unless you change your memcpy() as follows:

memcpy(&buffer,bytes,sizeof(bytes));