memcpy from unsigned char * to unsigned int

Question

I am not sure what the following C++ code does (I have replaced parts that don't matter with dots):

unsigned char* p = ...;
unsigned int metaNum;
memcpy( &metaNum, p, sizeof( unsigned int ) );
p += sizeof( unsigned int );

for ( unsigned int m = 0; m < metaNum; m++ ) {

...

}

I know for memcpy that:

The underlying type of the objects pointed to by both the source and destination pointers are irrelevant for this function; The result is a binary copy of the data.

Still I am not sure what is the interpretation of metaNum. Is it something obvious or it has to do with the implementation?

score 2 · Accepted Answer · answered Jun 23 '17 at 23:59

This is not implementation-specific: metaNum is interpreted as a sequence of sizeof(unsigned int) bytes, into which the content from the same number of bytes pointed to by p is copied.

The only thing in play here is endianness: the same sequence of bytes copied into metaNum will be interpreted differently depending on how the hardware is storing multi-byte values, such as unsigned int. If the sequence comes from the same hardware, interpretation would be the same. Otherwise, you would need to use ntoh/hton functions to bring byte sequences into proper order.

memcpy from unsigned char * to unsigned int

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