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tmarques4

Memory adressing.

4 posts in this topic

I'm wondering if I'm doing this the right way.

Suppose I have a 16bit data type, short in C, and I want it to represent two characters, 'H' and 'U', in a way, this short would act like a char vector of length 2, char[2].

This is the code:

[CODE]
short hum;
memset(&hum, 0, sizeof(short));
memcpy(&hum+0, "H", sizeof(char));
memcpy(&hum+1, "U", sizeof(char));
[/CODE]

I don't know if memory addresses can be handled in this byte by byte approach, ignoring the real data type size, this would be better than using bit shift and OR operator, I believe.

Thanks for the help in advance.
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If I'm not mistaken, using +1 since it's a short the compiler will assume that you want to increment by 2 bytes. What you'd need to do is cast it to a char* and then do the +1 like this
[CODE]
short hum;
memset(&hum, 0, sizeof(short));
memcpy(&hum, "H", sizeof(char));
memcpy(((char*)&hum)+1, "U", sizeof(char));
[/CODE]

or you could do something like this

[CODE]
short hum = 0;
char* humPtr = (char*)&hum;
*humPtr = 'H';
*(humPtr+1) = 'U';
[/CODE] Edited by Dancin_Fool
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[quote name='TMarques' timestamp='1340223566' post='4951108']I don't know if memory addresses can be handled in this byte by byte approach, ignoring the real data type size, this would be better than using bit shift and OR operator, I believe.[/quote]
Read up on [url=http://www.learncpp.com/cpp-tutorial/68-pointers-arrays-and-pointer-arithmetic/]pointer arithmetic[/url]. [tt]&hum+1[/tt] is the next [tt]short[/tt] if [tt]&hum[/tt] is an array, therefore, [tt]&hum+1[/tt] is [tt]sizeof(short)[/tt] bytes after [tt]&hum[/tt]. However, [tt]((char*)&hum) + 1[/tt] is the next byte after [tt]&hum[/tt]...

Personally, I think using memcpy to copy a single byte from a string is a waste. I'd prefer any one of the following:
1. Directly reinterpreting from a string:
[code]
short hum = *(const short*)"HU";

printf( "hum=%d\n", hum );
[/code]

2. Bit-wise operations
[code]
short hum = ('H' << 8) | 'U';

printf( "hum=%d\n", hum );
[/code]

3. Unions
[code]
union {
short hum;
char buf[sizeof(short)];
} _;

_.buf[0] = 'H';
_.buf[1] = 'U';

printf( "hum=%d\n", _.hum );
[/code] Edited by fastcall22
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It's quite confusing since there are only exemples of memset() and memcpy() involving "char" types, so I appreciate the article fastcall22. About the third parameter in both memset() and memcpy(), is it right to assume it's the size of a char and not the size of the data type, so they set or copy n bytes from the pointer specified?
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