hello friend
noticing that my previous post was too short and complicated to understand i
have decided to give a better reply to you. unions save a lot of time and
handle complicated tasks easily and urs seems similar to that so here is a
simple example of unions for you with its output:
unions.c:
========================================
#include <stdio.h>
union vals {
unsigned short x;
unsigned char y[1];
} myvals;
int main() {
myvals.y[0] = 1; //lsb -- 00000001
myvals.y[1] = 21; //msb --00010101
printf("%d", myvals.x); // displaying 16 bit uint 0001010100000001
printf("\n %d", sizeof(myvals));
}
==========================================
and its output is :
5377
2
The size of a union is equal to the size of it's largest data member. in this
case it is 2 bytes bcoz of 16bit uint. and the uint and the array of char share
the same memory space. hence using array of 2 1 byte uchars we can easily
access the lsb and msb of the uint.
i hope you understand this time what i am trying to say and this works for you
too. sorry for the last post which was not very useful. i was using borland
turbo c++ 3.0's c compiler on my machine.
bdrmachine <[EMAIL PROTECTED]> wrote: Can
someone give me a C example of how to combine 2 unsigned chars (msb
and lsb) and store as a unsigned int (L)? It seams the technique I'm
using is more involved then it should be. Would "L= msb<<8 + lsb;"
work? I'm working on a program for a 8 bit microcontroller. Unsigned
char's are 8 bits long and unsigned int's are 16 bits.
Thanks
Brian
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