Using the C programming language, it could be done as follows assuming that the two integers are less than 65535.
void take2IntegersAsOne(int x)
{
// int1 is stored in the bottom half of x, so take just that part.
int int1 = x & 0xFFFF;
// int2 is stored in the top half of x, so slide that part of the number
// into the bottom half, and take just that part.
int int2 = (x >> 16) & 0xFFFF
// use int1 and int2 here. They must both be less than 0xFFFF or 65535 in decimal
}
void pass2()
{
int int1 = 345;
int int2 = 2342;
take2Integers( int1 | (int2 << 16) );
}
This relies on the fact that in C an integer is stored in 4 bytes. So, the example uses the first two bytes to store one of the integers, and the next two bytes for the second. This does impose the limit though that each of the integers must have a small enough value so that they will each fit into just 2 bytes.
The shift operators << and >> are used to slide the bits of an integer up and down. Shifting by 16, moves the bits by two bytes (as there are 8 bits per byte).
Using 0xFFFF represents the bit pattern where all of the bits in the lower two bytes of the number are 1s So, ANDing (with with & operator) causes all the bits that are not in these bottom two bytes to be switched off (back to zero). This can be used to remove any parts of the 'other integer' from the one you're currently extracting.
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