Pushing Nibbles onto an integer stack in C - c

Question

I have an unsigned integer, and I want to push nibbles into it. For example, if I have nibbles with values 1, 2, 3, 4, 5, 6, 7 & 8, I want to be able to push the first nibble into my integer to make:
0x10000000 (268435456)
After the second push, I will have:
0x12000000 (301989888)
After the third push, I will have:
0x12300000 (305135616)
And so forth. Does anyone have a neat and cunning idea for how I might achieve this? The solution will need to be able to be given any number as a starting point and push onto the first available zero. So providing int 301989888 as a starting point and pushing 3 will result in 305135616. Pushing from MSB or LSB would be useful as well.
My apologies. It sounds like an exam question. It isn't - I just want to try an experiment, and I'm stuck before I start!
The answer ticked is perfect! I've modified it slightly as follows (just to make it self contained), and I'm as a happy as a tick!
#define left 0
#define right 1
void push(unsigned* number, int nibble,int direction){
int i, shift;
if (direction){
for (i = 28; i >= 0; i -= 4){
if (!(*number & (0xfU << i)))
shift = i;
}
}
else{
for (i = 0; i <= 28; i += 4){
if (!(*number & (0xfU << i)))
shift = i;
}
}
*number|=nibble<<shift;
}
Called as follows:
push(&x,nibble,left);
My apologies for the formatting.

Answer 1

There are two things you need to do - detect where to put the next nibble, and then put it there. For detection, you can mask & shift:
int nextLocation(uint32_t x)
{
int i;
for (i = 28; i >= 0; i -= 4)
{
if (!(x & (0xfU << i)))
return i;
}
return -1;
}
This function will return the amount of upshift you need to "push" your next nibble (or -1 if your integer is already full).
Then, you need to put in the new value (assuming that x is the value you want to push into and nibble is the value you want to push):
int shiftAmount = nextLocation(x);
x |= nibble << shiftAmount;
To push the other direction, you can just change the direction of the for loop in the nextLocation function:
for (i = 0; i <= 28; i += 4)

Answer 2

Here is an overly simple example which does what you want. It "pushes," but not in an automated way (if that's what you're looking for). But this demonstrates the concept (note: I combined each nibble into a byte).
#include <iostream>
using namespace std;
int main()
{
int x = (0x12 << 24) | (0x34 << 16) | (0x56 << 8) | (0x78);
cout<< hex << x << endl;
return 0;
}