Setting a bit in an uint64_t in c - c

Question

Hello I would like to set a bit ( value ) at a pos ( from left to right ). Here is my code that doesn't work for uint64_t ( here it should return 0 and not 1 ) but when I change values to make it works ( with the same logic ) with uint8_t it works. Any ideas? Please.
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <stdbool.h>
#include <inttypes.h>
int main()
{
uint64_t x = (uint64_t)1;
int pos = 63;
bool value = 0;
uint8_t z = x;
z = z >> (64-pos);
z = z << (64-pos);
uint64_t a = (uint64_t)1;
uint64_t d = (uint64_t)0;
uint64_t y;
uint64_t c = d;
uint64_t b = x;
b = b << (pos+1);
b = b >> (pos+1);
if (value == 1)
{
y = a;
y = y << (63-pos);
}
else
{
y=d;
}
c = c|z|y|b;
printf("%lld",c);
return c;
}
EDIT : I think there is a misunderstood ( I haven't be clear enough, my bad ), actually I have x that is an uint64_t and I have a int pos that is the position of one bit in x, and I have value that is a boolean ( 1 or 0 ) and if value is 1 the bit at the pos in x must become/stay 1 and if value is 0 the bit at the pos in x must become/stay 0.

Answer 1

To set the bit pos in value, you use
value |= ((uint64_t)1) << pos;
What is the rest of your code supposed to do?
For the updated question, see this answer.

Answer 2

//sets the bit pos to the value. if value is == 0 it zeroes the bit if value !=0 it sets the bit
void setbit(uint64_t *val, int bit, int value)
{
if(value)
{
*val |= ((uint64_t)1 << bit);
}
else
{
*val &= ~((uint64_t)1 << bit);
}
}
//sets nbits at location pos to the value of the first nbits of the value parameter.pos + nbits < 63
void setbits(uint64_t *val, int pos, int nbits, uint64_t value)
{
uint64_t mask = ((uint64_t)((uint64_t)1 << nbits) - 1);
*val &= ~(mask << pos);
*val |= (val & mask) << pos;
}
And usage with bool
#include <stdbool.h>
uint64_t obj;
/* ... */
setbit(&obj, 5, true); //sets the 5th bit to 1
setbit(&obj, 7, false); //sets the 7th bit to 0