assembly - How can I store 1byte value in 64bit register only using 64bit registers?

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I need to write a pixelization
assembly code ONLY using %rax, %rbx, %rcx, %rdx, %rsi, and %rdi (also %rsp and
%rbp)

So I've first wrote code in C and changed
any other registers into 64 bit registers, but at the point below when I change the
register it gives Segmentation default

C
code:

*temp =
            b;

*(temp + 1) = g; 
*(temp + 2) =
            r;

Assembly code By
gcc:

movq -48(%rbp), %rax
            
movl %eax, %edx
movq -16(%rbp), %rax 
movb %dl, (%rax)
            

movq -16(%rbp), %rax 
addq $1, %rax
movq
            -56(%rbp), %rdx 
movb %dl, (%rax)
movq -16(%rbp),
            %rax
addq $2, %rax
movq -64(%rbp), %rdx 
movb %dl,
            (%rax)

Changed
%dl to %rdx:

movq -16(%rbp),
            %rax
movq -48(%rbp), %rdx
movzbq (%rdx), %rbx
movq %rbx,
            (%rax)
movq -16(%rbp), %rax
addq $1, %rax
movq -56(%rbp),
            %rdx
movzbq (%rdx), %rbx

movq %rbx, (%rax)
movq
            -16(%rbp), %rax
addq $2, %rax
movq -64(%rbp), %rdx
movzbq
            (%rdx), %rbx
movq %rbx, (%rax)

Answer

I think you want to do something
like:

 t = r &
            0xff;
 u = *temp & ~ 0xfful;
 *temp = u | t;
 t = (g
            & 0xff) << 8;
 u = *temp & ~ 0xff00ul;
 *temp = u |
            t;
 t = (b & 0xff) << 16; 
 u = *temp &
            ~0xff00000ull;
 *temp = u |
            t;

You
should be able to write this with 64bit regs only. You should also be able to find a
whole bunch of ways to make this way smaller than this.