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

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.