How does a 64-bit computer change one byte in memory?

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If a C program changes one byte in a byte array, what machine instructions take place? Does the hardware need read 8 bytes, change a byte, and store it (using 2 memory operations)?

Edit: Specifically on x86-64 architecture

x86-64 assembly memory cpu-architecture c
2 Answers

On x86-64, the hardware will read one cache line, modify the byte in cache, and eventually that cache line will be written back to memory.

The main reason for the write-back to happen is that the CPU needs the cache line for other data. There are explicit instructions to force the write-back, but a C compiler would be unlikely to use those. It slows down the CPU to force an unnecessary write.

MSalters_154160

It all depends on the compiler, optimisations and etc. Just try to compile and to disassemble. As an example we will compile the following code:

#include <stdio.h>

int main() {
    char a[] = "01234567890";
    a[5] = 'A';
    printf("%s\n", a);
}

// gcc -o main -std=c11 -Wall -Wextra -O0 main.c

We get the disassemby by objdump:

 6c1:   48 b8 30 31 32 33 34    movabs rax,0x3736353433323130
 6c8:   35 36 37 
 6cb:   48 89 45 ec             mov    QWORD PTR [rbp-0x14],rax
 6cf:   c7 45 f4 38 39 30 00    mov    DWORD PTR [rbp-0xc],0x303938
 6d6:   c6 45 f1 41             mov    BYTE PTR [rbp-0xf],0x41
 6da:   48 8d 45 ec             lea    rax,[rbp-0x14]
 6de:   48 89 c7                mov    rdi,rax

// objdump -d ./main -Mintel | less
user_92102550
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