为何GCC编译的代码调用memmove而非memcpy？

在Linux系统使用GCC 12.2编译器，添加-nostdlib参数编译时，编译器提示缺少memcpy和memmove。我用汇编实现了一个简易memcpy，同时让memmove调用abort，目的是让代码始终使用memcpy。之后尝试用C实现自定义内存拷贝函数mymemcpy，通过#define memcpy mymemcpy替换标准memcpy，但运行时程序终止，发现GCC实际调用了memmove而非我实现的memcpy。Clang编译时会调用memcpy，但GCC的优化效果更好，我需要用它做优化构建。

可复现代码 #

dummy.cpp #

extern "C" {
void*malloc() { return 0; }
int read() { return 0; }
int write() { return 0; }
int memcpy() { return 0; }
int memmove() { return 0; }
}

main.cpp #

#include <unistd.h>
#include <cstdlib>
struct MyVector {
    void*p;
    long long position, length;
};

__attribute__ ((access(write_only, 1))) __attribute__((nonnull(1, 2)))
void mymemcpy(void *__restrict__ dest, const void *__restrict__ src, int size)
{
    const unsigned char *s = (const unsigned char*)src;
    unsigned char *d = (unsigned char*)dest;
    while(size--) *d++ = *s++;
}

//__attribute__ ((noinline))
int func(const char*file_from_disk, MyVector*v)
{
    if (v->position + 5 <= v->length ) {
        mymemcpy(v->p, file_from_disk, 5);
    }
    return 0;
}

char buf[4096];
extern "C"
int _start() {
    MyVector v{malloc(1024),0,1024};
    v.position += read(0, v.p, 1024-5);
    int len = read(0, buf, 4096);
    func(buf, &v);
    write(1, v.p, v.position);
}

编译命令 #

g++ -march=native -nostdlib -static -fno-exceptions -fno-rtti -O2 main.cpp dummy.cpp

编译结果验证（objdump输出） #

401040: e8 db 00 00 00          call   401120 <memmove>
40108d: e8 4e 00 00 00          call   4010e0 <malloc>
4010a3: e8 48 00 00 00          call   4010f0 <read>
4010ba: e8 31 00 00 00          call   4010f0 <read>
4010c5: e8 56 ff ff ff          call   401020 <_Z4funcPKcP8MyVector>
4010d5: e8 26 00 00 00          call   401100 <write>
402023: ff 11                   call   *(%rcx)

内容的提问来源于stack exchange，提问作者Stan