c - sret 到底是什么意思？

Question

我比较了 2 个返回结构的 C 函数。我们知道，在 ABI 级别，大型结构将通过指针作为第一个函数参数传递。

struct S {
    int words[8];
};

struct S fsret() {
    struct S s;
    s.words[0] = 1;
    return s;
}

void fout(struct S* s) {
    s->words[0] = 1;
}

对于这些函数，我检查了 x86_64 Linux 和 Windows 的程序集。 fsret 声明为 void @fsret(%struct.S* sret %s)。

比较这两个变体，在被调用方没有区别。然而，在函数内部，fsret 额外地将它的第一个参数(指向结构的指针)复制到 RAX 寄存器。为什么？

Answer 1

原因在于this审查差异:

if (Subtarget->is64Bit() || Subtarget->isTargetKnownWindowsMSVC()) {
  for (unsigned i = 0, e = ArgLocs.size(); i != e; ++i) {
    // The x86-64 ABIs require that for returning structs by value we copy
    // the sret argument into %rax/%eax (depending on ABI) for the return.
    // Win32 requires us to put the sret argument to %eax as well.
    // Save the argument into a virtual register so that we can access it
    // from the return points.

因此被调用者必须填充调用者提供的内存并返回传递给它的指针。

x86_64 证实了这一点r252 System V ABI 文档

Returning of Values The returning of values is done according to the following algorithm:

1. Classify the return type with the classification algorithm.
2. If the type has class MEMORY (ndMarco: i.e. big stuff), then the caller provides space for the return value and passes the address of this storage in %rdi as if it were the first argument to the function. In effect, this address becomes a “hidden” first argument. This storage must not overlap any data visible to the callee through other names than this argument. On return %rax will contain the address that has been passed in by the caller in %rdi.