ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<assert_bc_insn_size>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rdi
  xor    %eax,%eax
  mov    %sil,0x5(%rsp)
  xor    %esi,%esi
  mov    %ax,0x6(%rsp)
  call   <bc_insn_size@plt>
  mov    %r12,%rdx
  mov    $0x10,%esi
  lea    D5(%rip),%rdi
  mov    %eax,%ecx
  call   <__assert_int_equals@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type_infered: string

tool: D5: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/headius/jato/refs/heads/master/test/unit/vm/bytecodes-test.c

assistant: ```c
void assert_bc_insn_size(unsigned long expected_len, unsigned char opc) {
  unsigned char code[] = {opc, 0x00, 0x00};
  __assert_int_equals("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/headius/jato/refs/heads/master/test/unit/vm/bytecodes-test.c", 16, (expected_len), (bc_insn_size(code, 0)));
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<assert_bc_insn_size>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rdi
  xor    %eax,%eax
  mov    %sil,0x5(%rsp)
  xor    %esi,%esi
  mov    %ax,0x6(%rsp)
  call   <bc_insn_size@plt>
  mov    %r12,%rdx
  mov    $0x10,%esi
  lea    D4(%rip),%rdi
  mov    %eax,%ecx
  call   <__assert_int_equals@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D4
type: string

tool: D4: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/headius/jato/refs/heads/master/test/unit/vm/bytecodes-test.c

assistant: ```c
void assert_bc_insn_size(unsigned long expected_len, unsigned char opc) {
  unsigned char code[] = {opc, 0x00, 0x00};
  __assert_int_equals("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/headius/jato/refs/heads/master/test/unit/vm/bytecodes-test.c", 16, (expected_len), (bc_insn_size(code, 0)));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<redirect_io>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L8
  movl   $0x180,-0x24(%rbp)
L8:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <fflush@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D23(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0xffffffff,-0x8(%rbp)
  jne    L36
  mov    $0xffffffff,%eax
  jmp    L2
L36:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <fileno@plt>
  mov    %eax,%edi
  call   <dup@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <fileno@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <dup2@plt>
  cmp    $0xffffffff,%eax
  jne    L15
  mov    $0xfffffffe,%eax
  jmp    L2
L15:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    -0x4(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int redirect_io(const char *filename, FILE *fh, mode_t mask) {
  if (mask == 0)
    mask = 0600;
  fflush(fh);
  int fp = open(filename, O_RDWR | O_CREAT | O_APPEND, mask);
  if (fp == -1)
    return -1;
  int save_out = dup(fileno(fh));
  if (dup2(fp, fileno(fh)) == -1)
    return -2;
  close(fp);
  return save_out;
}
```
user: What is the c source code of the assembly code below?

```asm
<redirect_io>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %edx,%ebx
  test   %edx,%edx
  mov    $0x180,%eax
  cmove  %eax,%ebx
  mov    %rsi,%rdi
  call   <fflush@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%esi
  mov    D25(%rip),%rax
  or     (%rax),%esi
  mov    D24(%rip),%rax
  or     (%rax),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <open@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L33
  mov    %rbp,%rdi
  call   <fileno@plt>
  mov    %eax,%edi
  call   <dup@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <fileno@plt>
  mov    %eax,%esi
  mov    %ebx,%edi
  call   <dup2@plt>
  cmp    $0xffffffff,%eax
  je     L27
  mov    %ebx,%edi
  call   <close@plt>
L10:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  mov    %eax,%r12d
  jmp    L10
L27:
  mov    $0xfffffffe,%r12d
  jmp    L10
```

assistant: ```c
int redirect_io(const char *filename, FILE *fh, mode_t mask) {
  if (mask == 0)
    mask = 0600;
  fflush(fh);
  int fp = open(filename, O_RDWR | O_CREAT | O_APPEND, mask);
  if (fp == -1)
    return -1;
  int save_out = dup(fileno(fh));
  if (dup2(fp, fileno(fh)) == -1)
    return -2;
  close(fp);
  return save_out;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<redirect_io>:
  endbr64
  push   %r13
  test   %edx,%edx
  mov    %rdi,%r13
  mov    $0x180,%eax
  push   %r12
  mov    %rsi,%rdi
  mov    %edx,%r12d
  push   %rbp
  cmove  %eax,%r12d
  mov    %rsi,%rbp
  call   <fflush@plt>
  mov    D1(%rip),%rax
  mov    %r12d,%edx
  mov    %r13,%rdi
  mov    (%rax),%esi
  mov    D11(%rip),%rax
  or     (%rax),%esi
  mov    D4(%rip),%rax
  or     (%rax),%esi
  call   <open@plt>
  cmp    $0xffffffff,%eax
  je     L39
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <fileno@plt>
  mov    %eax,%edi
  call   <dup@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <fileno@plt>
  mov    %r12d,%edi
  mov    %eax,%esi
  call   <dup2@plt>
  cmp    $0xffffffff,%eax
  je     L17
  mov    %r12d,%edi
  call   <close@plt>
L5:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L39:
  mov    $0xffffffff,%r13d
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L17:
  mov    $0xfffffffe,%r13d
  jmp    L5
```

assistant: ```c
int redirect_io(const char *filename, FILE *fh, mode_t mask) {
  if (mask == 0)
    mask = 0600;
  fflush(fh);
  int fp = open(filename, O_RDWR | O_CREAT | O_APPEND, mask);
  if (fp == -1)
    return -1;
  int save_out = dup(fileno(fh));
  if (dup2(fp, fileno(fh)) == -1)
    return -2;
  close(fp);
  return save_out;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<redirect_io>:
  endbr64
  push   %r13
  test   %edx,%edx
  mov    %rdi,%r13
  mov    $0x180,%eax
  push   %r12
  mov    %rsi,%rdi
  mov    %edx,%r12d
  push   %rbp
  cmove  %eax,%r12d
  mov    %rsi,%rbp
  call   <fflush@plt>
  mov    D13(%rip),%rax
  mov    %r12d,%edx
  mov    %r13,%rdi
  mov    (%rax),%esi
  mov    D29(%rip),%rax
  or     (%rax),%esi
  mov    D24(%rip),%rax
  or     (%rax),%esi
  call   <open@plt>
  cmp    $0xffffffff,%eax
  je     L10
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <fileno@plt>
  mov    %eax,%edi
  call   <dup@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <fileno@plt>
  mov    %r12d,%edi
  mov    %eax,%esi
  call   <dup2@plt>
  cmp    $0xffffffff,%eax
  je     L23
  mov    %r12d,%edi
  call   <close@plt>
L30:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L10:
  mov    $0xffffffff,%r13d
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L23:
  mov    $0xfffffffe,%r13d
  jmp    L30
```

assistant: ```c
int redirect_io(const char *filename, FILE *fh, mode_t mask) {
  if (mask == 0)
    mask = 0600;
  fflush(fh);
  int fp = open(filename, O_RDWR | O_CREAT | O_APPEND, mask);
  if (fp == -1)
    return -1;
  int save_out = dup(fileno(fh));
  if (dup2(fp, fileno(fh)) == -1)
    return -2;
  close(fp);
  return save_out;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<table_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <util_list_start@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <util_list_next@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L2
L11:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <util_list_remove@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <target_free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <util_list_next@plt>
  mov    %rax,-0x8(%rbp)
L2:
  cmpq   $0x0,-0x10(%rbp)
  jne    L11
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <util_list_free@plt>
  nop
  leave
  ret
```

assistant: ```c
void table_free(struct util_list *table) {
  struct target *target, *n;
  for (target = util_list_start(table), n = util_list_next(table, target); target != NULL; target = n, n = util_list_next(table, target)) {
    util_list_remove(table, target);
    target_free(target);
  }
  util_list_free(table);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<table_free>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <util_list_start@plt>
  mov    %rax,%rbp
  mov    %rax,%rsi
  mov    %r12,%rdi
  call   <util_list_next@plt>
  mov    %rax,%rbx
  test   %rbp,%rbp
  je     L12
L9:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <util_list_remove@plt>
  mov    %rbp,%rdi
  call   <target_free@plt>
  mov    %rbx,%rbp
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <util_list_next@plt>
  mov    %rax,%rbx
  test   %rbp,%rbp
  jne    L9
L12:
  mov    %r12,%rdi
  call   <util_list_free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void table_free(struct util_list *table) {
  struct target *target, *n;
  for (target = util_list_start(table), n = util_list_next(table, target); target != NULL; target = n, n = util_list_next(table, target)) {
    util_list_remove(table, target);
    target_free(target);
  }
  util_list_free(table);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<table_free>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <util_list_start@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <util_list_next@plt>
  test   %rbp,%rbp
  je     L10
  mov    %rax,%rbx
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <util_list_remove@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rbp
  call   <target_free@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <util_list_next@plt>
  mov    %rax,%rbx
  test   %rbp,%rbp
  jne    L13
L10:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <util_list_free@plt>
```

assistant: ```c
void table_free(struct util_list *table) {
  struct target *target, *n;
  for (target = util_list_start(table), n = util_list_next(table, target); target != NULL; target = n, n = util_list_next(table, target)) {
    util_list_remove(table, target);
    target_free(target);
  }
  util_list_free(table);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<table_free>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <util_list_start@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <util_list_next@plt>
  test   %rbp,%rbp
  je     L14
  mov    %rax,%rbx
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <util_list_remove@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rbp
  call   <target_free@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <util_list_next@plt>
  mov    %rax,%rbx
  test   %rbp,%rbp
  jne    L6
L14:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <util_list_free@plt>
```

assistant: ```c
void table_free(struct util_list *table) {
  struct target *target, *n;
  for (target = util_list_start(table), n = util_list_next(table, target); target != NULL; target = n, n = util_list_next(table, target)) {
    util_list_remove(table, target);
    target_free(target);
  }
  util_list_free(table);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<flushWindows>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L5
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <flushWindows>
  jmp    L11
L5:
  nop
L11:
  leave
  ret
```

assistant: ```c
void flushWindows(depth) {
  if (depth == 0)
    return;
  flushWindows(depth - 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<flushWindows>:
  endbr64
  test   %edi,%edi
  jne    L0
  ret
L0:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  mov    $0x0,%eax
  call   <flushWindows>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void flushWindows(depth) {
  if (depth == 0)
    return;
  flushWindows(depth - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<flushWindows>:
  endbr64
  ret
```

assistant: ```c
void flushWindows(depth) {
  if (depth == 0)
    return;
  flushWindows(depth - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<flushWindows>:
  endbr64
  ret
```

assistant: ```c
void flushWindows(depth) {
  if (depth == 0)
    return;
  flushWindows(depth - 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<p1788_fwd_pow_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x4(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <mpfr_pow@plt>
  leave
  ret
```

assistant: ```c
int p1788_fwd_pow_2(mpfr_ptr rop, mpfr_srcptr op, const void *env) {
  mpfr_srcptr op1 = (mpfr_srcptr)env;
  return mpfr_pow(rop, op1, op, MPFR_RNDN);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<p1788_fwd_pow_2>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%eax
  mov    %rdx,%rsi
  mov    D1(%rip),%rdx
  mov    (%rdx),%ecx
  mov    %eax,%edx
  call   <mpfr_pow@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int p1788_fwd_pow_2(mpfr_ptr rop, mpfr_srcptr op, const void *env) {
  mpfr_srcptr op1 = (mpfr_srcptr)env;
  return mpfr_pow(rop, op1, op, MPFR_RNDN);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<p1788_fwd_pow_2>:
  endbr64
  mov    D4(%rip),%rax
  mov    %rdx,%r8
  mov    %esi,%edx
  mov    %r8d,%esi
  mov    (%rax),%ecx
  jmp    1050 <mpfr_pow@plt>
```

assistant: ```c
int p1788_fwd_pow_2(mpfr_ptr rop, mpfr_srcptr op, const void *env) {
  mpfr_srcptr op1 = (mpfr_srcptr)env;
  return mpfr_pow(rop, op1, op, MPFR_RNDN);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<p1788_fwd_pow_2>:
  endbr64
  mov    D8(%rip),%rax
  mov    %rdx,%r8
  mov    %esi,%edx
  mov    %r8d,%esi
  mov    (%rax),%ecx
  jmp    1050 <mpfr_pow@plt>
```

assistant: ```c
int p1788_fwd_pow_2(mpfr_ptr rop, mpfr_srcptr op, const void *env) {
  mpfr_srcptr op1 = (mpfr_srcptr)env;
  return mpfr_pow(rop, op1, op, MPFR_RNDN);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ring_rot_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L16
L17:
  mov    -0x10(%rbp),%eax
  rol    $0x4,%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  xor    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L16:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned ring_rot_hash(unsigned char *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++) {
    h = (h << 4) ^ (h >> 28) ^ p[i];
  }
  return h;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ring_rot_hash>:
  endbr64
  test   %esi,%esi
  jle    L7
  mov    %rdi,%rdx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rsi
  mov    $0x0,%eax
L4:
  rol    $0x4,%eax
  movzbl (%rdx),%ecx
  xor    %ecx,%eax
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L4
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned ring_rot_hash(unsigned char *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++) {
    h = (h << 4) ^ (h >> 28) ^ p[i];
  }
  return h;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ring_rot_hash>:
  endbr64
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl (%rdi),%edx
  rol    $0x4,%eax
  add    $0x1,%rdi
  xor    %edx,%eax
  cmp    %rcx,%rdi
  jne    L18
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned ring_rot_hash(unsigned char *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++) {
    h = (h << 4) ^ (h >> 28) ^ p[i];
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ring_rot_hash>:
  endbr64
  test   %esi,%esi
  jle    L14
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl (%rdi),%edx
  rol    $0x4,%eax
  add    $0x1,%rdi
  xor    %edx,%eax
  cmp    %rdi,%rcx
  jne    L15
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned ring_rot_hash(unsigned char *key, int len) {
  unsigned char *p = key;
  unsigned h = 0;
  int i;
  for (i = 0; i < len; i++) {
    h = (h << 4) ^ (h >> 28) ^ p[i];
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  mov    $0x4,%edi
  call   <omp_set_num_threads@plt>
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  cmp    $0xf423f,%eax
  jg     L14
  movslq %eax,%rdx
  movsd  (%rsp,%rdx,8),%xmm1
  mov    %eax,%edx
  movsd  D43(%rip),%xmm6
  movsd  D9(%rip),%xmm5
  movsd  D4(%rip),%xmm4
  movsd  D15(%rip),%xmm3
L22:
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  addsd  %xmm6,%xmm0
  mulsd  %xmm5,%xmm0
  mulsd  %xmm0,%xmm0
  addsd  %xmm4,%xmm0
  movapd %xmm3,%xmm2
  divsd  %xmm0,%xmm2
  addsd  %xmm2,%xmm1
  add    $0x4,%edx
  cmp    $0xf423f,%edx
  jle    L22
  cltq
  movsd  %xmm1,(%rsp,%rax,8)
L14:
  pxor   %xmm0,%xmm0
  addsd  (%rsp),%xmm0
  addsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  addsd  0x18(%rsp),%xmm0
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x38,%rsp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D43
type_guessed: f64
name: retrieve_data
arguments: tag: D9
type_guessed: byte[8]
name: retrieve_data
arguments: tag: D4
type_guessed: f64
name: retrieve_data
arguments: tag: D15
type_guessed: byte[8]
name: retrieve_data
arguments: tag: D39
type_guessed: string

tool: D43: 0.5

tool: {"D9": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D4: 1

tool: {"D15": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xd0,0x3e]}

tool: D39: pi = %6.12f


assistant: ```c
void main() {
  const int NUM_THREADS = 4;
  const long num_steps = 1000000;
  double step = 1.0 / ((double)num_steps);
  double sum[NUM_THREADS];
  for (int i = 0; i < NUM_THREADS; i++)
    sum[i] = 0;
  omp_set_num_threads(NUM_THREADS);
#pragma omp parallel
  {
    int id = omp_get_thread_num();
    for (int i = id; i < num_steps; i += NUM_THREADS) {
      double x = (i + 0.5) * step;
      sum[id] += 4.0 * step / (1.0 + x * x);
    }
  }
  double pi = 0;
  for (int i = 0; i < NUM_THREADS; i++)
    pi += sum[i];
  printf("pi = %6.12f\n", pi);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x4,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  call   <omp_set_num_threads@plt>
  xor    %eax,%eax
  call   <omp_get_thread_num@plt>
  cmp    $0xf423f,%eax
  jg     L21
  movslq %eax,%rdx
  movsd  D23(%rip),%xmm6
  movsd  D29(%rip),%xmm5
  movsd  (%rsp,%rdx,8),%xmm1
  movsd  D7(%rip),%xmm4
  movsd  D33(%rip),%xmm3
  nopl   0x0(%rax,%rax,1)
L12:
  pxor   %xmm0,%xmm0
  movapd %xmm3,%xmm2
  cvtsi2sd %eax,%xmm0
  add    $0x4,%eax
  addsd  %xmm6,%xmm0
  mulsd  %xmm5,%xmm0
  mulsd  %xmm0,%xmm0
  addsd  %xmm4,%xmm0
  divsd  %xmm0,%xmm2
  addsd  %xmm2,%xmm1
  cmp    $0xf423f,%eax
  jle    L12
  movsd  %xmm1,(%rsp,%rdx,8)
L21:
  pxor   %xmm0,%xmm0
  addsd  (%rsp),%xmm0
  addsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  addsd  0x18(%rsp),%xmm0
  lea    D16(%rip),%rdi
  add    $0x38,%rsp
  jmp    10a0 <printf@plt>
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D23
type: f64
name: parse_data
arguments: data_tag: D29
type: byte[8]
name: parse_data
arguments: data_tag: D7
type: f64
name: parse_data
arguments: data_tag: D33
type: byte[8]
name: parse_data
arguments: data_tag: D16
type: string

tool: D23: 0.5

tool: {"D29": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D7: 1

tool: {"D33": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xd0,0x3e]}

tool: D16: pi = %6.12f


assistant: ```c
void main() {
  const int NUM_THREADS = 4;
  const long num_steps = 1000000;
  double step = 1.0 / ((double)num_steps);
  double sum[NUM_THREADS];
  for (int i = 0; i < NUM_THREADS; i++)
    sum[i] = 0;
  omp_set_num_threads(NUM_THREADS);
#pragma omp parallel
  {
    int id = omp_get_thread_num();
    for (int i = id; i < num_steps; i += NUM_THREADS) {
      double x = (i + 0.5) * step;
      sum[id] += 4.0 * step / (1.0 + x * x);
    }
  }
  double pi = 0;
  for (int i = 0; i < NUM_THREADS; i++)
    pi += sum[i];
  printf("pi = %6.12f\n", pi);
}
```