ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x34(%rsp),%rax
  push   %rax
  lea    0x38(%rsp),%rax
  push   %rax
  lea    0x3c(%rsp),%rax
  push   %rax
  lea    0x40(%rsp),%rax
  push   %rax
  lea    0x44(%rsp),%rax
  push   %rax
  lea    0x48(%rsp),%rax
  push   %rax
  lea    0x4c(%rsp),%r9
  lea    0x48(%rsp),%r8
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  add    $0x30,%rsp
  mov    0x14(%rsp),%eax
  add    0xc(%rsp),%eax
  add    0x1c(%rsp),%eax
  add    0x24(%rsp),%eax
  add    0x2c(%rsp),%eax
  add    0x34(%rsp),%eax
  lea    (%rax,%rax,2),%edx
  mov    0x18(%rsp),%eax
  add    0x10(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x28(%rsp),%eax
  add    0x30(%rsp),%eax
  lea    -0x1(%rdx,%rax,1),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %edx,%eax
  lea    0x9(%rax),%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x48,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string
name: retrieve_data
arguments: tag: D16
type: string
name: retrieve_data
arguments: tag: D20
type: string

tool: D6: Enter first 11 digits of a UPC: 

tool: D16: %1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d

tool: D20: Check digit: %d


assistant: ```c
int main(void) {
  int d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, first_sum, second_sum, total;
  printf("Enter first 11 digits of a UPC: ");
  scanf("%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d", &d, &i1, &i2, &i3, &i4, &i5, &j1, &j2, &j3, &j4, &j5);
  first_sum = d + i2 + i4 + j1 + j3 + j5;
  second_sum = i1 + i3 + i5 + j2 + j4;
  total = (3 * first_sum) + second_sum;
  printf("Check digit: %d\n", 9 - ((total - 1) % 10));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  lea    D1(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x34(%rsp),%rax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  push   %rax
  lea    D19(%rip),%rdi
  lea    0x38(%rsp),%rax
  push   %rax
  lea    0x3c(%rsp),%rax
  push   %rax
  lea    0x40(%rsp),%rax
  push   %rax
  lea    0x44(%rsp),%rax
  push   %rax
  lea    0x48(%rsp),%rax
  push   %rax
  lea    0x4c(%rsp),%r9
  lea    0x48(%rsp),%r8
  call   <scanf@plt>
  mov    0x44(%rsp),%eax
  add    0x3c(%rsp),%eax
  add    $0x30,%rsp
  add    0x1c(%rsp),%eax
  add    0x24(%rsp),%eax
  lea    D16(%rip),%rdi
  add    0x2c(%rsp),%eax
  add    0x34(%rsp),%eax
  lea    (%rax,%rax,2),%edx
  mov    0x18(%rsp),%eax
  add    0x10(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x28(%rsp),%eax
  add    0x30(%rsp),%eax
  lea    -0x1(%rdx,%rax,1),%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %edx,%eax
  lea    0x9(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D1
data_type: string
name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D16
data_type: string

tool: D1: Enter first 11 digits of a UPC: 

tool: D19: %1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d

tool: D16: Check digit: %d


assistant: ```c
int main(void) {
  int d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, first_sum, second_sum, total;
  printf("Enter first 11 digits of a UPC: ");
  scanf("%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d", &d, &i1, &i2, &i3, &i4, &i5, &j1, &j2, &j3, &j4, &j5);
  first_sum = d + i2 + i4 + j1 + j3 + j5;
  second_sum = i1 + i3 + i5 + j2 + j4;
  total = (3 * first_sum) + second_sum;
  printf("Check digit: %d\n", 9 - ((total - 1) % 10));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x34(%rsp),%rax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  push   %rax
  lea    D8(%rip),%rdi
  lea    0x38(%rsp),%rax
  push   %rax
  lea    0x3c(%rsp),%rax
  push   %rax
  lea    0x40(%rsp),%rax
  push   %rax
  lea    0x44(%rsp),%rax
  push   %rax
  lea    0x48(%rsp),%rax
  push   %rax
  lea    0x4c(%rsp),%r9
  lea    0x48(%rsp),%r8
  call   <scanf@plt>
  mov    0x44(%rsp),%eax
  add    0x3c(%rsp),%eax
  add    $0x30,%rsp
  add    0x1c(%rsp),%eax
  add    0x24(%rsp),%eax
  lea    D10(%rip),%rdi
  add    0x2c(%rsp),%eax
  add    0x34(%rsp),%eax
  lea    (%rax,%rax,2),%edx
  mov    0x18(%rsp),%eax
  add    0x10(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x28(%rsp),%eax
  add    0x30(%rsp),%eax
  lea    -0x1(%rdx,%rax,1),%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %edx,%eax
  lea    0x9(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D16
data_type: string
name: retrieve_data
arguments: label: D8
data_type: string
name: retrieve_data
arguments: label: D10
data_type: string

tool: D16: Enter first 11 digits of a UPC: 

tool: D8: %1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d

tool: D10: Check digit: %d


assistant: ```c
int main(void) {
  int d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, first_sum, second_sum, total;
  printf("Enter first 11 digits of a UPC: ");
  scanf("%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d%1d", &d, &i1, &i2, &i3, &i4, &i5, &j1, &j2, &j3, &j4, &j5);
  first_sum = d + i2 + i4 + j1 + j3 + j5;
  second_sum = i1 + i3 + i5 + j2 + j4;
  total = (3 * first_sum) + second_sum;
  printf("Check digit: %d\n", 9 - ((total - 1) % 10));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mpath_recv_reply>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mpath_recv_reply_len@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jg     L4
  mov    -0x8(%rbp),%rax
  jmp    L52
L4:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L55
  mov    $0xffffffff,%eax
  jmp    L52
L55:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rsi
  mov    -0x18(%rbp),%ecx
  mov    -0x8(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <mpath_recv_reply_data@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L53
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x20(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0xffffffff,%eax
  jmp    L52
L53:
  mov    $0x0,%eax
L52:
  leave
  ret
```

assistant: ```c
int mpath_recv_reply(int fd, char **reply, unsigned int timeout) {
  int err;
  ssize_t len;
  *reply = NULL;
  len = mpath_recv_reply_len(fd, timeout);
  if (len <= 0)
    return len;
  *reply = malloc(len);
  if (!*reply)
    return -1;
  err = mpath_recv_reply_data(fd, *reply, len, timeout);
  if (err) {
    free(*reply);
    *reply = NULL;
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mpath_recv_reply>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %edx,%r13d
  movq   $0x0,(%rsi)
  mov    %edx,%esi
  call   <mpath_recv_reply_len@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jle    L46
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rsi
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L49
  mov    %r13d,%ecx
  mov    %rbx,%rdx
  mov    %r12d,%edi
  call   <mpath_recv_reply_data@plt>
  test   %eax,%eax
  jne    L43
L46:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L43:
  mov    0x0(%rbp),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  mov    $0xffffffff,%eax
  jmp    L46
L49:
  mov    $0xffffffff,%eax
  jmp    L46
```

assistant: ```c
int mpath_recv_reply(int fd, char **reply, unsigned int timeout) {
  int err;
  ssize_t len;
  *reply = NULL;
  len = mpath_recv_reply_len(fd, timeout);
  if (len <= 0)
    return len;
  *reply = malloc(len);
  if (!*reply)
    return -1;
  err = mpath_recv_reply_data(fd, *reply, len, timeout);
  if (err) {
    free(*reply);
    *reply = NULL;
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mpath_recv_reply>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movq   $0x0,(%rsi)
  mov    %edx,%esi
  call   <mpath_recv_reply_len@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jle    L20
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rsi
  test   %rax,%rax
  je     L36
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %ebp,%edi
  call   <mpath_recv_reply_data@plt>
  test   %eax,%eax
  jne    L46
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  mov    (%rbx),%rdi
  call   <free@plt>
  movq   $0x0,(%rbx)
  mov    $0xffffffff,%eax
  jmp    L20
L36:
  or     $0xffffffff,%eax
  jmp    L20
```

assistant: ```c
int mpath_recv_reply(int fd, char **reply, unsigned int timeout) {
  int err;
  ssize_t len;
  *reply = NULL;
  len = mpath_recv_reply_len(fd, timeout);
  if (len <= 0)
    return len;
  *reply = malloc(len);
  if (!*reply)
    return -1;
  err = mpath_recv_reply_data(fd, *reply, len, timeout);
  if (err) {
    free(*reply);
    *reply = NULL;
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<mpath_recv_reply>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movq   $0x0,(%rsi)
  mov    %edx,%esi
  call   <mpath_recv_reply_len@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jle    L48
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rsi
  test   %rax,%rax
  je     L22
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %ebp,%edi
  call   <mpath_recv_reply_data@plt>
  test   %eax,%eax
  jne    L20
L48:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  mov    (%rbx),%rdi
  call   <free@plt>
  movq   $0x0,(%rbx)
  mov    $0xffffffff,%eax
  jmp    L48
L22:
  or     $0xffffffff,%eax
  jmp    L48
```

assistant: ```c
int mpath_recv_reply(int fd, char **reply, unsigned int timeout) {
  int err;
  ssize_t len;
  *reply = NULL;
  len = mpath_recv_reply_len(fd, timeout);
  if (len <= 0)
    return len;
  *reply = malloc(len);
  if (!*reply)
    return -1;
  err = mpath_recv_reply_data(fd, *reply, len, timeout);
  if (err) {
    free(*reply);
    *reply = NULL;
    return -1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<assign_option>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <parse_option@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L11
  mov    $0x0,%eax
  call   <parse_error@plt>
  jmp    L16
L11:
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
L16:
  nop
  leave
  ret
```

assistant: ```c
void assign_option(int *target, const char *val, char *options) {
  int t = parse_option(val, options);
  if (t == -1) {
    parse_error();
  } else {
    *target = t;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<assign_option>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  call   <parse_option@plt>
  cmp    $0xffffffff,%eax
  je     L10
  mov    %eax,(%rbx)
L6:
  pop    %rbx
  ret
L10:
  mov    $0x0,%eax
  call   <parse_error@plt>
  jmp    L6
```

assistant: ```c
void assign_option(int *target, const char *val, char *options) {
  int t = parse_option(val, options);
  if (t == -1) {
    parse_error();
  } else {
    *target = t;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<assign_option>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  call   <parse_option@plt>
  cmp    $0xffffffff,%eax
  je     L2
  mov    %eax,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L2:
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <parse_error@plt>
```

assistant: ```c
void assign_option(int *target, const char *val, char *options) {
  int t = parse_option(val, options);
  if (t == -1) {
    parse_error();
  } else {
    *target = t;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<assign_option>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  call   <parse_option@plt>
  cmp    $0xffffffff,%eax
  je     L4
  mov    %eax,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L4:
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <parse_error@plt>
```

assistant: ```c
void assign_option(int *target, const char *val, char *options) {
  int t = parse_option(val, options);
  if (t == -1) {
    parse_error();
  } else {
    *target = t;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<analyzeResult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L14
L46:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L46
  cmpl   $0x0,-0x10(%rbp)
  jne    L50
  movss  D23(%rip),%xmm0
  jmp    L65
L50:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x10(%rbp),%xmm1
  movsd  D29(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L40
L16:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L84
L40:
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  test   %rax,%rax
  jne    L16
  jmp    L91
L84:
  nop
L91:
  mov    -0x8(%rbp),%rax
  test   %rax,%rax
  js     L96
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  jmp    L2
L96:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2ss %rdx,%xmm0
  addss  %xmm0,%xmm0
L2:
  movss  D37(%rip),%xmm2
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  movss  D1(%rip),%xmm0
  subss  %xmm1,%xmm0
L65:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D23
type_guessed: byte[4]
name: retrieve_data
arguments: data_tag: D29
type_guessed: byte[8]
name: retrieve_data
arguments: data_tag: D37
type_guessed: f32
name: retrieve_data
arguments: data_tag: D1
type_guessed: f32

tool: {"D23": [0x00,0x32,0xc0,0xc6]}

tool: {"D29": [0xa0,0x99,0x99,0x99,0x99,0x99,0xa9,0x3f]}

tool: D37: 100

tool: D1: 64.82

assistant: ```c
float analyzeResult(uint32_t *Array, size_t len) {
  uint32_t elems;
  int32_t upper;
  size_t i;
  elems = 0;
  for (i = 0; i < len; i++)
    elems += Array[i];
  if (elems == 0)
    return -24601;
  upper = (int32_t)ceil(elems * (1. - 0.95));
  for (i = len; i-- > 0;) {
    if ((upper -= Array[i]) <= 0)
      break;
  }
  return (float)((float)64.82 - (float)i / (float)100.);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<analyzeResult>:
  endbr64
  test   %rsi,%rsi
  je     L32
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdi,%rax
  lea    (%rdi,%rsi,4),%rcx
  mov    $0x0,%edx
L66:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L66
  movss  D3(%rip),%xmm0
  test   %edx,%edx
  je     L37
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  mulsd  D22(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  lea    -0x1(%rbx),%rdx
L21:
  cmp    $0xffffffffffffffff,%rdx
  je     L36
  sub    0x0(%rbp,%rdx,4),%eax
  lea    -0x1(%rdx),%rcx
  test   %eax,%eax
  jle    L36
  mov    %rcx,%rdx
  jmp    L21
L36:
  test   %rdx,%rdx
  js     L73
  pxor   %xmm1,%xmm1
  cvtsi2ss %rdx,%xmm1
L15:
  divss  D44(%rip),%xmm1
  movss  D19(%rip),%xmm0
  subss  %xmm1,%xmm0
L37:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L73:
  mov    %rdx,%rax
  shr    %rax
  and    $0x1,%edx
  or     %rdx,%rax
  pxor   %xmm1,%xmm1
  cvtsi2ss %rax,%xmm1
  addss  %xmm1,%xmm1
  jmp    L15
L32:
  movss  D3(%rip),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: byte[4]
name: unpack_data
arguments: data_label: D22
data_type: byte[8]
name: unpack_data
arguments: data_label: D44
data_type: f32
name: unpack_data
arguments: data_label: D19
data_type: f32

tool: {"D3": [0x00,0x32,0xc0,0xc6]}

tool: {"D22": [0xa0,0x99,0x99,0x99,0x99,0x99,0xa9,0x3f]}

tool: D44: 100

tool: D19: 64.82

assistant: ```c
float analyzeResult(uint32_t *Array, size_t len) {
  uint32_t elems;
  int32_t upper;
  size_t i;
  elems = 0;
  for (i = 0; i < len; i++)
    elems += Array[i];
  if (elems == 0)
    return -24601;
  upper = (int32_t)ceil(elems * (1. - 0.95));
  for (i = len; i-- > 0;) {
    if ((upper -= Array[i]) <= 0)
      break;
  }
  return (float)((float)64.82 - (float)i / (float)100.);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<analyzeResult>:
  endbr64
  test   %rsi,%rsi
  je     L30
  push   %rbp
  mov    %rdi,%rax
  mov    %rdi,%rbp
  lea    (%rdi,%rsi,4),%rcx
  push   %rbx
  xor    %edx,%edx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L7:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L7
  movss  D45(%rip),%xmm0
  test   %edx,%edx
  je     L29
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  mulsd  D0(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  lea    -0x1(%rbx),%rdx
  nopl   0x0(%rax)
L27:
  sub    0x0(%rbp,%rdx,4),%eax
  test   %eax,%eax
  jle    L41
  sub    $0x1,%rdx
  jae    L27
  movss  D35(%rip),%xmm0
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L41:
  movss  D47(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %rdx,%xmm1
  divss  D49(%rip),%xmm1
  add    $0x8,%rsp
  subss  %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L30:
  movss  D45(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D45
type: byte[4]
name: disassemble_data
arguments: data_label: D0
type: byte[8]
name: disassemble_data
arguments: data_label: D35
type: byte[4]
name: disassemble_data
arguments: data_label: D47
type: f32
name: disassemble_data
arguments: data_label: D49
type: f32

tool: {"D45": [0x00,0x32,0xc0,0xc6]}

tool: {"D0": [0xa0,0x99,0x99,0x99,0x99,0x99,0xa9,0x3f]}

tool: {"D35": [0x0a,0xd7,0x23,0xdc]}

tool: D47: 64.82

tool: D49: 100

assistant: ```c
float analyzeResult(uint32_t *Array, size_t len) {
  uint32_t elems;
  int32_t upper;
  size_t i;
  elems = 0;
  for (i = 0; i < len; i++)
    elems += Array[i];
  if (elems == 0)
    return -24601;
  upper = (int32_t)ceil(elems * (1. - 0.95));
  for (i = len; i-- > 0;) {
    if ((upper -= Array[i]) <= 0)
      break;
  }
  return (float)((float)64.82 - (float)i / (float)100.);
}
```
user: What is the c source code of the assembly code below:

```asm
<analyzeResult>:
  endbr64
  test   %rsi,%rsi
  je     L37
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    -0x1(%rsi),%rbx
  sub    $0x8,%rsp
  cmp    $0x2,%rbx
  jbe    L21
  mov    %rsi,%rdx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%rdx
  shl    $0x4,%rdx
  add    %rdi,%rdx
  nopl   0x0(%rax)
L18:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rdx,%rax
  jne    L18
  movdqa %xmm0,%xmm1
  mov    %rsi,%rdx
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffc,%rdx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L14
L93:
  lea    0x1(%rdx),%rdi
  lea    0x0(,%rdx,4),%rcx
  add    0x0(%rbp,%rdx,4),%eax
  cmp    %rdi,%rsi
  jbe    L14
  add    $0x2,%rdx
  add    0x4(%rbp,%rcx,1),%eax
  cmp    %rdx,%rsi
  jbe    L14
  add    0x8(%rbp,%rcx,1),%eax
L14:
  movss  D44(%rip),%xmm0
  test   %eax,%eax
  je     L87
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D38(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <ceil@plt>
  nopw   0x0(%rax,%rax,1)
L82:
  sub    0x0(%rbp,%rbx,4),%eax
  test   %eax,%eax
  jle    L24
  sub    $0x1,%rbx
  jae    L82
  movss  D18(%rip),%xmm0
L87:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L24:
  movss  D54(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %rbx,%xmm1
  divss  D49(%rip),%xmm1
  add    $0x8,%rsp
  subss  %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L37:
  movss  D44(%rip),%xmm0
  ret
L21:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L93
```

tool_calls: name: retrieve_data
arguments: data_label: D44
data_type: byte[4]
name: retrieve_data
arguments: data_label: D38
data_type: byte[8]
name: retrieve_data
arguments: data_label: D18
data_type: byte[4]
name: retrieve_data
arguments: data_label: D54
data_type: f32
name: retrieve_data
arguments: data_label: D49
data_type: f32

tool: {"D44": [0x00,0x32,0xc0,0xc6]}

tool: {"D38": [0xa0,0x99,0x99,0x99,0x99,0x99,0xa9,0x3f]}

tool: {"D18": [0x0a,0xd7,0x23,0xdc]}

tool: D54: 64.82

tool: D49: 100

assistant: ```c
float analyzeResult(uint32_t *Array, size_t len) {
  uint32_t elems;
  int32_t upper;
  size_t i;
  elems = 0;
  for (i = 0; i < len; i++)
    elems += Array[i];
  if (elems == 0)
    return -24601;
  upper = (int32_t)ceil(elems * (1. - 0.95));
  for (i = len; i-- > 0;) {
    if ((upper -= Array[i]) <= 0)
      break;
  }
  return (float)((float)64.82 - (float)i / (float)100.);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<socket_endwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L11
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <shutdown@plt>
  jmp    L12
L11:
  nop
L12:
  leave
  ret
```

assistant: ```c
void socket_endwrite(int fd) {
  if (fd < 0)
    return;
  shutdown(fd, SHUT_WR);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<socket_endwrite>:
  endbr64
  test   %edi,%edi
  js     L8
  sub    $0x8,%rsp
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  call   <shutdown@plt>
  add    $0x8,%rsp
  ret
L8:
  ret
```

assistant: ```c
void socket_endwrite(int fd) {
  if (fd < 0)
    return;
  shutdown(fd, SHUT_WR);
}
```
user: What is the c source code of the assembly code below:

```asm
<socket_endwrite>:
  endbr64
  test   %edi,%edi
  js     L2
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  jmp    1050 <shutdown@plt>
  cs nopw 0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void socket_endwrite(int fd) {
  if (fd < 0)
    return;
  shutdown(fd, SHUT_WR);
}
```