ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<my_get_big_nbr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <my_strlen@plt>
  xor    %r8d,%r8d
  cmp    $0xa,%eax
  jg     L49
  test   %eax,%eax
  jle    L49
  lea    -0x1(%rax),%ecx
  movslq %ecx,%rdx
  movzbl (%rbx,%rdx,1),%edx
  sub    $0x30,%edx
  cmp    $0x9,%dl
  ja     L49
  movsbl %dl,%r8d
  test   %ecx,%ecx
  je     L49
  lea    -0x2(%rax),%ecx
  movslq %ecx,%rdx
  movzbl (%rbx,%rdx,1),%edx
  sub    $0x30,%edx
  cmp    $0x9,%dl
  ja     L100
  movsbl %dl,%edx
  lea    (%rdx,%rdx,4),%edx
  lea    (%r8,%rdx,2),%edx
  xor    %r8d,%r8d
  test   %ecx,%ecx
  je     L35
  lea    -0x3(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0x64,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  lea    -0x4(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0x3e8,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  lea    -0x5(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0x2710,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  lea    -0x6(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L100
  movsbl %cl,%ecx
  imul   $0x186a0,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  lea    -0x7(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0xf4240,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  lea    -0x8(%rax),%esi
  movslq %esi,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0x989680,%ecx,%ecx
  add    %ecx,%edx
  test   %esi,%esi
  je     L35
  sub    $0x9,%eax
  movslq %eax,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  ja     L49
  movsbl %cl,%ecx
  imul   $0x5f5e100,%ecx,%ecx
  add    %ecx,%edx
  cmp    $0x1,%eax
  jne    L35
  movzbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L100
  movsbl %al,%r8d
  imul   $0x3b9aca00,%r8d,%r8d
  add    %edx,%r8d
L49:
  mov    %r8d,%eax
  pop    %rbx
  ret
L35:
  mov    %edx,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
L100:
  xor    %r8d,%r8d
  jmp    L49
```

assistant: ```c
unsigned int my_get_big_nbr(char *str) {
  int i;
  int pow;
  unsigned int pnb;
  unsigned int nb;
  i = my_strlen(str);
  if (i > 10)
    return (0);
  nb = 0;
  pow = 1;
  while (i > 0) {
    i = i - 1;
    if (str[i] < '0' || str[i] > '9')
      return (0);
    pnb = nb;
    nb = nb + (str[i] - '0') * pow;
    pow = pow * 10;
    if (nb < pnb)
      return (0);
  }
  return (nb);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x18(%rbp),%xmm1
  movss  -0x14(%rbp),%xmm0
  addss  %xmm0,%xmm1
  movss  -0x10(%rbp),%xmm0
  addss  %xmm0,%xmm1
  movss  -0xc(%rbp),%xmm0
  addss  %xmm0,%xmm1
  movss  %xmm1,-0x24(%rbp)
  movss  -0xc(%rbp),%xmm2
  movss  -0x10(%rbp),%xmm1
  movss  -0x14(%rbp),%xmm0
  mov    -0x18(%rbp),%eax
  movaps %xmm2,%xmm3
  movaps %xmm1,%xmm2
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  mov    $0x4,%edi
  call   <multiMin@plt>
  movaps %xmm0,%xmm1
  movss  -0x24(%rbp),%xmm0
  subss  %xmm1,%xmm0
  movss  D15(%rip),%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
type_guessed: string
name: retrieve_data
arguments: data_tag: D15
type_guessed: byte[4]
name: retrieve_data
arguments: data_tag: D5
type_guessed: string

tool: D16: %f
%f
%f
%f


tool: {"D15": [0x00,0x00,0x40,0x40]}

tool: D5: %f

assistant: ```c
int main(void) {
  float a, b, c, d;
  scanf("%f\n%f\n%f\n%f\n", &a, &b, &c, &d);
  printf("%f", (a + b + c + d - multiMin(4, a, b, c, d)) / 3);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rcx
  lea    0x1c(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  lea    0x24(%rsp),%r8
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  movss  0x18(%rsp),%xmm4
  movss  0x1c(%rsp),%xmm1
  movss  0x20(%rsp),%xmm2
  movss  0x24(%rsp),%xmm3
  movaps %xmm4,%xmm0
  addss  %xmm1,%xmm0
  addss  %xmm2,%xmm0
  addss  %xmm3,%xmm0
  movss  %xmm0,0xc(%rsp)
  movaps %xmm4,%xmm0
  mov    $0x4,%edi
  call   <multiMin@plt>
  movaps %xmm0,%xmm1
  movss  0xc(%rsp),%xmm0
  subss  %xmm1,%xmm0
  divss  D15(%rip),%xmm0
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D12
data_type: string
name: unpack_data
arguments: label: D15
data_type: byte[4]
name: unpack_data
arguments: label: D21
data_type: string

tool: D12: %f
%f
%f
%f


tool: {"D15": [0x00,0x00,0x40,0x40]}

tool: D21: %f

assistant: ```c
int main(void) {
  float a, b, c, d;
  scanf("%f\n%f\n%f\n%f\n", &a, &b, &c, &d);
  printf("%f", (a + b + c + d - multiMin(4, a, b, c, d)) / 3);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D14(%rip),%rdi
  lea    0x20(%rsp),%rcx
  lea    0x1c(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    0x24(%rsp),%r8
  call   <scanf@plt>
  movss  0x18(%rsp),%xmm0
  mov    $0x4,%edi
  movss  0x1c(%rsp),%xmm1
  movss  0x20(%rsp),%xmm2
  movss  0x24(%rsp),%xmm3
  movaps %xmm0,%xmm4
  addss  %xmm1,%xmm4
  addss  %xmm2,%xmm4
  addss  %xmm3,%xmm4
  movss  %xmm4,0xc(%rsp)
  call   <multiMin@plt>
  movss  0xc(%rsp),%xmm4
  lea    D17(%rip),%rdi
  subss  %xmm0,%xmm4
  movaps %xmm4,%xmm0
  divss  D4(%rip),%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
type: string
name: parse_data
arguments: data_tag: D17
type: string
name: parse_data
arguments: data_tag: D4
type: byte[4]

tool: D14: %f
%f
%f
%f


tool: D17: %f

tool: {"D4": [0x00,0x00,0x40,0x40]}

assistant: ```c
int main(void) {
  float a, b, c, d;
  scanf("%f\n%f\n%f\n%f\n", &a, &b, &c, &d);
  printf("%f", (a + b + c + d - multiMin(4, a, b, c, d)) / 3);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D19(%rip),%rdi
  lea    0x20(%rsp),%rcx
  lea    0x1c(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    0x24(%rsp),%r8
  call   <scanf@plt>
  movss  0x18(%rsp),%xmm0
  mov    $0x4,%edi
  movss  0x1c(%rsp),%xmm1
  movss  0x20(%rsp),%xmm2
  movss  0x24(%rsp),%xmm3
  movaps %xmm0,%xmm4
  addss  %xmm1,%xmm4
  addss  %xmm2,%xmm4
  addss  %xmm3,%xmm4
  movss  %xmm4,0xc(%rsp)
  call   <multiMin@plt>
  movss  0xc(%rsp),%xmm4
  lea    D23(%rip),%rdi
  subss  %xmm0,%xmm4
  movaps %xmm4,%xmm0
  divss  D17(%rip),%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D19
type_infered: string
name: read_data
arguments: tag: D23
type_infered: string
name: read_data
arguments: tag: D17
type_infered: byte[4]

tool: D19: %f
%f
%f
%f


tool: D23: %f

tool: {"D17": [0x00,0x00,0x40,0x40]}

assistant: ```c
int main(void) {
  float a, b, c, d;
  scanf("%f\n%f\n%f\n%f\n", &a, &b, &c, &d);
  printf("%f", (a + b + c + d - multiMin(4, a, b, c, d)) / 3);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<digit_from_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %al,-0x14(%rbp)
  cmpb   $0x2f,-0x14(%rbp)
  jle    L30
  cmpb   $0x39,-0x14(%rbp)
  jg     L30
  movzbl -0x14(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  cmp    %rax,-0x20(%rbp)
  setg   %al
  movzbl %al,%eax
  jmp    L1
L30:
  cmpq   $0xa,-0x20(%rbp)
  jg     L44
  mov    $0x0,%eax
  jmp    L1
L44:
  cmpb   $0x40,-0x14(%rbp)
  jle    L2
  cmpb   $0x5a,-0x14(%rbp)
  jg     L2
  movzbl -0x14(%rbp),%eax
  sub    $0x41,%eax
  mov    %al,-0x1(%rbp)
  jmp    L62
L2:
  cmpb   $0x60,-0x14(%rbp)
  jle    L48
  cmpb   $0x7a,-0x14(%rbp)
  jg     L48
  movzbl -0x14(%rbp),%eax
  sub    $0x61,%eax
  mov    %al,-0x1(%rbp)
  jmp    L62
L48:
  mov    $0x0,%eax
  jmp    L1
L62:
  movzbl -0x1(%rbp),%eax
  add    $0xa,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  movsbq -0x1(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  setg   %al
  movzbl %al,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
int digit_from_char(char c, long base, char *digit) {
  if (c >= '0' && c <= '9') {
    *digit = c - '0';
    return (*digit < base);
  }
  if (base <= 10) {
    return 0;
  }
  char p;
  if (c >= 'A' && c <= 'Z') {
    p = c - 'A';
  } else if (c >= 'a' && c <= 'z') {
    p = c - 'a';
  } else {
    return 0;
  }
  *digit = p + 10;
  return (p < base);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<digit_from_char>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L15
  mov    $0x0,%eax
  cmp    $0xa,%rsi
  jle    L19
  lea    -0x41(%rdi),%eax
  cmp    $0x19,%al
  jbe    L2
  lea    -0x61(%rdi),%ecx
  mov    $0x0,%eax
  cmp    $0x19,%cl
  ja     L19
  mov    %ecx,%eax
L2:
  lea    0xa(%rax),%ecx
  mov    %cl,(%rdx)
  movsbq %al,%rax
  cmp    %rsi,%rax
  setl   %al
  movzbl %al,%eax
L19:
  ret
L15:
  mov    %al,(%rdx)
  movsbq %al,%rax
  cmp    %rsi,%rax
  setl   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int digit_from_char(char c, long base, char *digit) {
  if (c >= '0' && c <= '9') {
    *digit = c - '0';
    return (*digit < base);
  }
  if (base <= 10) {
    return 0;
  }
  char p;
  if (c >= 'A' && c <= 'Z') {
    p = c - 'A';
  } else if (c >= 'a' && c <= 'z') {
    p = c - 'a';
  } else {
    return 0;
  }
  *digit = p + 10;
  return (p < base);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<digit_from_char>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L9
  xor    %r8d,%r8d
  cmp    $0xa,%rsi
  jle    L8
  lea    -0x41(%rdi),%eax
  cmp    $0x19,%al
  jbe    L25
  lea    -0x61(%rdi),%eax
  cmp    $0x19,%al
  ja     L8
L25:
  lea    0xa(%rax),%ecx
  movsbq %al,%rax
  xor    %r8d,%r8d
  mov    %cl,(%rdx)
  cmp    %rsi,%rax
  setl   %r8b
L8:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L9:
  mov    %al,(%rdx)
  movzbl %al,%eax
  xor    %r8d,%r8d
  cmp    %rsi,%rax
  setl   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int digit_from_char(char c, long base, char *digit) {
  if (c >= '0' && c <= '9') {
    *digit = c - '0';
    return (*digit < base);
  }
  if (base <= 10) {
    return 0;
  }
  char p;
  if (c >= 'A' && c <= 'Z') {
    p = c - 'A';
  } else if (c >= 'a' && c <= 'z') {
    p = c - 'a';
  } else {
    return 0;
  }
  *digit = p + 10;
  return (p < base);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<digit_from_char>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L22
  xor    %r8d,%r8d
  cmp    $0xa,%rsi
  jle    L4
  lea    -0x41(%rdi),%eax
  cmp    $0x19,%al
  jbe    L23
  lea    -0x61(%rdi),%eax
  cmp    $0x19,%al
  ja     L4
L23:
  lea    0xa(%rax),%ecx
  movsbq %al,%rax
  xor    %r8d,%r8d
  mov    %cl,(%rdx)
  cmp    %rsi,%rax
  setl   %r8b
L4:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L22:
  mov    %al,(%rdx)
  movzbl %al,%eax
  xor    %r8d,%r8d
  cmp    %rsi,%rax
  setl   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int digit_from_char(char c, long base, char *digit) {
  if (c >= '0' && c <= '9') {
    *digit = c - '0';
    return (*digit < base);
  }
  if (base <= 10) {
    return 0;
  }
  char p;
  if (c >= 'A' && c <= 'Z') {
    p = c - 'A';
  } else if (c >= 'a' && c <= 'z') {
    p = c - 'a';
  } else {
    return 0;
  }
  *digit = p + 10;
  return (p < base);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<iobuf_queuev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L47
L51:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %rax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L47:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L51
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <iobuf_reserve@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L65
  mov    $0xffffffff,%eax
  jmp    L55
L65:
  movl   $0x0,-0x14(%rbp)
  jmp    L19
L7:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L76
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    0x8(%rax),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %rax,-0x8(%rbp)
  jmp    L49
L76:
  nop
L49:
  addl   $0x1,-0x14(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L7
  mov    $0x0,%eax
L55:
  leave
  ret
```

assistant: ```c
int iobuf_queuev(struct iobuf *io, const struct iovec *iov, int iovcnt) {
  int i;
  size_t len = 0;
  char *buf;
  for (i = 0; i < iovcnt; i++)
    len += iov[i].iov_len;
  if ((buf = iobuf_reserve(io, len)) == NULL)
    return (-1);
  for (i = 0; i < iovcnt; i++) {
    if (iov[i].iov_len == 0)
      continue;
    memmove(buf, iov[i].iov_base, iov[i].iov_len);
    buf += iov[i].iov_len;
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<iobuf_queuev>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L15
  mov    %rsi,%rbx
  mov    %edx,%r12d
  mov    %rsi,%rax
  lea    -0x1(%rdx),%edx
  shl    $0x4,%rdx
  lea    0x10(%rsi,%rdx,1),%rdx
  mov    $0x0,%esi
L16:
  add    (%rax),%rsi
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L16
  call   <iobuf_reserve@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L58
  mov    $0x0,%ebp
  jmp    L81
L15:
  mov    $0x0,%esi
  call   <iobuf_reserve@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  jmp    L42
L61:
  add    $0x1,%ebp
  add    $0x10,%rbx
  cmp    %ebp,%r12d
  jle    L85
L81:
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L61
  mov    0x8(%rbx),%esi
  mov    %r13,%rdi
  call   <memmove@plt>
  add    (%rbx),%r13
  jmp    L61
L85:
  mov    $0x0,%eax
L42:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L58:
  mov    $0xffffffff,%eax
  jmp    L42
```

assistant: ```c
int iobuf_queuev(struct iobuf *io, const struct iovec *iov, int iovcnt) {
  int i;
  size_t len = 0;
  char *buf;
  for (i = 0; i < iovcnt; i++)
    len += iov[i].iov_len;
  if ((buf = iobuf_reserve(io, len)) == NULL)
    return (-1);
  for (i = 0; i < iovcnt; i++) {
    if (iov[i].iov_len == 0)
      continue;
    memmove(buf, iov[i].iov_base, iov[i].iov_len);
    buf += iov[i].iov_len;
  }
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<iobuf_queuev>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L64
  lea    -0x1(%rdx),%eax
  mov    %rsi,%rbx
  shl    $0x4,%rax
  lea    0x10(%rsi,%rax,1),%rbp
  mov    %rsi,%rax
  xor    %esi,%esi
L55:
  add    (%rax),%rsi
  add    $0x10,%rax
  cmp    %rbp,%rax
  jne    L55
  call   <iobuf_reserve@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  nopl   0x0(%rax)
L35:
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L61
  mov    0x8(%rbx),%esi
  mov    %r12,%rdi
  call   <memmove@plt>
  add    (%rbx),%r12
L61:
  add    $0x10,%rbx
  cmp    %rbp,%rbx
  jne    L35
L68:
  xor    %eax,%eax
L65:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L64:
  xor    %esi,%esi
  call   <iobuf_reserve@plt>
  test   %rax,%rax
  jne    L68
L3:
  mov    $0xffffffff,%eax
  jmp    L65
```

assistant: ```c
int iobuf_queuev(struct iobuf *io, const struct iovec *iov, int iovcnt) {
  int i;
  size_t len = 0;
  char *buf;
  for (i = 0; i < iovcnt; i++)
    len += iov[i].iov_len;
  if ((buf = iobuf_reserve(io, len)) == NULL)
    return (-1);
  for (i = 0; i < iovcnt; i++) {
    if (iov[i].iov_len == 0)
      continue;
    memmove(buf, iov[i].iov_base, iov[i].iov_len);
    buf += iov[i].iov_len;
  }
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<iobuf_queuev>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edx,%edx
  jle    L3
  lea    -0x1(%rdx),%r12d
  mov    %rsi,%rbx
  cmp    $0x1,%r12d
  jbe    L32
  mov    %r12d,%ecx
  mov    %rsi,%rax
  pxor   %xmm1,%xmm1
  shr    %ecx
  sub    $0x1,%ecx
  shl    $0x5,%rcx
  lea    0x20(%rsi,%rcx,1),%rcx
  nopl   0x0(%rax)
L26:
  movdqu (%rax),%xmm0
  movdqu 0x10(%rax),%xmm2
  add    $0x20,%rax
  punpcklqdq %xmm2,%xmm0
  paddq  %xmm0,%xmm1
  cmp    %rcx,%rax
  jne    L26
  movdqa %xmm1,%xmm0
  mov    %r12d,%eax
  psrldq $0x8,%xmm0
  and    $0xfffffffe,%eax
  paddq  %xmm0,%xmm1
  movq   %xmm1,%rsi
L31:
  movslq %eax,%rcx
  add    $0x1,%eax
  shl    $0x4,%rcx
  add    (%rbx,%rcx,1),%rsi
  cmp    %eax,%edx
  jle    L13
  add    0x10(%rbx,%rcx,1),%rsi
L13:
  call   <iobuf_reserve@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L90
  shl    $0x4,%r12
  lea    0x10(%rbx,%r12,1),%r12
  nop
L77:
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L79
  mov    0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <memmove@plt>
  add    (%rbx),%rbp
L79:
  add    $0x10,%rbx
  cmp    %r12,%rbx
  jne    L77
L61:
  xor    %eax,%eax
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L3:
  xor    %esi,%esi
  call   <iobuf_reserve@plt>
  test   %rax,%rax
  jne    L61
L90:
  mov    $0xffffffff,%eax
  jmp    L42
  nopl   0x0(%rax,%rax,1)
L32:
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L31
```

assistant: ```c
int iobuf_queuev(struct iobuf *io, const struct iovec *iov, int iovcnt) {
  int i;
  size_t len = 0;
  char *buf;
  for (i = 0; i < iovcnt; i++)
    len += iov[i].iov_len;
  if ((buf = iobuf_reserve(io, len)) == NULL)
    return (-1);
  for (i = 0; i < iovcnt; i++) {
    if (iov[i].iov_len == 0)
      continue;
    memmove(buf, iov[i].iov_base, iov[i].iov_len);
    buf += iov[i].iov_len;
  }
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<T>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  movsd  %xmm0,-0x10(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L27
  cmpl   $0x1,-0x4(%rbp)
  je     L43
  jmp    L4
L27:
  movsd  D8(%rip),%xmm0
  jmp    L8
L43:
  movsd  -0x10(%rbp),%xmm0
  jmp    L8
L4:
  movsd  -0x10(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  mov    %edx,%edi
  call   <T>
  mulsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  mov    %edx,%edi
  call   <T>
  movapd %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  subsd  %xmm1,%xmm0
L8:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double T(unsigned int k, double x) {
  switch (k) {
  case 0:
    return 1;
  case 1:
    return x;
  }
  return 2 * x * T(k - 1, x) - T(k - 2, x);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<T>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    D9(%rip),%rbp
  test   %edi,%edi
  je     L11
  mov    %edi,%ebx
  movq   %xmm0,%rbp
  cmp    $0x1,%edi
  je     L11
  lea    -0x1(%rdi),%edi
  call   <T>
  movapd %xmm0,%xmm1
  movq   %rbp,%xmm0
  addsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rsp)
  lea    -0x2(%rbx),%edi
  movq   %rbp,%xmm0
  call   <T>
  movsd  0x8(%rsp),%xmm2
  subsd  %xmm0,%xmm2
  movq   %xmm2,%rbp
L11:
  movq   %rbp,%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double T(unsigned int k, double x) {
  switch (k) {
  case 0:
    return 1;
  case 1:
    return x;
  }
  return 2 * x * T(k - 1, x) - T(k - 2, x);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<T>:
  endbr64
  test   %edi,%edi
  je     L16
  push   %rbx
  movapd %xmm0,%xmm1
  mov    %edi,%ebx
  sub    $0x10,%rsp
  cmp    $0x1,%edi
  je     L2
  lea    -0x1(%rdi),%edi
  movsd  %xmm0,0x8(%rsp)
  call   <T>
  movsd  0x8(%rsp),%xmm1
  lea    -0x2(%rbx),%edi
  movapd %xmm0,%xmm2
  movapd %xmm1,%xmm0
  addsd  %xmm1,%xmm0
  mulsd  %xmm0,%xmm2
  movapd %xmm1,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <T>
  movsd  0x8(%rsp),%xmm2
  subsd  %xmm0,%xmm2
  movapd %xmm2,%xmm1
L2:
  add    $0x10,%rsp
  movapd %xmm1,%xmm0
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L16:
  movsd  D8(%rip),%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double T(unsigned int k, double x) {
  switch (k) {
  case 0:
    return 1;
  case 1:
    return x;
  }
  return 2 * x * T(k - 1, x) - T(k - 2, x);
}
```