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
<test_to_run>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  mov    %rdi,%rbp
  lea    D9(%rip),%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L7
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <strcmp@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  test   %r8d,%r8d
  je     L7
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <strcmp@plt>
  mov    %eax,%r8d
  mov    $0x2,%eax
  test   %r8d,%r8d
  je     L7
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  call   <strcmp@plt>
  cmp    $0x1,%eax
  sbb    %eax,%eax
  and    $0x4,%eax
  sub    $0x1,%eax
L7:
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_infered: string
name: extract_symbolic_data
arguments: tag: D28
type_infered: string
name: extract_symbolic_data
arguments: tag: D2
type_infered: string
name: extract_symbolic_data
arguments: tag: D20
type_infered: string

tool: D9: Write

tool: D28: Read

tool: D2: Copy

tool: D20: ServerCopy

assistant: ```c
int test_to_run(char *s) {
  if (!strcmp("Write", s)) {
    return 0;
  }
  if (!strcmp("Read", s)) {
    return 1;
  }
  if (!strcmp("Copy", s)) {
    return 2;
  }
  if (!strcmp("ServerCopy", s)) {
    return 3;
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<retrieveValue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x64(%rbp),%eax
  mov    %eax,-0x50(%rbp)
  movl   $0x1,-0x4c(%rbp)
  movl   $0x2,-0x48(%rbp)
  movl   $0x3,-0x44(%rbp)
  movl   $0x4,-0x40(%rbp)
  movl   $0x5,-0x3c(%rbp)
  movl   $0x6,-0x38(%rbp)
  movl   $0x7,-0x34(%rbp)
  movl   $0x8,-0x30(%rbp)
  movl   $0x9,-0x2c(%rbp)
  movl   $0xa,-0x28(%rbp)
  mov    -0x70(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x50(%rbp),%rcx
  mov    $0xb,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  cmp    $0xb,%eax
  jne    L44
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    -0x70(%rbp),%rax
  mov    (%rax),%eax
  lea    -0xb(%rbp),%rcx
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  cmp    $0x2,%eax
  jne    L47
  movzbl -0xb(%rbp),%eax
  movsbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0xa(%rbp),%eax
  movsbl %al,%eax
  or     %edx,%eax
  mov    %eax,-0x54(%rbp)
  mov    -0x54(%rbp),%eax
  jmp    L44
L47:
  mov    $0xffffffff,%eax
  jmp    L44
L44:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int retrieveValue(int call, int *pointfile) {
  unsigned int cmd[16];
  cmd[0] = call;
  cmd[1] = 1;
  cmd[2] = 2;
  cmd[3] = 3;
  cmd[4] = 4;
  cmd[5] = 5;
  cmd[6] = 6;
  cmd[7] = 7;
  cmd[8] = 8;
  cmd[9] = 9;
  cmd[10] = 10;
  if (write(*pointfile, cmd, 11) == 11) {
    usleep(10000);
    char buf[3];
    if (read(*pointfile, buf, 2) == 2) {
      int resultant = buf[0] << 8 | buf[1];
      return resultant;
    } else {
      return -1;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<retrieveValue>:
  endbr64
  push   %rbx
  sub    $0x50,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %edi,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x7,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  movl   $0x9,0x24(%rsp)
  movl   $0xa,0x28(%rsp)
  mov    %rsp,%rsi
  mov    $0xb,%edx
  mov    (%rbx),%edi
  call   <write@plt>
  cmp    $0xb,%eax
  je     L27
L26:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x50,%rsp
  pop    %rbx
  ret
L27:
  mov    $0x2710,%edi
  call   <usleep@plt>
  lea    0x45(%rsp),%rsi
  mov    $0x2,%edx
  mov    (%rbx),%edi
  call   <read@plt>
  mov    %eax,%edx
  mov    $0xffffffff,%eax
  cmp    $0x2,%edx
  jne    L26
  movsbl 0x45(%rsp),%eax
  shl    $0x8,%eax
  movsbl 0x46(%rsp),%edx
  or     %edx,%eax
  jmp    L26
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int retrieveValue(int call, int *pointfile) {
  unsigned int cmd[16];
  cmd[0] = call;
  cmd[1] = 1;
  cmd[2] = 2;
  cmd[3] = 3;
  cmd[4] = 4;
  cmd[5] = 5;
  cmd[6] = 6;
  cmd[7] = 7;
  cmd[8] = 8;
  cmd[9] = 9;
  cmd[10] = 10;
  if (write(*pointfile, cmd, 11) == 11) {
    usleep(10000);
    char buf[3];
    if (read(*pointfile, buf, 2) == 2) {
      int resultant = buf[0] << 8 | buf[1];
      return resultant;
    } else {
      return -1;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<retrieveValue>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0xb,%edx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %edi,(%rsp)
  mov    (%rbx),%edi
  mov    %rsp,%rsi
  movabs $0x200000001,%rax
  mov    %rax,0x4(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0xc(%rsp)
  movabs $0x600000005,%rax
  mov    %rax,0x14(%rsp)
  movabs $0x800000007,%rax
  mov    %rax,0x1c(%rsp)
  movabs $0xa00000009,%rax
  mov    %rax,0x24(%rsp)
  call   <write@plt>
  cmp    $0xb,%eax
  je     L2
L27:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x50,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    (%rbx),%edi
  lea    0x45(%rsp),%rsi
  mov    $0x2,%edx
  call   <read@plt>
  mov    %eax,%r8d
  mov    $0xffffffff,%eax
  cmp    $0x2,%r8d
  jne    L27
  movsbl 0x45(%rsp),%eax
  movsbl 0x46(%rsp),%edx
  shl    $0x8,%eax
  or     %edx,%eax
  jmp    L27
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int retrieveValue(int call, int *pointfile) {
  unsigned int cmd[16];
  cmd[0] = call;
  cmd[1] = 1;
  cmd[2] = 2;
  cmd[3] = 3;
  cmd[4] = 4;
  cmd[5] = 5;
  cmd[6] = 6;
  cmd[7] = 7;
  cmd[8] = 8;
  cmd[9] = 9;
  cmd[10] = 10;
  if (write(*pointfile, cmd, 11) == 11) {
    usleep(10000);
    char buf[3];
    if (read(*pointfile, buf, 2) == 2) {
      int resultant = buf[0] << 8 | buf[1];
      return resultant;
    } else {
      return -1;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<retrieveValue>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0xb,%edx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %edi,(%rsp)
  mov    (%rbx),%edi
  mov    %rsp,%rsi
  movabs $0x200000001,%rax
  mov    %rax,0x4(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0xc(%rsp)
  movabs $0x600000005,%rax
  mov    %rax,0x14(%rsp)
  movabs $0x800000007,%rax
  mov    %rax,0x1c(%rsp)
  movabs $0xa00000009,%rax
  mov    %rax,0x24(%rsp)
  call   <write@plt>
  cmp    $0xb,%eax
  je     L38
L31:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x50,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L38:
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    (%rbx),%edi
  lea    0x45(%rsp),%rsi
  mov    $0x2,%edx
  call   <read@plt>
  mov    %eax,%r8d
  mov    $0xffffffff,%eax
  cmp    $0x2,%r8d
  jne    L31
  movsbl 0x45(%rsp),%eax
  movsbl 0x46(%rsp),%edx
  shl    $0x8,%eax
  or     %edx,%eax
  jmp    L31
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int retrieveValue(int call, int *pointfile) {
  unsigned int cmd[16];
  cmd[0] = call;
  cmd[1] = 1;
  cmd[2] = 2;
  cmd[3] = 3;
  cmd[4] = 4;
  cmd[5] = 5;
  cmd[6] = 6;
  cmd[7] = 7;
  cmd[8] = 8;
  cmd[9] = 9;
  cmd[10] = 10;
  if (write(*pointfile, cmd, 11) == 11) {
    usleep(10000);
    char buf[3];
    if (read(*pointfile, buf, 2) == 2) {
      int resultant = buf[0] << 8 | buf[1];
      return resultant;
    } else {
      return -1;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<add_sub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <mult_div_mod@plt>
  mov    %eax,-0x8(%rbp)
  jmp    L0
L49:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L48:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L49
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x2b,-0x9(%rbp)
  je     L11
  cmpb   $0x2d,-0x9(%rbp)
  je     L11
  mov    -0x8(%rbp),%eax
  jmp    L45
L11:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <mult_div_mod@plt>
  mov    %eax,-0x4(%rbp)
  movsbl -0x9(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <do_op@plt>
  mov    %eax,-0x8(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L48
  mov    -0x8(%rbp),%eax
L45:
  leave
  ret
```

assistant: ```c
int add_sub(char **str) {
  int a;
  int b;
  char op;
  a = mult_div_mod(str);
  while (**str) {
    while (**str == ' ')
      *str += 1;
    op = **str;
    if (op != '+' && op != '-')
      return (a);
    *str += 1;
    b = mult_div_mod(str);
    a = do_op(a, b, op);
  }
  return (a);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<add_sub>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <mult_div_mod@plt>
  mov    %eax,%r12d
  mov    0x0(%rbp),%rax
  cmpb   $0x0,(%rax)
  jne    L46
L12:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  mov    %rax,%rdx
  mov    %rax,0x0(%rbp)
  add    $0x1,%rax
  movzbl -0x1(%rax),%ebx
  cmp    $0x20,%bl
  je     L16
L42:
  lea    -0x2b(%rbx),%eax
  test   $0xfd,%al
  jne    L12
  add    $0x1,%rdx
  mov    %rdx,0x0(%rbp)
  mov    %rbp,%rdi
  call   <mult_div_mod@plt>
  mov    %eax,%esi
  movsbl %bl,%edx
  mov    %r12d,%edi
  call   <do_op@plt>
  mov    %eax,%r12d
  mov    0x0(%rbp),%rax
  cmpb   $0x0,(%rax)
  je     L12
L46:
  mov    0x0(%rbp),%rdx
  movzbl (%rdx),%ebx
  lea    0x1(%rdx),%rax
  cmp    $0x20,%bl
  je     L16
  jmp    L42
```

assistant: ```c
int add_sub(char **str) {
  int a;
  int b;
  char op;
  a = mult_div_mod(str);
  while (**str) {
    while (**str == ' ')
      *str += 1;
    op = **str;
    if (op != '+' && op != '-')
      return (a);
    *str += 1;
    b = mult_div_mod(str);
    a = do_op(a, b, op);
  }
  return (a);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<add_sub>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <mult_div_mod@plt>
  mov    0x0(%rbp),%rdx
  mov    %eax,%r12d
  movzbl (%rdx),%ebx
  test   %bl,%bl
  je     L41
  xchg   %ax,%ax
L7:
  lea    0x1(%rdx),%rcx
  cmp    $0x20,%bl
  jne    L23
  nopl   0x0(%rax)
L13:
  mov    %rcx,0x0(%rbp)
  mov    %rcx,%rdx
  movzbl (%rcx),%ebx
  add    $0x1,%rcx
  cmp    $0x20,%bl
  je     L13
L23:
  lea    -0x2b(%rbx),%eax
  test   $0xfd,%al
  jne    L41
  add    $0x1,%rdx
  mov    %rbp,%rdi
  mov    %rdx,0x0(%rbp)
  call   <mult_div_mod@plt>
  movsbl %bl,%edx
  mov    %r12d,%edi
  mov    %eax,%esi
  call   <do_op@plt>
  mov    0x0(%rbp),%rdx
  mov    %eax,%r12d
  movzbl (%rdx),%ebx
  test   %bl,%bl
  jne    L7
L41:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int add_sub(char **str) {
  int a;
  int b;
  char op;
  a = mult_div_mod(str);
  while (**str) {
    while (**str == ' ')
      *str += 1;
    op = **str;
    if (op != '+' && op != '-')
      return (a);
    *str += 1;
    b = mult_div_mod(str);
    a = do_op(a, b, op);
  }
  return (a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add_sub>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <mult_div_mod@plt>
  mov    0x0(%rbp),%rdx
  mov    %eax,%r12d
  movzbl (%rdx),%ebx
  test   %bl,%bl
  je     L9
  xchg   %ax,%ax
L46:
  lea    0x1(%rdx),%rcx
  cmp    $0x20,%bl
  jne    L34
  nopl   0x0(%rax)
L41:
  mov    %rcx,0x0(%rbp)
  mov    %rcx,%rdx
  movzbl (%rcx),%ebx
  add    $0x1,%rcx
  cmp    $0x20,%bl
  je     L41
L34:
  lea    -0x2b(%rbx),%eax
  test   $0xfd,%al
  jne    L9
  add    $0x1,%rdx
  mov    %rbp,%rdi
  mov    %rdx,0x0(%rbp)
  call   <mult_div_mod@plt>
  movsbl %bl,%edx
  mov    %r12d,%edi
  mov    %eax,%esi
  call   <do_op@plt>
  mov    0x0(%rbp),%rdx
  mov    %eax,%r12d
  movzbl (%rdx),%ebx
  test   %bl,%bl
  jne    L46
L9:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int add_sub(char **str) {
  int a;
  int b;
  char op;
  a = mult_div_mod(str);
  while (**str) {
    while (**str == ' ')
      *str += 1;
    op = **str;
    if (op != '+' && op != '-')
      return (a);
    *str += 1;
    b = mult_div_mod(str);
    a = do_op(a, b, op);
  }
  return (a);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pi_even>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movss  %xmm0,-0x4(%rbp)
  movss  %xmm1,-0x8(%rbp)
  movss  -0x4(%rbp),%xmm0
  comiss -0x8(%rbp),%xmm0
  jb     L9
  pxor   %xmm0,%xmm0
  jmp    L19
L9:
  pxor   %xmm0,%xmm0
  cvtss2sd -0x4(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movsd  D21(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movss  -0x4(%rbp),%xmm1
  movss  D18(%rip),%xmm0
  addss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movss  -0x8(%rbp),%xmm0
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <pi_odd@plt>
  cvtss2sd %xmm0,%xmm0
  addsd  -0x10(%rbp),%xmm0
  cvtsd2ss %xmm0,%xmm0
L19:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_infered: f64
name: retrieve_data
arguments: data_tag: D21
type_infered: f64
name: retrieve_data
arguments: data_tag: D18
type_infered: byte[4]

tool: D3: 1

tool: D21: 4

tool: {"D18": [0x00,0x00,0x80,0x3f]}

assistant: ```c
float pi_even(float i, float num_its) {
  if (i >= num_its)
    return 0.0;
  return (4.0 / (1.0 + (i * 2.0))) + pi_odd(i + 1.0, num_its);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pi_even>:
  endbr64
  comiss %xmm1,%xmm0
  jb     L0
  pxor   %xmm0,%xmm0
  ret
L0:
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  addss  D24(%rip),%xmm0
  call   <pi_odd@plt>
  movaps %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  addsd  %xmm0,%xmm0
  addsd  D5(%rip),%xmm0
  movsd  D12(%rip),%xmm1
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  add    $0x18,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D24
type: f32
name: disassemble_data
arguments: label: D5
type: f32[2]
name: disassemble_data
arguments: label: D12
type: f64

tool: D24: 1

tool: {"D5": [0.0,1.875]}

tool: D12: 4

assistant: ```c
float pi_even(float i, float num_its) {
  if (i >= num_its)
    return 0.0;
  return (4.0 / (1.0 + (i * 2.0))) + pi_odd(i + 1.0, num_its);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pi_even>:
  endbr64
  comiss %xmm1,%xmm0
  jb     L2
  pxor   %xmm0,%xmm0
  ret
  xchg   %ax,%ax
L2:
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  addss  D8(%rip),%xmm0
  call   <pi_odd@plt>
  movss  0xc(%rsp),%xmm2
  movaps %xmm0,%xmm1
  movsd  D7(%rip),%xmm0
  cvtss2sd %xmm2,%xmm2
  addsd  %xmm2,%xmm2
  addsd  D19(%rip),%xmm2
  cvtss2sd %xmm1,%xmm1
  add    $0x18,%rsp
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
datatype: f32
name: read_data
arguments: data_tag: D7
datatype: f64
name: read_data
arguments: data_tag: D19
datatype: f32[2]

tool: D8: 1

tool: D7: 4

tool: {"D19": [0.0,1.875]}

assistant: ```c
float pi_even(float i, float num_its) {
  if (i >= num_its)
    return 0.0;
  return (4.0 / (1.0 + (i * 2.0))) + pi_odd(i + 1.0, num_its);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pi_even>:
  endbr64
  comiss %xmm1,%xmm0
  jb     L5
  pxor   %xmm0,%xmm0
  ret
  xchg   %ax,%ax
L5:
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  addss  D15(%rip),%xmm0
  call   <pi_odd@plt>
  movss  0xc(%rsp),%xmm2
  movaps %xmm0,%xmm1
  movsd  D2(%rip),%xmm0
  cvtss2sd %xmm2,%xmm2
  addsd  %xmm2,%xmm2
  addsd  D17(%rip),%xmm2
  cvtss2sd %xmm1,%xmm1
  add    $0x18,%rsp
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
data_type: f32
name: disassemble_data
arguments: label: D2
data_type: f64
name: disassemble_data
arguments: label: D17
data_type: f32[2]

tool: D15: 1

tool: D2: 4

tool: {"D17": [0.0,1.875]}

assistant: ```c
float pi_even(float i, float num_its) {
  if (i >= num_its)
    return 0.0;
  return (4.0 / (1.0 + (i * 2.0))) + pi_odd(i + 1.0, num_its);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L19:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L19
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sum(int *arr, int n) {
  int sum = 0;
  for (int i = 0; i < n; ++i)
    sum += arr[i];
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L12
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L21:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L21
L16:
  mov    %edx,%eax
  ret
L12:
  mov    $0x0,%edx
  jmp    L16
```

assistant: ```c
int sum(int *arr, int n) {
  int sum = 0;
  for (int i = 0; i < n; ++i)
    sum += arr[i];
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L7:
  add    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L7
  ret
  nopl   0x0(%rax)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sum(int *arr, int n) {
  int sum = 0;
  for (int i = 0; i < n; ++i)
    sum += arr[i];
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sum>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L54
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L52
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L68:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L68
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L41
L47:
  movslq %ecx,%rdi
  lea    0x0(,%rdi,4),%r8
  add    (%rdx,%rdi,4),%eax
  lea    0x1(%rcx),%edi
  cmp    %edi,%esi
  jle    L58
  add    $0x2,%ecx
  add    0x4(%rdx,%r8,1),%eax
  cmp    %ecx,%esi
  jle    L58
  add    0x8(%rdx,%r8,1),%eax
  ret
  nopl   0x0(%rax)
L54:
  xor    %eax,%eax
L58:
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  ret
L52:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L47
```

assistant: ```c
int sum(int *arr, int n) {
  int sum = 0;
  for (int i = 0; i < n; ++i)
    sum += arr[i];
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<estrdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x5,%rax
  mov    %rax,%rdi
  call   <emalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *estrdup(const char *oldstring) {
  char *s;
  s = emalloc(strlen(oldstring) + 4 + 1);
  strcpy(s, oldstring);
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<estrdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x5(%rax),%rdi
  call   <emalloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *estrdup(const char *oldstring) {
  char *s;
  s = emalloc(strlen(oldstring) + 4 + 1);
  strcpy(s, oldstring);
  return s;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<estrdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x5(%rax),%rdi
  call   <emalloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *estrdup(const char *oldstring) {
  char *s;
  s = emalloc(strlen(oldstring) + 4 + 1);
  strcpy(s, oldstring);
  return s;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<estrdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x5(%rax),%rdi
  call   <emalloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *estrdup(const char *oldstring) {
  char *s;
  s = emalloc(strlen(oldstring) + 4 + 1);
  strcpy(s, oldstring);
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ucShift>:
  endbr64
  push   %rbx
  test   %edx,%edx
  je     L20
  mov    %edx,%r8d
  test   %esi,%esi
  je     L20
  cmp    %edx,%esi
  jle    L193
  neg    %edx
  cmp    %edx,%esi
  jle    L193
  cmp    $0x1,%r8d
  je     L100
  cmp    $0xffffffff,%r8d
  je     L74
  mov    $0x0,%ebx
  test   %r8d,%r8d
  jg     L122
  jmp    L204
L193:
  cmp    $0x1,%r8d
  je     L155
  mov    $0x0,%ebx
  cmp    $0xffffffff,%r8d
  je     L180
L168:
  call   <vSetZero@plt>
L208:
  mov    %ebx,%eax
  pop    %rbx
  ret
L155:
  lea    -0x1(%rsi),%eax
  movzbl (%rdi,%rax,1),%ebx
  jmp    L168
L180:
  movzbl (%rdi),%ebx
  jmp    L168
L100:
  lea    -0x1(%rsi),%eax
  movzbl (%rdi,%rax,1),%ebx
L122:
  mov    %esi,%edx
  sub    $0x1,%edx
  js     L208
  lea    -0x2(%rsi),%eax
  movslq %edx,%rdx
  add    %rdi,%rdx
  mov    $0x1,%esi
  sub    %r8d,%esi
  jmp    L156
L74:
  movzbl (%rdi),%ebx
L204:
  lea    0x1(%rsi),%r9d
  mov    $0x1,%eax
  movslq %edx,%rdx
  lea    (%rdi,%rdx,1),%rcx
  test   %esi,%esi
  jg     L162
  jmp    L208
L98:
  movb   $0x0,(%rdx)
  test   %eax,%eax
  js     L208
L151:
  sub    $0x1,%eax
  sub    $0x1,%rdx
L156:
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r8d
  jg     L98
  lea    (%rsi,%rax,1),%ecx
  movslq %ecx,%rcx
  movzbl (%rdi,%rcx,1),%ecx
  mov    %cl,(%rdx)
  jmp    L151
L192:
  movb   $0x0,-0x1(%rdi,%rax,1)
L145:
  add    $0x1,%rax
  cmp    %r9,%rax
  je     L208
L162:
  mov    %eax,%edx
  sub    %r8d,%edx
  sub    $0x1,%edx
  cmp    %edx,%esi
  jle    L192
  movzbl -0x1(%rcx,%rax,1),%edx
  mov    %dl,-0x1(%rdi,%rax,1)
  jmp    L145
L20:
  mov    $0x0,%ebx
  jmp    L208
```

assistant: ```c
unsigned char ucShift(unsigned char *psA, unsigned int uiLen, int iShift) {
  unsigned char ucRet;
  int i;
  if (iShift == 0 || uiLen == 0)
    return (0);
  if (iShift >= (int)uiLen || -iShift >= (int)uiLen) {
    if (iShift == 1)
      ucRet = psA[uiLen - 1];
    else if (iShift == -1)
      ucRet = psA[0];
    else
      ucRet = 0;
    vSetZero(psA, uiLen);
    return (ucRet);
  }
  if (iShift == 1)
    ucRet = psA[uiLen - 1];
  else if (iShift == -1)
    ucRet = psA[0];
  else
    ucRet = 0;
  if (iShift > 0) {
    for (i = uiLen - 1; i >= 0; i--) {
      if (i >= iShift)
        psA[i] = psA[i - iShift];
      else
        psA[i] = 0;
    }
  } else {
    iShift = -iShift;
    for (i = 0; i < (int)uiLen; i++) {
      if (i + iShift < (int)uiLen)
        psA[i] = psA[i + iShift];
      else
        psA[i] = 0;
    }
  }
  return (ucRet);
}
```