ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<checkizero>:
  endbr64
  test   %rdi,%rdi
  je     L2
  ret
L2:
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <xlFmtError@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D1: division by zero

assistant: ```c
void checkizero(long iarg) {
  if (iarg == 0)
    xlFmtError("division by zero");
}
```
user: What is the c source code of the assembly code below:

```asm
<checkizero>:
  endbr64
  test   %rdi,%rdi
  je     L7
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  lea    D6(%rip),%rdi
  jmp    1050 <xlFmtError@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D6: division by zero

assistant: ```c
void checkizero(long iarg) {
  if (iarg == 0)
    xlFmtError("division by zero");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<checkizero>:
  endbr64
  test   %rdi,%rdi
  je     L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  lea    D4(%rip),%rdi
  jmp    1050 <xlFmtError@plt>
```

tool_calls: name: parse_data
arguments: tag: D4
type_infered: string

tool: D4: division by zero

assistant: ```c
void checkizero(long iarg) {
  if (iarg == 0)
    xlFmtError("division by zero");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dctsub4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  movsd  D19(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x34(%rbp),%eax
  sar    %eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  jne    L3
  movsd  -0x20(%rbp),%xmm1
  movsd  D2(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x20(%rbp),%xmm1
  movsd  D20(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mov    -0x40(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm1
  mulsd  -0x18(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x18(%rbp),%xmm1
  mov    -0x40(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x40(%rbp),%rax
  add    $0x18,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
L3:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mulsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D19
type: f64
name: unpack_data
arguments: tag: D2
type: f64
name: unpack_data
arguments: tag: D20
type: f64

tool: D19: 0.7071067811865476

tool: D2: 0.3826834323650898

tool: D20: 0.9238795325112867

assistant: ```c
void dctsub4(int n, double *a) {
  int m;
  double wki, wdr, wdi, xr;
  wki = 0.707106781186547524400844362104849039284835937688;
  m = n >> 1;
  if (m == 2) {
    wdr = wki * 0.382683432365089771728459984030398866761344562485;
    wdi = wki * 0.923879532511286756128183189396788286822416625863;
    xr = wdi * a[1] - wdr * a[3];
    a[1] = wdr * a[1] + wdi * a[3];
    a[3] = xr;
  }
  a[m] *= wki;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dctsub4>:
  endbr64
  sar    %edi
  cmp    $0x2,%edi
  je     L14
L8:
  movslq %edi,%rdi
  lea    (%rsi,%rdi,8),%rax
  movsd  D3(%rip),%xmm0
  mulsd  (%rax),%xmm0
  movsd  %xmm0,(%rax)
  ret
L14:
  movsd  0x8(%rsi),%xmm0
  movsd  0x18(%rsi),%xmm1
  movsd  D13(%rip),%xmm3
  movapd %xmm0,%xmm2
  mulsd  %xmm3,%xmm2
  movsd  D22(%rip),%xmm4
  movapd %xmm1,%xmm5
  mulsd  %xmm4,%xmm5
  addsd  %xmm5,%xmm2
  movsd  %xmm2,0x8(%rsi)
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x18(%rsi)
  jmp    L8
```

tool_calls: name: unpack_data
arguments: label: D3
type_guessed: f64
name: unpack_data
arguments: label: D13
type_guessed: byte[8]
name: unpack_data
arguments: label: D22
type_guessed: byte[8]

tool: D3: 0.7071067811865476

tool: {"D13": [0x95,0x38,0xdb,0x7b,0x7a,0x51,0xd1,0x3f]}

tool: {"D22": [0xfc,0xf0,0x44,0x91,0xae,0xe7,0xe4,0x3f]}

assistant: ```c
void dctsub4(int n, double *a) {
  int m;
  double wki, wdr, wdi, xr;
  wki = 0.707106781186547524400844362104849039284835937688;
  m = n >> 1;
  if (m == 2) {
    wdr = wki * 0.382683432365089771728459984030398866761344562485;
    wdi = wki * 0.923879532511286756128183189396788286822416625863;
    xr = wdi * a[1] - wdr * a[3];
    a[1] = wdr * a[1] + wdi * a[3];
    a[3] = xr;
  }
  a[m] *= wki;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dctsub4>:
  endbr64
  sar    %edi
  cmp    $0x2,%edi
  jne    L0
  movsd  0x8(%rsi),%xmm0
  movsd  0x18(%rsi),%xmm1
  movsd  D7(%rip),%xmm3
  movsd  D11(%rip),%xmm4
  movapd %xmm0,%xmm2
  movapd %xmm1,%xmm5
  mulsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm5
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  addsd  %xmm5,%xmm2
  subsd  %xmm1,%xmm0
  movsd  %xmm2,0x8(%rsi)
  movsd  %xmm0,0x18(%rsi)
L0:
  movslq %edi,%rdi
  movsd  D22(%rip),%xmm0
  lea    (%rsi,%rdi,8),%rax
  mulsd  (%rax),%xmm0
  movsd  %xmm0,(%rax)
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: byte[8]
name: retrieve_data
arguments: data_label: D11
datatype: byte[8]
name: retrieve_data
arguments: data_label: D22
datatype: f64

tool: {"D7": [0x95,0x38,0xdb,0x7b,0x7a,0x51,0xd1,0x3f]}

tool: {"D11": [0xfc,0xf0,0x44,0x91,0xae,0xe7,0xe4,0x3f]}

tool: D22: 0.7071067811865476

assistant: ```c
void dctsub4(int n, double *a) {
  int m;
  double wki, wdr, wdi, xr;
  wki = 0.707106781186547524400844362104849039284835937688;
  m = n >> 1;
  if (m == 2) {
    wdr = wki * 0.382683432365089771728459984030398866761344562485;
    wdi = wki * 0.923879532511286756128183189396788286822416625863;
    xr = wdi * a[1] - wdr * a[3];
    a[1] = wdr * a[1] + wdi * a[3];
    a[3] = xr;
  }
  a[m] *= wki;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dctsub4>:
  endbr64
  sar    %edi
  cmp    $0x2,%edi
  jne    L7
  movsd  0x8(%rsi),%xmm0
  movsd  0x18(%rsi),%xmm1
  movsd  D8(%rip),%xmm3
  movsd  D23(%rip),%xmm4
  movapd %xmm0,%xmm2
  movapd %xmm1,%xmm5
  mulsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm5
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  addsd  %xmm5,%xmm2
  subsd  %xmm1,%xmm0
  movsd  %xmm2,0x8(%rsi)
  movsd  %xmm0,0x18(%rsi)
L7:
  movslq %edi,%rdi
  movsd  D19(%rip),%xmm0
  lea    (%rsi,%rdi,8),%rax
  mulsd  (%rax),%xmm0
  movsd  %xmm0,(%rax)
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type: byte[8]
name: retrieve_data
arguments: data_tag: D23
type: byte[8]
name: retrieve_data
arguments: data_tag: D19
type: f64

tool: {"D8": [0x95,0x38,0xdb,0x7b,0x7a,0x51,0xd1,0x3f]}

tool: {"D23": [0xfc,0xf0,0x44,0x91,0xae,0xe7,0xe4,0x3f]}

tool: D19: 0.7071067811865476

assistant: ```c
void dctsub4(int n, double *a) {
  int m;
  double wki, wdr, wdi, xr;
  wki = 0.707106781186547524400844362104849039284835937688;
  m = n >> 1;
  if (m == 2) {
    wdr = wki * 0.382683432365089771728459984030398866761344562485;
    wdi = wki * 0.923879532511286756128183189396788286822416625863;
    xr = wdi * a[1] - wdr * a[3];
    a[1] = wdr * a[1] + wdi * a[3];
    a[3] = xr;
  }
  a[m] *= wki;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Show_Definition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type: string

tool: D7: struct Show {
  int (*show)(var, var, int);
  int (*look)(var, var, int);
};


assistant: ```c
const char *Show_Definition(void) {
  return "struct Show {\n"
         "  int (*show)(var, var, int);\n"
         "  int (*look)(var, var, int);\n"
         "};\n";
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Show_Definition>:
  endbr64
  lea    D0(%rip),%rax
  ret
```

tool_calls: name: parse_data
arguments: label: D0
data_type: string

tool: D0: struct Show {
  int (*show)(var, var, int);
  int (*look)(var, var, int);
};


assistant: ```c
const char *Show_Definition(void) {
  return "struct Show {\n"
         "  int (*show)(var, var, int);\n"
         "  int (*look)(var, var, int);\n"
         "};\n";
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Show_Definition>:
  endbr64
  lea    D2(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_infered: string

tool: D2: struct Show {
  int (*show)(var, var, int);
  int (*look)(var, var, int);
};


assistant: ```c
const char *Show_Definition(void) {
  return "struct Show {\n"
         "  int (*show)(var, var, int);\n"
         "  int (*look)(var, var, int);\n"
         "};\n";
}
```
user: What is the c source code of the assembly code below:

```asm
<Show_Definition>:
  endbr64
  lea    D6(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type: string

tool: D6: struct Show {
  int (*show)(var, var, int);
  int (*look)(var, var, int);
};


assistant: ```c
const char *Show_Definition(void) {
  return "struct Show {\n"
         "  int (*show)(var, var, int);\n"
         "  int (*look)(var, var, int);\n"
         "};\n";
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<IntToStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  lea    D3(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_guessed: string

tool: D3: %i

assistant: ```c
char *IntToStr(int v) {
  char *buffer = (char *)malloc(32);
  sprintf(buffer, "%i", v);
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<IntToStr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %ebp,%edx
  lea    D0(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
data_type: string

tool: D0: %i

assistant: ```c
char *IntToStr(int v) {
  char *buffer = (char *)malloc(32);
  sprintf(buffer, "%i", v);
  return buffer;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<IntToStr>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x20,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%edx
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type_guessed: string

tool: D5: %i

assistant: ```c
char *IntToStr(int v) {
  char *buffer = (char *)malloc(32);
  sprintf(buffer, "%i", v);
  return buffer;
}
```
user: What is the c source code of the assembly code below:

```asm
<IntToStr>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x20,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13d,%edx
  lea    D2(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string

tool: D2: %i

assistant: ```c
char *IntToStr(int v) {
  char *buffer = (char *)malloc(32);
  sprintf(buffer, "%i", v);
  return buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<brute_force>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L61
L33:
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L19:
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L11
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  add    %ecx,%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L19
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L68
  mov    -0x8(%rbp),%eax
  jmp    L29
L68:
  addl   $0x1,-0x8(%rbp)
L61:
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L33
  mov    $0xffffffff,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int brute_force(char *needle, int n, char *haystack, int h) {
  int i, j;
  for (i = 0; i <= h - n; ++i) {
    for (j = 0; j < n && needle[j] == haystack[j + i]; ++j)
      ;
    if (j >= n) {
      return i;
    }
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<brute_force>:
  endbr64
  sub    %esi,%ecx
  js     L56
  mov    $0x0,%r10d
  mov    %esi,%r9d
  jmp    L17
L41:
  cmp    %eax,%esi
  jle    L89
  add    $0x1,%r10d
  add    $0x1,%rdx
  cmp    %ecx,%r10d
  jg     L81
L17:
  test   %esi,%esi
  jle    L89
  mov    $0x0,%eax
L63:
  movzbl (%rdx,%rax,1),%r11d
  cmp    %r11b,(%rdi,%rax,1)
  jne    L41
  add    $0x1,%rax
  cmp    %r9,%rax
  jne    L63
L89:
  mov    %r10d,%eax
  ret
L56:
  mov    $0xffffffff,%r10d
  jmp    L89
L81:
  mov    $0xffffffff,%r10d
  jmp    L89
```

assistant: ```c
int brute_force(char *needle, int n, char *haystack, int h) {
  int i, j;
  for (i = 0; i <= h - n; ++i) {
    for (j = 0; j < n && needle[j] == haystack[j + i]; ++j)
      ;
    if (j >= n) {
      return i;
    }
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<brute_force>:
  endbr64
  xor    %r9d,%r9d
  movslq %esi,%r8
  sub    %esi,%ecx
  js     L32
  xchg   %ax,%ax
L57:
  test   %esi,%esi
  jle    L5
  xor    %eax,%eax
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L4:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L5
L38:
  movzbl (%rdx,%rax,1),%r10d
  cmp    %r10b,(%rdi,%rax,1)
  je     L4
  add    $0x1,%r9d
  add    $0x1,%rdx
  cmp    %ecx,%r9d
  jle    L57
L32:
  mov    $0xffffffff,%r9d
L5:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int brute_force(char *needle, int n, char *haystack, int h) {
  int i, j;
  for (i = 0; i <= h - n; ++i) {
    for (j = 0; j < n && needle[j] == haystack[j + i]; ++j)
      ;
    if (j >= n) {
      return i;
    }
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<brute_force>:
  endbr64
  xor    %r9d,%r9d
  movslq %esi,%r8
  sub    %esi,%ecx
  js     L42
  xchg   %ax,%ax
L47:
  test   %esi,%esi
  jle    L3
  xor    %eax,%eax
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L18:
  add    $0x1,%rax
  cmp    %rax,%r8
  je     L3
L58:
  movzbl (%rdx,%rax,1),%r10d
  cmp    %r10b,(%rdi,%rax,1)
  je     L18
  add    $0x1,%r9d
  add    $0x1,%rdx
  cmp    %ecx,%r9d
  jle    L47
L42:
  mov    $0xffffffff,%r9d
L3:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int brute_force(char *needle, int n, char *haystack, int h) {
  int i, j;
  for (i = 0; i <= h - n; ++i) {
    for (j = 0; j < n && needle[j] == haystack[j + i]; ++j)
      ;
    if (j >= n) {
      return i;
    }
  }
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<osn_memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L3
L8:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L3:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  test   %eax,%eax
  jne    L8
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void osn_memcpy(void *dst, const void *src, int size) {
  while (size--) {
    *(unsigned char *)dst = *((const unsigned char *)src);
    dst = (unsigned char *)dst + 1;
    src = (const char *)src + 1;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<osn_memcpy>:
  endbr64
  test   %edx,%edx
  je     L14
  mov    %edx,%edx
  mov    $0x0,%eax
L19:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L19
L14:
  ret
```

assistant: ```c
void osn_memcpy(void *dst, const void *src, int size) {
  while (size--) {
    *(unsigned char *)dst = *((const unsigned char *)src);
    dst = (unsigned char *)dst + 1;
    src = (const char *)src + 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<osn_memcpy>:
  endbr64
  test   %edx,%edx
  je     L12
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L17:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L12:
  ret
```

assistant: ```c
void osn_memcpy(void *dst, const void *src, int size) {
  while (size--) {
    *(unsigned char *)dst = *((const unsigned char *)src);
    dst = (unsigned char *)dst + 1;
    src = (const char *)src + 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<osn_memcpy>:
  endbr64
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%edi
  test   %edx,%edx
  je     L119
  lea    0x1(%rsi),%r8
  mov    %rcx,%rax
  sub    %r8,%rax
  cmp    $0xe,%rax
  jbe    L140
  cmp    $0x6,%edi
  jbe    L140
  cmp    $0xe,%edi
  jbe    L71
  mov    %edx,%r8d
  xor    %eax,%eax
  shr    $0x4,%r8d
  shl    $0x4,%r8
  nopl   (%rax)
L18:
  movdqu (%rsi,%rax,1),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%r8
  jne    L18
  mov    %edx,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%r8d
  sub    %eax,%edi
  lea    (%rcx,%r8,1),%r9
  add    %rsi,%r8
  cmp    %eax,%edx
  je     L53
L89:
  sub    %eax,%edx
  lea    -0x1(%rdx),%r10d
  cmp    $0x6,%r10d
  jbe    L80
  mov    (%rsi,%rax,1),%rsi
  mov    %rsi,(%rcx,%rax,1)
  mov    %edx,%eax
  and    $0xfffffff8,%eax
  mov    %eax,%ecx
  sub    %eax,%edi
  add    %rcx,%r9
  add    %rcx,%r8
  cmp    %eax,%edx
  je     L119
L80:
  movzbl (%r8),%eax
  mov    %al,(%r9)
  test   %edi,%edi
  je     L119
  movzbl 0x1(%r8),%eax
  mov    %al,0x1(%r9)
  cmp    $0x1,%edi
  je     L119
  movzbl 0x2(%r8),%eax
  mov    %al,0x2(%r9)
  cmp    $0x2,%edi
  je     L119
  movzbl 0x3(%r8),%eax
  mov    %al,0x3(%r9)
  cmp    $0x3,%edi
  je     L119
  movzbl 0x4(%r8),%eax
  mov    %al,0x4(%r9)
  cmp    $0x4,%edi
  je     L119
  movzbl 0x5(%r8),%eax
  mov    %al,0x5(%r9)
  cmp    $0x5,%edi
  je     L119
  movzbl 0x6(%r8),%eax
  mov    %al,0x6(%r9)
  ret
  nopl   (%rax)
L140:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L42:
  movzbl (%rsi,%rax,1),%edi
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L42
L119:
  ret
L71:
  mov    %rsi,%r8
  mov    %rcx,%r9
  xor    %eax,%eax
  jmp    L89
L53:
  ret
```

assistant: ```c
void osn_memcpy(void *dst, const void *src, int size) {
  while (size--) {
    *(unsigned char *)dst = *((const unsigned char *)src);
    dst = (unsigned char *)dst + 1;
    src = (const char *)src + 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xf,-0xc(%rbp)
  movl   $0xa,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  sub    -0x8(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  imul   -0x8(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  movd   %ecx,%xmm0
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_guessed: string
name: extract_symbolic_data
arguments: tag: D22
type_guessed: string
name: extract_symbolic_data
arguments: tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: tag: D15
type_guessed: string
name: extract_symbolic_data
arguments: tag: D17
type_guessed: string

tool: D7: %d + %d = %f


tool: D22: %d - %d = %f


tool: D5: %d * %d = %f


tool: D15: %d / %d = %f


tool: D17: (%d / %d)[remainder] = %f


assistant: ```c
main() {
  int num1 = 15, num2 = 10;
  float result = 0;
  result = num1 + num2;
  printf("%d + %d = %f\n", num1, num2, result);
  result = num1 - num2;
  printf("%d - %d = %f\n", num1, num2, result);
  result = num1 * num2;
  printf("%d * %d = %f\n", num1, num2, result);
  result = num1 / num2;
  printf("%d / %d = %f\n", num1, num2, result);
  result = num1 % num2;
  printf("(%d / %d)[remainder] = %f\n", num1, num2, result);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D65(%rip),%xmm0
  mov    $0xa,%edx
  mov    $0xf,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  movss  D48(%rip),%xmm0
  mov    $0xa,%edx
  mov    $0xf,%esi
  lea    D83(%rip),%rdi
  call   <printf@plt>
  movss  D39(%rip),%xmm0
  mov    $0xa,%edx
  mov    $0xf,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  movss  D50(%rip),%xmm0
  mov    $0xa,%edx
  mov    $0xf,%esi
  lea    D91(%rip),%rdi
  call   <printf@plt>
  movss  D48(%rip),%xmm0
  mov    $0xa,%edx
  mov    $0xf,%esi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D65
type_infered: byte[4]
name: unpack_data
arguments: data_tag: D6
type_infered: string
name: unpack_data
arguments: data_tag: D48
type_infered: byte[4]
name: unpack_data
arguments: data_tag: D83
type_infered: string
name: unpack_data
arguments: data_tag: D39
type_infered: byte[4]
name: unpack_data
arguments: data_tag: D8
type_infered: string
name: unpack_data
arguments: data_tag: D50
type_infered: byte[4]
name: unpack_data
arguments: data_tag: D91
type_infered: string
name: unpack_data
arguments: data_tag: D34
type_infered: string

tool: {"D65": [0x00,0x00,0xc8,0x41]}

tool: D6: %d + %d = %f


tool: {"D48": [0x00,0x00,0xa0,0x40]}

tool: D83: %d - %d = %f


tool: {"D39": [0x00,0x00,0x16,0x43]}

tool: D8: %d * %d = %f


tool: {"D50": [0x00,0x00,0x80,0x3f]}

tool: D91: %d / %d = %f


tool: D34: (%d / %d)[remainder] = %f


assistant: ```c
main() {
  int num1 = 15, num2 = 10;
  float result = 0;
  result = num1 + num2;
  printf("%d + %d = %f\n", num1, num2, result);
  result = num1 - num2;
  printf("%d - %d = %f\n", num1, num2, result);
  result = num1 * num2;
  printf("%d * %d = %f\n", num1, num2, result);
  result = num1 / num2;
  printf("%d / %d = %f\n", num1, num2, result);
  result = num1 % num2;
  printf("(%d / %d)[remainder] = %f\n", num1, num2, result);
}
```