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
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%ecx
  add    $0x45,%eax
  test   %cl,%cl
  je     L63
  xor    %edx,%edx
  jmp    L50
  cs nopw 0x0(%rax,%rax,1)
L58:
  lea    (%rax,%rax,1),%edi
  cmp    %dil,%dl
  jae    L51
  shl    $0x2,%eax
L15:
  add    $0x2,%edx
  cmp    %dl,%cl
  jbe    L63
L50:
  cmp    %al,%dl
  jae    L58
  add    $0x2,%edx
  xor    %eax,%eax
  cmp    %dl,%cl
  ja     L50
L63:
  add    $0x3f,%eax
  mov    %al,(%rsi)
  ret
L51:
  mov    %edi,%eax
  imul   %edi,%eax
  jmp    L15
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] > local1) {
        state[0UL] = state[local1] - state[0UL];
      } else {
        state[0UL] += state[0UL];
      }
      if (state[0UL] <= local1) {
        state[local1] = state[0UL] * state[0UL];
      } else {
        state[local1] += state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)63;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sanitize_file_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L6
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L38
L20:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isgraph@plt>
  test   %eax,%eax
  je     L42
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L23
L42:
  mov    -0x8(%rbp),%rax
  movb   $0x5f,(%rax)
L23:
  addq   $0x1,-0x8(%rbp)
L38:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
L6:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *sanitize_file_string(char *str) {
  char *p;
  if (str) {
    for (p = str; *p; p++) {
      if (!isgraph((int)*p) || (*p == '/'))
        *p = '_';
    }
  }
  return (str);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sanitize_file_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L11
  movzbl (%rdi),%ebx
  test   %bl,%bl
  je     L11
  mov    %rdi,%rbp
  jmp    L24
L37:
  movb   $0x5f,0x0(%rbp)
L21:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  je     L11
L24:
  movsbl %bl,%edi
  call   <isgraph@plt>
  cmp    $0x2f,%bl
  je     L37
  test   %eax,%eax
  je     L37
  jmp    L21
L11:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *sanitize_file_string(char *str) {
  char *p;
  if (str) {
    for (p = str; *p; p++) {
      if (!isgraph((int)*p) || (*p == '/'))
        *p = '_';
    }
  }
  return (str);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sanitize_file_string>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L12
  movzbl (%rdi),%ebx
  test   %bl,%bl
  je     L12
  mov    %rdi,%rbp
  nopw   0x0(%rax,%rax,1)
L45:
  movsbl %bl,%edi
  call   <isgraph@plt>
  test   %eax,%eax
  je     L29
  cmp    $0x2f,%bl
  jne    L44
L29:
  movb   $0x5f,0x0(%rbp)
L44:
  movzbl 0x1(%rbp),%ebx
  add    $0x1,%rbp
  test   %bl,%bl
  jne    L45
L12:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *sanitize_file_string(char *str) {
  char *p;
  if (str) {
    for (p = str; *p; p++) {
      if (!isgraph((int)*p) || (*p == '/'))
        *p = '_';
    }
  }
  return (str);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sanitize_file_string>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L5
  movzbl (%rdi),%ebx
  test   %bl,%bl
  je     L5
  mov    %rdi,%rbp
  nopw   0x0(%rax,%rax,1)
L22:
  movsbl %bl,%edi
  call   <isgraph@plt>
  test   %eax,%eax
  je     L17
  cmp    $0x2f,%bl
  je     L17
  movzbl 0x1(%rbp),%ebx
  add    $0x1,%rbp
  test   %bl,%bl
  jne    L22
L5:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L17:
  movb   $0x5f,0x0(%rbp)
  movzbl 0x1(%rbp),%ebx
  add    $0x1,%rbp
  test   %bl,%bl
  jne    L22
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *sanitize_file_string(char *str) {
  char *p;
  if (str) {
    for (p = str; *p; p++) {
      if (!isgraph((int)*p) || (*p == '/'))
        *p = '_';
    }
  }
  return (str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L12:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_strdup(char *str) {
  char *copy;
  int i;
  copy = malloc(sizeof(char) * ft_strlen(str) + 1);
  i = 0;
  while (str[i]) {
    copy[i] = str[i];
    i++;
  }
  copy[i] = '\0';
  return (copy);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ft_strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L24
  mov    $0x0,%edx
L9:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L9
L28:
  movb   $0x0,(%rax,%rdx,1)
  pop    %rbx
  ret
L24:
  mov    $0x0,%edx
  jmp    L28
```

assistant: ```c
char *ft_strdup(char *str) {
  char *copy;
  int i;
  copy = malloc(sizeof(char) * ft_strlen(str) + 1);
  i = 0;
  while (str[i]) {
    copy[i] = str[i];
    i++;
  }
  copy[i] = '\0';
  return (copy);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L3
  xor    %edx,%edx
  xchg   %ax,%ax
L16:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L16
  add    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
```

assistant: ```c
char *ft_strdup(char *str) {
  char *copy;
  int i;
  copy = malloc(sizeof(char) * ft_strlen(str) + 1);
  i = 0;
  while (str[i]) {
    copy[i] = str[i];
    i++;
  }
  copy[i] = '\0';
  return (copy);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L8
  xor    %edx,%edx
  xchg   %ax,%ax
L3:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L3
  add    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
```

assistant: ```c
char *ft_strdup(char *str) {
  char *copy;
  int i;
  copy = malloc(sizeof(char) * ft_strlen(str) + 1);
  i = 0;
  while (str[i]) {
    copy[i] = str[i];
    i++;
  }
  copy[i] = '\0';
  return (copy);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  D6(%rip),%xmm0
  movss  %xmm0,-0x20(%rbp)
  movss  D13(%rip),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  movss  D21(%rip),%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  D13(%rip),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  D21(%rip),%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  D64(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <dump@plt>
  lea    -0x40(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <inv3x3@plt>
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <dump@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
datatype: byte[4]
name: extract_symbolic_data
arguments: data_label: D13
datatype: byte[4]
name: extract_symbolic_data
arguments: data_label: D21
datatype: byte[4]
name: extract_symbolic_data
arguments: data_label: D64
datatype: byte[4]
name: extract_symbolic_data
arguments: data_label: D20
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string

tool: {"D6": [0x00,0x00,0x80,0x3f]}

tool: {"D13": [0x00,0x00,0x00,0x40]}

tool: {"D21": [0x00,0x00,0x40,0x40]}

tool: {"D64": [0x00,0x00,0x80,0x40]}

tool: D20: m:

tool: D23: r:

assistant: ```c
int main() {
  float r[6];
  float m[6] = {1, 2, 3, 2, 3, 4};
  puts("m:");
  dump(m);
  inv3x3(m, r);
  puts("r:");
  dump(r);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x3f800000,0x20(%rsp)
  movss  D9(%rip),%xmm1
  movss  %xmm1,0x24(%rsp)
  movss  D30(%rip),%xmm0
  movss  %xmm0,0x28(%rsp)
  movss  %xmm1,0x2c(%rsp)
  movss  %xmm0,0x30(%rsp)
  movl   $0x40800000,0x34(%rsp)
  lea    D5(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%rbp
  mov    %rbp,%rdi
  call   <dump@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <inv3x3@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <dump@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
type_guessed: byte[4]
name: read_data
arguments: data_tag: D30
type_guessed: byte[4]
name: read_data
arguments: data_tag: D5
type_guessed: string
name: read_data
arguments: data_tag: D0
type_guessed: string

tool: {"D9": [0x00,0x00,0x00,0x40]}

tool: {"D30": [0x00,0x00,0x40,0x40]}

tool: D5: m:

tool: D0: r:

assistant: ```c
int main() {
  float r[6];
  float m[6] = {1, 2, 3, 2, 3, 4};
  puts("m:");
  dump(m);
  inv3x3(m, r);
  puts("r:");
  dump(r);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D9(%rip),%rdi
  push   %rbp
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x400000003f800000,%rax
  lea    0x20(%rsp),%r12
  mov    %rsp,%rbp
  mov    %rax,0x20(%rsp)
  add    $0xc00000,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x4080000040400000,%rax
  mov    %rax,0x30(%rsp)
  call   <puts@plt>
  mov    %r12,%rdi
  call   <dump@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <inv3x3@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <dump@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string
name: parse_data
arguments: tag: D10
type_infered: string

tool: D9: m:

tool: D10: r:

assistant: ```c
int main() {
  float r[6];
  float m[6] = {1, 2, 3, 2, 3, 4};
  puts("m:");
  dump(m);
  inv3x3(m, r);
  puts("r:");
  dump(r);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  lea    D30(%rip),%rdi
  push   %rbp
  sub    $0x48,%rsp
  movaps D2(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  lea    0x20(%rsp),%r12
  mov    %rsp,%rbp
  movaps %xmm0,0x20(%rsp)
  mov    %rax,0x30(%rsp)
  call   <puts@plt>
  mov    %r12,%rdi
  call   <dump@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <inv3x3@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <dump@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D30
type_guessed: string
name: unpack_data
arguments: data_tag: D2
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D3
type_guessed: byte[8]
name: unpack_data
arguments: data_tag: D8
type_guessed: string

tool: D30: m:

tool: {"D2": [0x00,0x00,0x80,0x3f,0x00,0x00,0x00,0x40,0x00,0x00,0x40,0x40,0x00,0x00,0x00,0x40]}

tool: {"D3": [0x00,0x00,0x40,0x40,0x00,0x00,0x80,0x40]}

tool: D8: r:

assistant: ```c
int main() {
  float r[6];
  float m[6] = {1, 2, 3, 2, 3, 4};
  puts("m:");
  dump(m);
  inv3x3(m, r);
  puts("r:");
  dump(r);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L16
L20:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L15
L35:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D24(%rip),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L15:
  cmpl   $0x2,-0x14(%rbp)
  jle    L35
  addl   $0x1,-0x18(%rbp)
L16:
  cmpl   $0x6,-0x18(%rbp)
  jle    L20
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D24
datatype: string
name: read_data
arguments: tag: D20
datatype: string
name: read_data
arguments: tag: D2
datatype: string

tool: D24: dnansumpw

tool: D20: # c::%s:len=%d


tool: D2: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnansumpw", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D17(%rip),%r15
  jmp    L20
L12:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L6
L20:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L32:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L32
  jmp    L12
L6:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D17
type: string
name: parse_data
arguments: label: D24
type: string
name: parse_data
arguments: label: D28
type: string

tool: D17: dnansumpw

tool: D24: # c::%s:len=%d


tool: D28: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnansumpw", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D20(%rip),%r15
  push   %r14
  lea    D7(%rip),%r14
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L15:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L18:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L18
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L15
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D20
type_infered: string
name: extract_symbolic_data
arguments: data_label: D7
type_infered: string
name: extract_symbolic_data
arguments: data_label: D17
type_infered: string

tool: D20: dnansumpw

tool: D7: # c::%s:len=%d


tool: D17: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnansumpw", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D23(%rip),%r15
  push   %r14
  lea    D20(%rip),%r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L9:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L0:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L0
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L9
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D23
type: string
name: retrieve_data
arguments: tag: D20
type: string
name: retrieve_data
arguments: tag: D7
type: string

tool: D23: dnansumpw

tool: D20: # c::%s:len=%d


tool: D7: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnansumpw", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```