ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D16(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <puts@plt>
  call   <fork@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r12d,%edx
  lea    D18(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D16
data_type: string
name: retrieve_data
arguments: tag: D1
data_type: string
name: retrieve_data
arguments: tag: D26
data_type: string
name: retrieve_data
arguments: tag: D18
data_type: string

tool: D16: 
I am a process my process id is %d


tool: D1: My parent's process id is       %d


tool: D26: Now i am going to create a child process by calling fork


tool: D18: My PID : %d, My parent PID : %d


assistant: ```c
int main() {
  printf("\nI am a process my process id is %d\n", getpid());
  printf("My parent's process id is       %d\n", getppid());
  printf("Now i am going to create a child process by calling fork\n\n");
  fork();
  printf("My PID : %d, My parent PID : %d\n", getpid(), getppid());
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D18(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  lea    D24(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  call   <fork@plt>
  xor    %eax,%eax
  call   <getppid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %r12d,%edx
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D24
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D5
type_guessed: string

tool: D18: 
I am a process my process id is %d


tool: D24: My parent's process id is       %d


tool: D28: Now i am going to create a child process by calling fork


tool: D5: My PID : %d, My parent PID : %d


assistant: ```c
int main() {
  printf("\nI am a process my process id is %d\n", getpid());
  printf("My parent's process id is       %d\n", getppid());
  printf("Now i am going to create a child process by calling fork\n\n");
  fork();
  printf("My PID : %d, My parent PID : %d\n", getpid(), getppid());
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<basefile_pattern>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L54
L13:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L35
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L35
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L27
L35:
  addl   $0x1,-0xc(%rbp)
L54:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L13
L27:
  cmpq   $0x0,-0x8(%rbp)
  jne    L26
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <basefile@plt>
  mov    %rax,-0x8(%rbp)
L26:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *basefile_pattern(char *path, char *pattern) {
  char *base = NULL;
  int i;
  for (i = 0; i < strlen(pattern); i++) {
    if (pattern[i] == '/' && pattern[i + 1] == '*') {
      base = strdup(path + i + 1);
      break;
    }
  }
  if (!base) {
    base = basefile(path);
  }
  return 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
<basefile_pattern>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  jmp    L25
L9:
  add    $0x1,%rbx
L25:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L46
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  jne    L9
  lea    0x1(%rbx),%rdi
  cmpb   $0x2a,0x1(%rbp,%rbx,1)
  jne    L9
  add    %r12,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L46
L57:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  mov    %r12,%rdi
  call   <basefile@plt>
  jmp    L57
```

assistant: ```c
char *basefile_pattern(char *path, char *pattern) {
  char *base = NULL;
  int i;
  for (i = 0; i < strlen(pattern); i++) {
    if (pattern[i] == '/' && pattern[i + 1] == '*') {
      base = strdup(path + i + 1);
      break;
    }
  }
  if (!base) {
    base = basefile(path);
  }
  return base;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<basefile_pattern>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %rdi,%rbx
L34:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L42
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%rdi
  jne    L33
  cmpb   $0x2a,0x1(%rbp,%rbx,1)
  jne    L33
  add    %r12,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L42
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <basefile@plt>
```

assistant: ```c
char *basefile_pattern(char *path, char *pattern) {
  char *base = NULL;
  int i;
  for (i = 0; i < strlen(pattern); i++) {
    if (pattern[i] == '/' && pattern[i + 1] == '*') {
      base = strdup(path + i + 1);
      break;
    }
  }
  if (!base) {
    base = basefile(path);
  }
  return base;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<basefile_pattern>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L41
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %rdi,%rbx
L41:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L43
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%rdi
  jne    L28
  cmpb   $0x2a,0x1(%rbp,%rbx,1)
  jne    L28
  add    %r12,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L43
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <basefile@plt>
```

assistant: ```c
char *basefile_pattern(char *path, char *pattern) {
  char *base = NULL;
  int i;
  for (i = 0; i < strlen(pattern); i++) {
    if (pattern[i] == '/' && pattern[i + 1] == '*') {
      base = strdup(path + i + 1);
      break;
    }
  }
  if (!base) {
    base = basefile(path);
  }
  return base;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L36
L32:
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L15
  mov    -0x8(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L15
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L15:
  addl   $0x1,-0x8(%rbp)
L36:
  cmpl   $0x64,-0x8(%rbp)
  jle    L32
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type: string
name: disassemble_data
arguments: data_label: D15
type: string

tool: D5: %d 


tool: D15: Total: %d

assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 100; i++) {
    if (i % 2 != 0 && i % 3 != 0) {
      printf("%d \n", i);
      sum++;
    }
  }
  printf("Total: %d", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  lea    D18(%rip),%r12
  jmp    L23
L9:
  add    $0x1,%ebx
  cmp    $0x65,%ebx
  je     L49
L23:
  test   $0x1,%bl
  je     L9
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  cmp    %eax,%ebx
  je     L9
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  jmp    L9
L49:
  mov    %ebp,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
type: string
name: extract_symbolic_data
arguments: data_label: D5
type: string

tool: D18: %d 


tool: D5: Total: %d

assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 100; i++) {
    if (i % 2 != 0 && i % 3 != 0) {
      printf("%d \n", i);
      sum++;
    }
  }
  printf("Total: %d", sum);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L29:
  add    $0x1,%ebx
  cmp    $0x65,%ebx
  je     L41
L3:
  test   $0x1,%bl
  je     L29
  imul   $0xaaaaaaab,%ebx,%eax
  cmp    $0x55555555,%eax
  jbe    L29
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    $0x65,%ebx
  jne    L3
L41:
  mov    %ebp,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
datatype: string
name: unpack_data
arguments: label: D5
datatype: string

tool: D7: %d 


tool: D5: Total: %d

assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 100; i++) {
    if (i % 2 != 0 && i % 3 != 0) {
      printf("%d \n", i);
      sum++;
    }
  }
  printf("Total: %d", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L16:
  add    $0x1,%ebx
  cmp    $0x65,%ebx
  je     L40
L7:
  test   $0x1,%bl
  je     L16
  imul   $0xaaaaaaab,%ebx,%eax
  cmp    $0x55555555,%eax
  jbe    L16
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    $0x65,%ebx
  jne    L7
L40:
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D7
type_guessed: string
name: read_data
arguments: data_tag: D0
type_guessed: string

tool: D7: %d 


tool: D0: Total: %d

assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 100; i++) {
    if (i % 2 != 0 && i % 3 != 0) {
      printf("%d \n", i);
      sum++;
    }
  }
  printf("Total: %d", sum);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<float_to_bigendian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movss  %xmm0,-0x1c(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    0xc(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  lea    0xc(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void float_to_bigendian(uint8_t *dst, float f) {
  uint8_t *b = (uint8_t *)&f;
  dst[0] = b[3];
  dst[1] = b[2];
  dst[2] = b[1];
  dst[3] = b[0];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<float_to_bigendian>:
  endbr64
  mov    0x8(%rsp),%eax
  mov    %eax,(%rdi)
  mov    0x4(%rsp),%eax
  mov    %eax,0x4(%rdi)
  mov    (%rsp),%eax
  mov    %eax,0x8(%rdi)
  movss  %xmm0,0xc(%rdi)
  ret
```

assistant: ```c
void float_to_bigendian(uint8_t *dst, float f) {
  uint8_t *b = (uint8_t *)&f;
  dst[0] = b[3];
  dst[1] = b[2];
  dst[2] = b[1];
  dst[3] = b[0];
}
```
user: What is the c source code of the assembly code below:

```asm
<float_to_bigendian>:
  endbr64
  mov    0x8(%rsp),%eax
  movss  %xmm0,0xc(%rdi)
  mov    %eax,(%rdi)
  mov    0x4(%rsp),%eax
  mov    %eax,0x4(%rdi)
  mov    (%rsp),%eax
  mov    %eax,0x8(%rdi)
  ret
```

assistant: ```c
void float_to_bigendian(uint8_t *dst, float f) {
  uint8_t *b = (uint8_t *)&f;
  dst[0] = b[3];
  dst[1] = b[2];
  dst[2] = b[1];
  dst[3] = b[0];
}
```
user: What is the c source code of the assembly code below?

```asm
<float_to_bigendian>:
  endbr64
  movss  %xmm0,-0xc(%rsp)
  movdqu -0xc(%rsp),%xmm1
  pshufd $0x1b,%xmm1,%xmm0
  movups %xmm0,(%rdi)
  ret
```

assistant: ```c
void float_to_bigendian(uint8_t *dst, float f) {
  uint8_t *b = (uint8_t *)&f;
  dst[0] = b[3];
  dst[1] = b[2];
  dst[2] = b[1];
  dst[3] = b[0];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<x1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type: string

tool: D8: Called mod1-x1 


assistant: ```c
void x1(void) {
  printf("Called mod1-x1 "
         ""
         "\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<x1>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D4
type: string

tool: D4: Called mod1-x1 


assistant: ```c
void x1(void) {
  printf("Called mod1-x1 "
         ""
         "\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<x1>:
  endbr64
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D8
type: string

tool: D8: Called mod1-x1 


assistant: ```c
void x1(void) {
  printf("Called mod1-x1 "
         ""
         "\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<x1>:
  endbr64
  lea    D6(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type_guessed: string

tool: D6: Called mod1-x1 


assistant: ```c
void x1(void) {
  printf("Called mod1-x1 "
         ""
         "\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<make_ver_output>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L14
  mov    $0x7c,%eax
  jmp    L7
L14:
  mov    $0x20,%eax
L7:
  mov    %al,-0x6(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L30
  mov    $0x7c,%eax
  jmp    L3
L30:
  mov    $0x20,%eax
L3:
  mov    %al,-0x5(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x30(%rbp),%rdx
  mov    %ecx,(%rdx)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x6(%rbp),%eax
  mov    %al,(%rdx)
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L43:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x30(%rbp),%rdx
  mov    %ecx,(%rdx)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  addl   $0x1,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L43
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void make_ver_output(int number, int size, const char *ver_map[], char *output, int *pos) {
  char sym1, sym2;
  int i;
  sym1 = (ver_map[number][0] == '1') ? '|' : ' ';
  sym2 = (ver_map[number][1] == '1') ? '|' : ' ';
  output[(*pos)++] = sym1;
  for (i = 0; i < size; i++) {
    output[(*pos)++] = ' ';
  }
  output[(*pos)] = sym2;
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<make_ver_output>:
  endbr64
  mov    %esi,%r9d
  mov    %rcx,%rsi
  movslq %edi,%rdi
  mov    (%rdx,%rdi,8),%rax
  cmpb   $0x31,(%rax)
  mov    $0x7c,%edx
  mov    $0x20,%ecx
  cmovne %ecx,%edx
  cmpb   $0x31,0x1(%rax)
  mov    $0x7c,%edi
  mov    $0x20,%eax
  cmovne %eax,%edi
  mov    (%r8),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,(%r8)
  cltq
  mov    %dl,(%rsi,%rax,1)
  test   %r9d,%r9d
  jle    L2
  mov    $0x0,%edx
L19:
  mov    (%r8),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,(%r8)
  cltq
  movb   $0x20,(%rsi,%rax,1)
  add    $0x1,%edx
  cmp    %edx,%r9d
  jne    L19
L2:
  movslq (%r8),%rax
  mov    %dil,(%rsi,%rax,1)
  ret
```

assistant: ```c
void make_ver_output(int number, int size, const char *ver_map[], char *output, int *pos) {
  char sym1, sym2;
  int i;
  sym1 = (ver_map[number][0] == '1') ? '|' : ' ';
  sym2 = (ver_map[number][1] == '1') ? '|' : ' ';
  output[(*pos)++] = sym1;
  for (i = 0; i < size; i++) {
    output[(*pos)++] = ' ';
  }
  output[(*pos)] = sym2;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<make_ver_output>:
  endbr64
  movslq %edi,%rdi
  mov    %esi,%r9d
  mov    %rcx,%rsi
  mov    $0x7c,%eax
  mov    (%rdx,%rdi,8),%rcx
  mov    $0x20,%edx
  mov    %edx,%r10d
  mov    %edx,%edi
  cmpb   $0x31,(%rcx)
  cmove  %eax,%r10d
  cmpb   $0x31,0x1(%rcx)
  cmove  %eax,%edi
  movslq (%r8),%rax
  lea    0x1(%rax),%edx
  mov    %edx,(%r8)
  mov    %r10b,(%rsi,%rax,1)
  test   %r9d,%r9d
  jle    L13
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L6:
  movslq (%r8),%rax
  add    $0x1,%edx
  lea    0x1(%rax),%ecx
  mov    %ecx,(%r8)
  movb   $0x20,(%rsi,%rax,1)
  cmp    %edx,%r9d
  jne    L6
L13:
  movslq (%r8),%rax
  mov    %dil,(%rsi,%rax,1)
  ret
```

assistant: ```c
void make_ver_output(int number, int size, const char *ver_map[], char *output, int *pos) {
  char sym1, sym2;
  int i;
  sym1 = (ver_map[number][0] == '1') ? '|' : ' ';
  sym2 = (ver_map[number][1] == '1') ? '|' : ' ';
  output[(*pos)++] = sym1;
  for (i = 0; i < size; i++) {
    output[(*pos)++] = ' ';
  }
  output[(*pos)] = sym2;
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<make_ver_output>:
  endbr64
  movslq %edi,%rdi
  mov    %esi,%r9d
  mov    %rcx,%rsi
  mov    $0x7c,%eax
  mov    (%rdx,%rdi,8),%rcx
  mov    $0x20,%edx
  mov    %edx,%r10d
  mov    %edx,%edi
  cmpb   $0x31,(%rcx)
  cmove  %eax,%r10d
  cmpb   $0x31,0x1(%rcx)
  cmove  %eax,%edi
  movslq (%r8),%rax
  lea    0x1(%rax),%edx
  mov    %edx,(%r8)
  mov    %r10b,(%rsi,%rax,1)
  test   %r9d,%r9d
  jle    L15
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L18:
  movslq (%r8),%rax
  add    $0x1,%edx
  lea    0x1(%rax),%ecx
  mov    %ecx,(%r8)
  movb   $0x20,(%rsi,%rax,1)
  cmp    %edx,%r9d
  jne    L18
L15:
  movslq (%r8),%rax
  mov    %dil,(%rsi,%rax,1)
  ret
```

assistant: ```c
void make_ver_output(int number, int size, const char *ver_map[], char *output, int *pos) {
  char sym1, sym2;
  int i;
  sym1 = (ver_map[number][0] == '1') ? '|' : ' ';
  sym2 = (ver_map[number][1] == '1') ? '|' : ' ';
  output[(*pos)++] = sym1;
  for (i = 0; i < size; i++) {
    output[(*pos)++] = ' ';
  }
  output[(*pos)] = sym2;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_int_arg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L6
  mov    $0xffffffff,%eax
  jmp    L8
L6:
  mov    $0x0,%eax
L8:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int get_int_arg(const char *arg, int *val) {
  char *endptr = (char *)arg;
  *val = strtol(arg, &endptr, 10);
  return (*endptr) ? -1 : 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_int_arg>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,(%rsp)
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    %eax,(%rbx)
  mov    (%rsp),%rax
  movzbl (%rax),%eax
  neg    %al
  sbb    %eax,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_int_arg(const char *arg, int *val) {
  char *endptr = (char *)arg;
  *val = strtol(arg, &endptr, 10);
  return (*endptr) ? -1 : 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_int_arg>:
  endbr64
  push   %rbx
  mov    $0xa,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    %rdi,(%rsp)
  call   <strtol@plt>
  mov    %eax,(%rbx)
  mov    (%rsp),%rax
  movzbl (%rax),%eax
  neg    %al
  sbb    %eax,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_int_arg(const char *arg, int *val) {
  char *endptr = (char *)arg;
  *val = strtol(arg, &endptr, 10);
  return (*endptr) ? -1 : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_int_arg>:
  endbr64
  push   %rbx
  mov    $0xa,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    %rdi,(%rsp)
  call   <strtol@plt>
  mov    %eax,(%rbx)
  mov    (%rsp),%rax
  movzbl (%rax),%eax
  neg    %al
  sbb    %eax,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_int_arg(const char *arg, int *val) {
  char *endptr = (char *)arg;
  *val = strtol(arg, &endptr, 10);
  return (*endptr) ? -1 : 0;
}
```