ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  movslq %esi,%rbp
  imul   $0xffffffff91a2b3c5,%rbp,%rbp
  shr    $0x20,%rbp
  add    %esi,%ebp
  sar    $0xb,%ebp
  mov    %esi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebp
  imul   $0xfffff1f0,%ebp,%eax
  add    %esi,%eax
  movslq %eax,%rbx
  imul   $0xffffffff88888889,%rbx,%rbx
  shr    $0x20,%rbx
  add    %eax,%ebx
  sar    $0x5,%ebx
  cltd
  sub    %edx,%ebx
  mov    %ebx,%ecx
  shl    $0x4,%ecx
  mov    %ebx,%edx
  sub    %ecx,%edx
  lea    (%rax,%rdx,4),%r12d
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D21
type_guessed: string
name: retrieve_data
arguments: data_label: D24
type_guessed: string
name: retrieve_data
arguments: data_label: D4
type_guessed: string

tool: D21: %d

tool: D24: %d 

tool: D4: s = %d h %d m %d s


assistant: ```c
int main(void) {
  int seconds;
  int time_in_hours, time_in_minutes, time_in_seconds;
  scanf("%d", &seconds);
  time_in_hours = seconds / 3600;
  time_in_minutes = (seconds - (time_in_hours * 3600)) / 60;
  time_in_seconds = (seconds - (time_in_hours * 3600) - (time_in_minutes * 60));
  printf("%d ", seconds);
  printf("s = %d h %d m %d s\n", time_in_hours, time_in_minutes, time_in_seconds);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D5(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rbp
  lea    D21(%rip),%rdi
  mov    %rbp,%rsi
  imul   $0xffffffff91a2b3c5,%rbp,%rbp
  mov    %esi,%eax
  sar    $0x1f,%eax
  shr    $0x20,%rbp
  add    %esi,%ebp
  sar    $0xb,%ebp
  sub    %eax,%ebp
  imul   $0xfffff1f0,%ebp,%eax
  add    %esi,%eax
  movslq %eax,%r12
  cltd
  imul   $0xffffffff88888889,%r12,%r12
  shr    $0x20,%r12
  add    %eax,%r12d
  sar    $0x5,%r12d
  sub    %edx,%r12d
  mov    %r12d,%ecx
  mov    %r12d,%edx
  shl    $0x4,%ecx
  sub    %ecx,%edx
  lea    (%rax,%rdx,4),%r13d
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
type: string
name: parse_data
arguments: data_label: D21
type: string
name: parse_data
arguments: data_label: D11
type: string

tool: D5: %d

tool: D21: %d 

tool: D11: s = %d h %d m %d s


assistant: ```c
int main(void) {
  int seconds;
  int time_in_hours, time_in_minutes, time_in_seconds;
  scanf("%d", &seconds);
  time_in_hours = seconds / 3600;
  time_in_minutes = (seconds - (time_in_hours * 3600)) / 60;
  time_in_seconds = (seconds - (time_in_hours * 3600) - (time_in_minutes * 60));
  printf("%d ", seconds);
  printf("s = %d h %d m %d s\n", time_in_hours, time_in_minutes, time_in_seconds);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D8(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rbp
  lea    D25(%rip),%rdi
  mov    %rbp,%rsi
  imul   $0xffffffff91a2b3c5,%rbp,%rbp
  mov    %esi,%eax
  sar    $0x1f,%eax
  shr    $0x20,%rbp
  add    %esi,%ebp
  sar    $0xb,%ebp
  sub    %eax,%ebp
  imul   $0xfffff1f0,%ebp,%eax
  add    %esi,%eax
  movslq %eax,%r12
  cltd
  imul   $0xffffffff88888889,%r12,%r12
  shr    $0x20,%r12
  add    %eax,%r12d
  sar    $0x5,%r12d
  sub    %edx,%r12d
  mov    %r12d,%ecx
  mov    %r12d,%edx
  shl    $0x4,%ecx
  sub    %ecx,%edx
  lea    (%rax,%rdx,4),%r13d
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type_guessed: string
name: disassemble_data
arguments: data_tag: D25
type_guessed: string
name: disassemble_data
arguments: data_tag: D11
type_guessed: string

tool: D8: %d

tool: D25: %d 

tool: D11: s = %d h %d m %d s


assistant: ```c
int main(void) {
  int seconds;
  int time_in_hours, time_in_minutes, time_in_seconds;
  scanf("%d", &seconds);
  time_in_hours = seconds / 3600;
  time_in_minutes = (seconds - (time_in_hours * 3600)) / 60;
  time_in_seconds = (seconds - (time_in_hours * 3600) - (time_in_minutes * 60));
  printf("%d ", seconds);
  printf("s = %d h %d m %d s\n", time_in_hours, time_in_minutes, time_in_seconds);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<merge>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %rdx,%r15
  mov    %ecx,%r13d
  lea    (%rsi,%rcx,1),%r14d
  movslq %r14d,%rdi
  shl    $0x2,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r14d,%r14d
  jle    L111
  test   %r12d,%r12d
  je     L28
  test   %r13d,%r13d
  je     L28
  mov    %r14d,%r8d
  mov    $0x1,%edx
  mov    $0x0,%ecx
  mov    $0x0,%eax
  jmp    L24
L91:
  mov    %esi,-0x4(%rbp,%rdx,4)
  add    $0x1,%ecx
L39:
  mov    %edx,%esi
  cmp    %r8,%rdx
  je     L34
  add    $0x1,%rdx
  cmp    %r12d,%eax
  je     L34
  cmp    %r13d,%ecx
  je     L34
L24:
  movslq %eax,%rsi
  mov    (%rbx,%rsi,4),%edi
  movslq %ecx,%rsi
  mov    (%r15,%rsi,4),%esi
  cmp    %esi,%edi
  jg     L91
  mov    %edi,-0x4(%rbp,%rdx,4)
  add    $0x1,%eax
  jmp    L39
L111:
  mov    $0x0,%ecx
  mov    $0x0,%eax
  mov    $0x0,%esi
L34:
  cmp    %eax,%r12d
  jle    L131
  cltq
  movslq %esi,%rsi
  sub    %rax,%rsi
  lea    0x0(%rbp,%rsi,4),%rcx
L116:
  mov    (%rbx,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r12d
  jg     L116
L53:
  test   %r14d,%r14d
  jle    L38
  mov    %r14d,%r14d
  mov    $0x0,%eax
L87:
  mov    0x0(%rbp,%rax,4),%edx
  mov    %edx,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r14
  jne    L87
L38:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  mov    $0x0,%ecx
  mov    $0x0,%eax
  mov    $0x0,%esi
  jmp    L34
L131:
  cmp    %ecx,%r13d
  jle    L30
  movslq %ecx,%rcx
  movslq %esi,%rsi
  sub    %rcx,%rsi
  lea    0x0(%rbp,%rsi,4),%rdx
L10:
  mov    (%r15,%rcx,4),%eax
  mov    %eax,(%rdx,%rcx,4)
  add    $0x1,%rcx
  cmp    %ecx,%r13d
  jg     L10
  jmp    L53
L30:
  lea    D14(%rip),%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L53
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
data_type: string
name: extract_symbolic_data
arguments: data_tag: D11
data_type: string

tool: D14: merge

tool: D11: %s: both a and b not done!


assistant: ```c
int *merge(int *a, int an, int *b, int bn) {
  int i, j, k;
  int *c;
  c = malloc(sizeof(int) * (an + bn));
  for (i = j = k = 0; i < an + bn; i++) {
    if ((j == an) || (k == bn))
      break;
    if (a[j] <= b[k]) {
      c[i] = a[j];
      j++;
    } else {
      c[i] = b[k];
      k++;
    }
  }
  if (j < an) {
    for (; j < an; j++, i++)
      c[i] = a[j];
  } else if (k < bn) {
    for (; k < bn; k++, i++)
      c[i] = b[k];
  } else {
    printf("%s: both a and b not done!\n", __func__);
  }
  for (i = 0; i < an + bn; i++) {
    a[i] = c[i];
  }
  free(c);
  return a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<merge>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  lea    (%rsi,%rcx,1),%ebx
  movslq %ebx,%r13
  lea    0x0(,%r13,4),%rdi
  sub    $0x18,%rsp
  mov    %ecx,0xc(%rsp)
  call   <malloc@plt>
  test   %ebx,%ebx
  mov    0xc(%rsp),%r8d
  mov    %rax,%rbp
  jle    L7
  test   %r14d,%r14d
  je     L7
  test   %r8d,%r8d
  je     L7
  mov    $0x1,%eax
  xor    %esi,%esi
  xor    %ecx,%ecx
  jmp    L118
  nop
L27:
  mov    %edx,-0x4(%rbp,%rax,4)
  add    $0x1,%ecx
  movslq %eax,%rdx
  cmp    %rax,%r13
  je     L84
L85:
  add    $0x1,%rax
  cmp    %r14d,%ecx
  je     L84
  cmp    %r8d,%esi
  je     L84
L118:
  movslq %ecx,%rdx
  movslq %esi,%rdi
  mov    (%r12,%rdx,4),%edx
  mov    (%r15,%rdi,4),%edi
  cmp    %edi,%edx
  jle    L27
  mov    %edi,%edx
  add    $0x1,%esi
  mov    %edx,-0x4(%rbp,%rax,4)
  movslq %eax,%rdx
  cmp    %rax,%r13
  jne    L85
L84:
  cmp    %r14d,%ecx
  jge    L35
  sub    $0x1,%r14d
  lea    0x0(%rbp,%rdx,4),%rdi
  sub    %ecx,%r14d
  movslq %ecx,%rcx
  lea    0x4(,%r14,4),%rdx
  lea    (%r12,%rcx,4),%rsi
  call   <memcpy@plt>
L111:
  test   %ebx,%ebx
  jle    L67
  lea    0x0(,%r13,4),%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
L67:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L35:
  cmp    %esi,%r8d
  jle    L20
  sub    $0x1,%r8d
  lea    0x0(%rbp,%rdx,4),%rdi
  sub    %esi,%r8d
  movslq %esi,%rsi
  lea    0x4(,%r8,4),%rdx
  lea    (%r15,%rsi,4),%rsi
  call   <memcpy@plt>
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L20:
  lea    D6(%rip),%rsi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %esi,%esi
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L84
```

tool_calls: name: read_data
arguments: label: D6
type_guessed: string
name: read_data
arguments: label: D9
type_guessed: string

tool: D6: merge

tool: D9: %s: both a and b not done!


assistant: ```c
int *merge(int *a, int an, int *b, int bn) {
  int i, j, k;
  int *c;
  c = malloc(sizeof(int) * (an + bn));
  for (i = j = k = 0; i < an + bn; i++) {
    if ((j == an) || (k == bn))
      break;
    if (a[j] <= b[k]) {
      c[i] = a[j];
      j++;
    } else {
      c[i] = b[k];
      k++;
    }
  }
  if (j < an) {
    for (; j < an; j++, i++)
      c[i] = a[j];
  } else if (k < bn) {
    for (; k < bn; k++, i++)
      c[i] = b[k];
  } else {
    printf("%s: both a and b not done!\n", __func__);
  }
  for (i = 0; i < an + bn; i++) {
    a[i] = c[i];
  }
  free(c);
  return a;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<merge>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  lea    (%rsi,%rcx,1),%ebx
  movslq %ebx,%r13
  lea    0x0(,%r13,4),%rdi
  sub    $0x18,%rsp
  mov    %ecx,0xc(%rsp)
  call   <malloc@plt>
  test   %ebx,%ebx
  mov    0xc(%rsp),%r8d
  mov    %rax,%rbp
  jle    L108
  test   %r14d,%r14d
  je     L108
  test   %r8d,%r8d
  je     L108
  mov    $0x1,%eax
  xor    %esi,%esi
  xor    %ecx,%ecx
  jmp    L20
  nop
L112:
  mov    %edx,-0x4(%rbp,%rax,4)
  add    $0x1,%ecx
  movslq %eax,%rdx
  cmp    %rax,%r13
  je     L92
L68:
  add    $0x1,%rax
  cmp    %r14d,%ecx
  je     L92
  cmp    %r8d,%esi
  je     L92
L20:
  movslq %ecx,%rdx
  movslq %esi,%rdi
  mov    (%r12,%rdx,4),%edx
  mov    (%r15,%rdi,4),%edi
  cmp    %edi,%edx
  jle    L112
  mov    %edi,%edx
  add    $0x1,%esi
  mov    %edx,-0x4(%rbp,%rax,4)
  movslq %eax,%rdx
  cmp    %rax,%r13
  jne    L68
L92:
  cmp    %r14d,%ecx
  jge    L12
  sub    $0x1,%r14d
  lea    0x0(%rbp,%rdx,4),%rdi
  sub    %ecx,%r14d
  movslq %ecx,%rcx
  lea    0x4(,%r14,4),%rdx
  lea    (%r12,%rcx,4),%rsi
  call   <memcpy@plt>
L138:
  test   %ebx,%ebx
  jle    L10
  lea    0x0(,%r13,4),%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
L10:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  cmp    %esi,%r8d
  jle    L42
  sub    $0x1,%r8d
  lea    0x0(%rbp,%rdx,4),%rdi
  sub    %esi,%r8d
  movslq %esi,%rsi
  lea    0x4(,%r8,4),%rdx
  lea    (%r15,%rsi,4),%rsi
  call   <memcpy@plt>
  jmp    L138
  nopw   0x0(%rax,%rax,1)
L42:
  lea    D19(%rip),%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L138
  nopw   0x0(%rax,%rax,1)
L108:
  xor    %esi,%esi
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L92
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
datatype: string
name: extract_symbolic_data
arguments: tag: D16
datatype: string

tool: D19: merge

tool: D16: %s: both a and b not done!


assistant: ```c
int *merge(int *a, int an, int *b, int bn) {
  int i, j, k;
  int *c;
  c = malloc(sizeof(int) * (an + bn));
  for (i = j = k = 0; i < an + bn; i++) {
    if ((j == an) || (k == bn))
      break;
    if (a[j] <= b[k]) {
      c[i] = a[j];
      j++;
    } else {
      c[i] = b[k];
      k++;
    }
  }
  if (j < an) {
    for (; j < an; j++, i++)
      c[i] = a[j];
  } else if (k < bn) {
    for (; k < bn; k++, i++)
      c[i] = b[k];
  } else {
    printf("%s: both a and b not done!\n", __func__);
  }
  for (i = 0; i < an + bn; i++) {
    a[i] = c[i];
  }
  free(c);
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
data_type: string
name: retrieve_data
arguments: label: D14
data_type: string

tool: D6: Usage  : %s <listen port> [buffer size (KB, default: 400] [send_count (default: 0)]


tool: D14: Example: %s 9000 200 100


assistant: ```c
void usage(char **argv, int exit_code) {
  printf("Usage  : %s <listen port> "
         "[buffer size (KB, default: 400] "
         "[send_count (default: 0)]\n",
         argv[0]);
  printf("Example: %s 9000 200 100\n", argv[0]);
  exit(exit_code);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<usage>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    (%rdi),%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type_guessed: string
name: unpack_data
arguments: data_tag: D10
type_guessed: string

tool: D19: Usage  : %s <listen port> [buffer size (KB, default: 400] [send_count (default: 0)]


tool: D10: Example: %s 9000 200 100


assistant: ```c
void usage(char **argv, int exit_code) {
  printf("Usage  : %s <listen port> "
         "[buffer size (KB, default: 400] "
         "[send_count (default: 0)]\n",
         argv[0]);
  printf("Example: %s 9000 200 100\n", argv[0]);
  exit(exit_code);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string

tool: D1: Usage  : %s <listen port> [buffer size (KB, default: 400] [send_count (default: 0)]


tool: D5: Example: %s 9000 200 100


assistant: ```c
void usage(char **argv, int exit_code) {
  printf("Usage  : %s <listen port> "
         "[buffer size (KB, default: 400] "
         "[send_count (default: 0)]\n",
         argv[0]);
  printf("Example: %s 9000 200 100\n", argv[0]);
  exit(exit_code);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <exit@plt>
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: string
name: unpack_data
arguments: tag: D7
type_infered: string

tool: D5: Usage  : %s <listen port> [buffer size (KB, default: 400] [send_count (default: 0)]


tool: D7: Example: %s 9000 200 100


assistant: ```c
void usage(char **argv, int exit_code) {
  printf("Usage  : %s <listen port> "
         "[buffer size (KB, default: 400] "
         "[send_count (default: 0)]\n",
         argv[0]);
  printf("Example: %s 9000 200 100\n", argv[0]);
  exit(exit_code);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<file_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <fstat@plt>
  test   %rax,%rax
  je     L9
  mov    $0xffffffffffffffff,%rax
  jmp    L7
L9:
  mov    -0x10(%rbp),%rax
L7:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
size_t file_size(int fd) {
  struct stat stbuf;
  if (fstat(fd, &stbuf) != 0) {
    return ((size_t)-1);
  }
  return stbuf.st_size;
}
```
user: What is the c source code of the assembly code below:

```asm
<file_size>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <fstat@plt>
  test   %rax,%rax
  mov    $0xffffffffffffffff,%rax
  cmove  (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t file_size(int fd) {
  struct stat stbuf;
  if (fstat(fd, &stbuf) != 0) {
    return ((size_t)-1);
  }
  return stbuf.st_size;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<file_size>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <fstat@plt>
  test   %rax,%rax
  mov    $0xffffffffffffffff,%rax
  cmove  (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t file_size(int fd) {
  struct stat stbuf;
  if (fstat(fd, &stbuf) != 0) {
    return ((size_t)-1);
  }
  return stbuf.st_size;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<file_size>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <fstat@plt>
  test   %rax,%rax
  mov    $0xffffffffffffffff,%rax
  cmove  (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t file_size(int fd) {
  struct stat stbuf;
  if (fstat(fd, &stbuf) != 0) {
    return ((size_t)-1);
  }
  return stbuf.st_size;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcmp2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L55
  cmpq   $0x0,-0x10(%rbp)
  jne    L8
L55:
  cmpq   $0x0,-0x8(%rbp)
  jne    L47
  cmpq   $0x0,-0x10(%rbp)
  jne    L47
  mov    $0x0,%eax
  jmp    L58
L47:
  cmpq   $0x0,-0x8(%rbp)
  jne    L25
  mov    $0xffffffff,%eax
  jmp    L58
L25:
  mov    $0x1,%eax
  jmp    L58
L8:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
L58:
  leave
  ret
```

assistant: ```c
int strcmp2(const char *s1, const char *s2) {
  if (s1 == NULL || s2 == NULL) {
    if (s1 == NULL && s2 == NULL) {
      return 0;
    } else if (s1 == NULL) {
      return -1;
    } else {
      return 1;
    }
  } else {
    return strcmp(s1, s2);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<strcmp2>:
  endbr64
  test   %rdi,%rdi
  je     L23
  test   %rsi,%rsi
  je     L23
  sub    $0x8,%rsp
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
L23:
  mov    $0x0,%eax
  or     %rdi,%rsi
  je     L27
  cmp    $0x1,%rdi
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
L27:
  ret
```

assistant: ```c
int strcmp2(const char *s1, const char *s2) {
  if (s1 == NULL || s2 == NULL) {
    if (s1 == NULL && s2 == NULL) {
      return 0;
    } else if (s1 == NULL) {
      return -1;
    } else {
      return 1;
    }
  } else {
    return strcmp(s1, s2);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcmp2>:
  endbr64
  test   %rdi,%rdi
  je     L3
  test   %rsi,%rsi
  je     L3
  jmp    1050 <strcmp@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  or     %rdi,%rsi
  je     L16
  cmp    $0x1,%rdi
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
  nopl   0x0(%rax)
L16:
  ret
```

assistant: ```c
int strcmp2(const char *s1, const char *s2) {
  if (s1 == NULL || s2 == NULL) {
    if (s1 == NULL && s2 == NULL) {
      return 0;
    } else if (s1 == NULL) {
      return -1;
    } else {
      return 1;
    }
  } else {
    return strcmp(s1, s2);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcmp2>:
  endbr64
  test   %rdi,%rdi
  je     L20
  test   %rsi,%rsi
  je     L20
  jmp    1050 <strcmp@plt>
  nopl   0x0(%rax,%rax,1)
L20:
  xor    %eax,%eax
  or     %rdi,%rsi
  je     L1
  cmp    $0x1,%rdi
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
  nopl   0x0(%rax)
L1:
  ret
```

assistant: ```c
int strcmp2(const char *s1, const char *s2) {
  if (s1 == NULL || s2 == NULL) {
    if (s1 == NULL && s2 == NULL) {
      return 0;
    } else if (s1 == NULL) {
      return -1;
    } else {
      return 1;
    }
  } else {
    return strcmp(s1, s2);
  }
}
```