ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x6e686f6a,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    D34(%rip),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x18(%rbp)
  lea    -0xd(%rbp),%rdx
  lea    -0x12(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L26
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L8
L26:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L8:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D34
type_guessed: string
name: unpack_data
arguments: tag: D33
type_guessed: string
name: unpack_data
arguments: tag: D37
type_guessed: string
name: unpack_data
arguments: tag: D44
type_guessed: string
name: unpack_data
arguments: tag: D24
type_guessed: string

tool: D34: john

tool: D33: %c

tool: D37: String 1 = %s


tool: D44: Strings are identical

tool: D24: Strings are not identical

assistant: ```c
main() {
  char str1[5];
  char str2[5] = "john";
  scanf("%c", "john");
  printf("String 1 = %s\n", str2);
  int same = 1;
  same = strcmp(str1, str2);
  if (same == 0) {
    printf("Strings are identical");
  } else {
    printf("Strings are not identical");
  }
  getchar();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x6e686f6a,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  lea    D15(%rip),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  lea    0x13(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xe(%rsp),%rdi
  mov    %rbx,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L27
  lea    D48(%rip),%rdi
  call   <printf@plt>
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L27:
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L17
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D24
data_type: string
name: extract_symbolic_data
arguments: tag: D38
data_type: string
name: extract_symbolic_data
arguments: tag: D48
data_type: string
name: extract_symbolic_data
arguments: tag: D46
data_type: string

tool: D15: john

tool: D24: %c

tool: D38: String 1 = %s


tool: D48: Strings are identical

tool: D46: Strings are not identical

assistant: ```c
main() {
  char str1[5];
  char str2[5] = "john";
  scanf("%c", "john");
  printf("String 1 = %s\n", str2);
  int same = 1;
  same = strcmp(str1, str2);
  if (same == 0) {
    printf("Strings are identical");
  } else {
    printf("Strings are not identical");
  }
  getchar();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rsi
  lea    D16(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x13(%rsp),%rbp
  movl   $0x6e686f6a,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xe(%rsp),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L24
  lea    D23(%rip),%rdi
  call   <printf@plt>
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L24:
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L18
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D1
data_type: string
name: parse_data
arguments: label: D16
data_type: string
name: parse_data
arguments: label: D37
data_type: string
name: parse_data
arguments: label: D23
data_type: string
name: parse_data
arguments: label: D9
data_type: string

tool: D1: john

tool: D16: %c

tool: D37: String 1 = %s


tool: D23: Strings are identical

tool: D9: Strings are not identical

assistant: ```c
main() {
  char str1[5];
  char str2[5] = "john";
  scanf("%c", "john");
  printf("String 1 = %s\n", str2);
  int same = 1;
  same = strcmp(str1, str2);
  if (same == 0) {
    printf("Strings are identical");
  } else {
    printf("Strings are not identical");
  }
  getchar();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rsi
  lea    D43(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x13(%rsp),%rbp
  movl   $0x6e686f6a,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xe(%rsp),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L17
  lea    D47(%rip),%rdi
  call   <printf@plt>
L27:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L17:
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L27
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D43
type_infered: string
name: parse_data
arguments: tag: D49
type_infered: string
name: parse_data
arguments: tag: D47
type_infered: string
name: parse_data
arguments: tag: D12
type_infered: string

tool: D7: john

tool: D43: %c

tool: D49: String 1 = %s


tool: D47: Strings are identical

tool: D12: Strings are not identical

assistant: ```c
main() {
  char str1[5];
  char str2[5] = "john";
  scanf("%c", "john");
  printf("String 1 = %s\n", str2);
  int same = 1;
  same = strcmp(str1, str2);
  if (same == 0) {
    printf("Strings are identical");
  } else {
    printf("Strings are not identical");
  }
  getchar();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fdmi_param_set>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x400,%edi
  call   <fdmi_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <putenv@plt>
  nop
  leave
  ret
```

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

tool: D5: %s=%s

assistant: ```c
void fdmi_param_set(char *name, char *value) {
  char *env;
  env = (char *)fdmi_malloc(1024);
  sprintf(env, "%s=%s", name, value);
  putenv(env);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fdmi_param_set>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x400,%edi
  call   <fdmi_malloc@plt>
  mov    %rax,%rbx
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <putenv@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
data_type: string

tool: D5: %s=%s

assistant: ```c
void fdmi_param_set(char *name, char *value) {
  char *env;
  env = (char *)fdmi_malloc(1024);
  sprintf(env, "%s=%s", name, value);
  putenv(env);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fdmi_param_set>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  mov    $0x400,%edi
  push   %rbp
  call   <fdmi_malloc@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D3(%rip),%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <putenv@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type_infered: string

tool: D3: %s=%s

assistant: ```c
void fdmi_param_set(char *name, char *value) {
  char *env;
  env = (char *)fdmi_malloc(1024);
  sprintf(env, "%s=%s", name, value);
  putenv(env);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fdmi_param_set>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  mov    $0x400,%edi
  push   %rbp
  call   <fdmi_malloc@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D8(%rip),%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <putenv@plt>
```

tool_calls: name: read_data
arguments: tag: D8
type_guessed: string

tool: D8: %s=%s

assistant: ```c
void fdmi_param_set(char *name, char *value) {
  char *env;
  env = (char *)fdmi_malloc(1024);
  sprintf(env, "%s=%s", name, value);
  putenv(env);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dbdqr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %rcx,0x10(%rsp)
  mov    %r8,0x18(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmpl   $0x1,(%rdi)
  jle    L16
  mov    %rdi,%r13
  mov    %rdx,%r15
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    $0x0,%r12d
  lea    0x30(%rsp),%r14
L24:
  lea    0x28(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  mov    %r14,%r8
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <dlartg_@plt>
  movsd  0x30(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rbx),%xmm0
  mulsd  0x28(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x8(%rbx),%xmm0
  mulsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x8(%rbx)
  add    $0x1,%r12d
  mov    0x0(%r13),%eax
  sub    $0x1,%eax
  add    $0x8,%rbx
  add    $0x8,%rbp
  cmp    %r12d,%eax
  jg     L24
  cltq
  shl    $0x3,%rax
  add    %rax,%r15
  mov    0x8(%rsp),%rbx
  add    %rax,%rbx
  lea    0x28(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x30(%rsp),%r8
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <dlartg_@plt>
  movsd  0x30(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  movq   $0x0,(%r15)
  movsd  0x28(%rsp),%xmm0
  mov    0x10(%rsp),%rax
  movsd  %xmm0,(%rax)
  movsd  0x20(%rsp),%xmm0
  mov    0x18(%rsp),%rax
  movsd  %xmm0,(%rax)
L16:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dbdqr(int *n, double *d, double *e, double *c1, double *c2) {
  int i;
  double cs, sn, r;
  if (*n < 2) {
    return;
  }
  for (i = 0; i < *n - 1; i++) {
    dlartg_(d + i, e + i, &cs, &sn, &r);
    d[i] = r;
    e[i] = sn * d[i + 1];
    d[i + 1] = cs * d[i + 1];
  }
  i = *n - 1;
  dlartg_(d + i, e + i, &cs, &sn, &r);
  d[i] = r;
  e[i] = 0.0;
  *c1 = sn;
  *c2 = cs;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dbdqr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rsi,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  mov    %rcx,0x20(%rsp)
  mov    %r8,0x28(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  cmpl   $0x1,(%rdi)
  jle    L3
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  xor    %r13d,%r13d
  lea    0x40(%rsp),%r8
  lea    0x38(%rsp),%r15
  lea    0x30(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r8,0x8(%rsp)
  add    $0x1,%r13d
  add    $0x8,%rbx
  add    $0x8,%rbp
  call   <dlartg_@plt>
  movsd  0x40(%rsp),%xmm0
  mov    (%r12),%eax
  mov    0x8(%rsp),%r8
  movsd  %xmm0,-0x8(%rbx)
  movsd  (%rbx),%xmm0
  mulsd  0x38(%rsp),%xmm0
  sub    $0x1,%eax
  movsd  %xmm0,-0x8(%rbp)
  movsd  (%rbx),%xmm0
  mulsd  0x30(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  cmp    %r13d,%eax
  jg     L13
  mov    0x18(%rsp),%rbx
  mov    0x10(%rsp),%r13
  cltq
  mov    %r15,%rcx
  shl    $0x3,%rax
  mov    %r14,%rdx
  add    %rax,%rbx
  add    %rax,%r13
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <dlartg_@plt>
  movsd  0x40(%rsp),%xmm0
  mov    0x20(%rsp),%rax
  movsd  %xmm0,0x0(%r13)
  movsd  0x38(%rsp),%xmm0
  movq   $0x0,(%rbx)
  movsd  %xmm0,(%rax)
  movsd  0x30(%rsp),%xmm0
  mov    0x28(%rsp),%rax
  movsd  %xmm0,(%rax)
L3:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dbdqr(int *n, double *d, double *e, double *c1, double *c2) {
  int i;
  double cs, sn, r;
  if (*n < 2) {
    return;
  }
  for (i = 0; i < *n - 1; i++) {
    dlartg_(d + i, e + i, &cs, &sn, &r);
    d[i] = r;
    e[i] = sn * d[i + 1];
    d[i + 1] = cs * d[i + 1];
  }
  i = *n - 1;
  dlartg_(d + i, e + i, &cs, &sn, &r);
  d[i] = r;
  e[i] = 0.0;
  *c1 = sn;
  *c2 = cs;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dbdqr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rsi,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  mov    %rcx,0x20(%rsp)
  mov    %r8,0x28(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  cmpl   $0x1,(%rdi)
  jle    L4
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  xor    %r13d,%r13d
  lea    0x40(%rsp),%r8
  lea    0x38(%rsp),%r15
  lea    0x30(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r8,0x8(%rsp)
  add    $0x1,%r13d
  add    $0x8,%rbx
  add    $0x8,%rbp
  call   <dlartg_@plt>
  movsd  0x40(%rsp),%xmm0
  mov    (%r12),%eax
  mov    0x8(%rsp),%r8
  movsd  %xmm0,-0x8(%rbx)
  movsd  (%rbx),%xmm0
  mulsd  0x38(%rsp),%xmm0
  sub    $0x1,%eax
  movsd  %xmm0,-0x8(%rbp)
  movsd  (%rbx),%xmm0
  mulsd  0x30(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  cmp    %r13d,%eax
  jg     L1
  mov    0x18(%rsp),%rbx
  mov    0x10(%rsp),%r13
  cltq
  mov    %r15,%rcx
  shl    $0x3,%rax
  mov    %r14,%rdx
  add    %rax,%rbx
  add    %rax,%r13
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <dlartg_@plt>
  movsd  0x40(%rsp),%xmm0
  mov    0x20(%rsp),%rax
  movsd  %xmm0,0x0(%r13)
  movsd  0x38(%rsp),%xmm0
  movq   $0x0,(%rbx)
  movsd  %xmm0,(%rax)
  movsd  0x30(%rsp),%xmm0
  mov    0x28(%rsp),%rax
  movsd  %xmm0,(%rax)
L4:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dbdqr(int *n, double *d, double *e, double *c1, double *c2) {
  int i;
  double cs, sn, r;
  if (*n < 2) {
    return;
  }
  for (i = 0; i < *n - 1; i++) {
    dlartg_(d + i, e + i, &cs, &sn, &r);
    d[i] = r;
    e[i] = sn * d[i + 1];
    d[i + 1] = cs * d[i + 1];
  }
  i = *n - 1;
  dlartg_(d + i, e + i, &cs, &sn, &r);
  d[i] = r;
  e[i] = 0.0;
  *c1 = sn;
  *c2 = cs;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<func2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L18:
  addl   $0x1,-0x4(%rbp)
L4:
  cmpl   $0xffffe,-0x4(%rbp)
  jle    L18
  nop
  pop    %rbp
  ret
```

assistant: ```c
void func2() {
  int count = 0;
  for (count = 0; count < 0XFFFFF; count++)
    ;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<func2>:
  endbr64
  mov    $0xfffff,%eax
L4:
  sub    $0x1,%eax
  jne    L4
  ret
```

assistant: ```c
void func2() {
  int count = 0;
  for (count = 0; count < 0XFFFFF; count++)
    ;
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<func2>:
  endbr64
  ret
```

assistant: ```c
void func2() {
  int count = 0;
  for (count = 0; count < 0XFFFFF; count++)
    ;
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<func2>:
  endbr64
  ret
```

assistant: ```c
void func2() {
  int count = 0;
  for (count = 0; count < 0XFFFFF; count++)
    ;
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<absolute_time_to_delta_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  test   %rax,%rax
  jns    L16
  mov    $0xffffffff,%eax
  jmp    L26
L16:
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  sub    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  sub    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jns    L19
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x3b9aca00(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
L19:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    $0x3b9ac9ff,%rax
  jle    L33
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    -0x3b9aca00(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
L33:
  mov    $0x0,%eax
L26:
  leave
  ret
```

assistant: ```c
int absolute_time_to_delta_time(struct timespec *delta, const struct timespec *absolute, clockid_t clockid) {
  if (clock_gettime(clockid, delta) < 0)
    return -1;
  delta->tv_sec = absolute->tv_sec - delta->tv_sec;
  delta->tv_nsec = absolute->tv_nsec - delta->tv_nsec;
  if (delta->tv_nsec < 0L) {
    delta->tv_nsec += 1000000000L;
    delta->tv_sec -= 1;
  }
  if (delta->tv_nsec >= 1000000000L) {
    delta->tv_nsec -= 1000000000L;
    delta->tv_sec += 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<absolute_time_to_delta_time>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%edi
  mov    %rbx,%rsi
  call   <clock_gettime@plt>
  test   %rax,%rax
  js     L54
  mov    0x8(%rbp),%rdx
  sub    0x8(%rbx),%rdx
  mov    %rdx,0x8(%rbx)
  mov    0x0(%rbp),%rax
  sub    (%rbx),%rax
  mov    %rax,(%rbx)
  js     L7
  cmp    $0x3b9ac9ff,%rax
  jle    L37
  sub    $0x3b9aca00,%rax
  mov    %rax,(%rbx)
  add    $0x1,%rdx
  mov    %rdx,0x8(%rbx)
  mov    $0x0,%eax
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  add    $0x3b9aca00,%rax
  mov    %rax,(%rbx)
  sub    $0x1,%rdx
  mov    %rdx,0x8(%rbx)
  mov    $0x0,%eax
  jmp    L26
L54:
  mov    $0xffffffff,%eax
  jmp    L26
L37:
  mov    $0x0,%eax
  jmp    L26
```

assistant: ```c
int absolute_time_to_delta_time(struct timespec *delta, const struct timespec *absolute, clockid_t clockid) {
  if (clock_gettime(clockid, delta) < 0)
    return -1;
  delta->tv_sec = absolute->tv_sec - delta->tv_sec;
  delta->tv_nsec = absolute->tv_nsec - delta->tv_nsec;
  if (delta->tv_nsec < 0L) {
    delta->tv_nsec += 1000000000L;
    delta->tv_sec -= 1;
  }
  if (delta->tv_nsec >= 1000000000L) {
    delta->tv_nsec -= 1000000000L;
    delta->tv_sec += 1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<absolute_time_to_delta_time>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %edx,%edi
  mov    %rbx,%rsi
  sub    $0x8,%rsp
  call   <clock_gettime@plt>
  test   %rax,%rax
  js     L45
  mov    0x8(%rbp),%rdx
  sub    0x8(%rbx),%rdx
  mov    %rdx,0x8(%rbx)
  mov    0x0(%rbp),%rax
  sub    (%rbx),%rax
  js     L23
  cmp    $0x3b9ac9ff,%rax
  jle    L40
  add    $0x1,%rdx
  sub    $0x3b9aca00,%rax
  mov    %rdx,0x8(%rbx)
  mov    %rax,(%rbx)
  xor    %eax,%eax
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L23:
  add    $0x3b9aca00,%rax
  sub    $0x1,%rdx
  mov    %rax,(%rbx)
  xor    %eax,%eax
  mov    %rdx,0x8(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  mov    %rax,(%rbx)
  xor    %eax,%eax
  jmp    L17
L45:
  mov    $0xffffffff,%eax
  jmp    L17
```

assistant: ```c
int absolute_time_to_delta_time(struct timespec *delta, const struct timespec *absolute, clockid_t clockid) {
  if (clock_gettime(clockid, delta) < 0)
    return -1;
  delta->tv_sec = absolute->tv_sec - delta->tv_sec;
  delta->tv_nsec = absolute->tv_nsec - delta->tv_nsec;
  if (delta->tv_nsec < 0L) {
    delta->tv_nsec += 1000000000L;
    delta->tv_sec -= 1;
  }
  if (delta->tv_nsec >= 1000000000L) {
    delta->tv_nsec -= 1000000000L;
    delta->tv_sec += 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<absolute_time_to_delta_time>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %edx,%edi
  mov    %rbx,%rsi
  sub    $0x8,%rsp
  call   <clock_gettime@plt>
  test   %rax,%rax
  js     L48
  mov    0x8(%rbp),%rdx
  sub    0x8(%rbx),%rdx
  mov    %rdx,0x8(%rbx)
  mov    0x0(%rbp),%rax
  sub    (%rbx),%rax
  js     L9
  cmp    $0x3b9ac9ff,%rax
  jle    L12
  add    $0x1,%rdx
  sub    $0x3b9aca00,%rax
  mov    %rdx,0x8(%rbx)
  mov    %rax,(%rbx)
  xor    %eax,%eax
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  add    $0x3b9aca00,%rax
  sub    $0x1,%rdx
  mov    %rax,(%rbx)
  xor    %eax,%eax
  mov    %rdx,0x8(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %rax,(%rbx)
  xor    %eax,%eax
  jmp    L11
L48:
  mov    $0xffffffff,%eax
  jmp    L11
```

assistant: ```c
int absolute_time_to_delta_time(struct timespec *delta, const struct timespec *absolute, clockid_t clockid) {
  if (clock_gettime(clockid, delta) < 0)
    return -1;
  delta->tv_sec = absolute->tv_sec - delta->tv_sec;
  delta->tv_nsec = absolute->tv_nsec - delta->tv_nsec;
  if (delta->tv_nsec < 0L) {
    delta->tv_nsec += 1000000000L;
    delta->tv_sec -= 1;
  }
  if (delta->tv_nsec >= 1000000000L) {
    delta->tv_nsec -= 1000000000L;
    delta->tv_sec += 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<unset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    D25(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L30
  mov    D25(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <min_putp@plt>
  jmp    L10
L30:
  orl    $0x1,-0x4(%rbp)
L10:
  mov    D49(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L9
  mov    D49(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <min_putp@plt>
  mov    D25(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <min_putp@plt>
  mov    -0x4(%rbp),%eax
  jmp    L4
L9:
  mov    $0x1,%eax
L4:
  leave
  ret
```

assistant: ```c
int unset() {
  int ret = 0;
  if (exit_alt_charset_mode) {
    min_putp(exit_alt_charset_mode);
  } else
    ret |= 1;
  if (exit_attribute_mode) {
    min_putp(exit_attribute_mode);
    min_putp(exit_alt_charset_mode);
    return ret;
  } else
    return 1;
}
```