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

```asm
<setObjAlt>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  call   <getSexComponents@plt>
  lea    0x10(%rsp),%rbx
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  lea    D8(%rip),%rdx
  mov    $0x10,%esi
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <setStandardProcedure@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x30,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
type_guessed: string

tool: D8: #:Sa%+02d*%02d#

assistant: ```c
int setObjAlt(double alt) {
  int d, m, s;
  char tempString[16];
  getSexComponents(alt, &d, &m, &s);
  snprintf(tempString, sizeof(tempString), "#:Sa%+02d*%02d#", d, m);
  return (setStandardProcedure(tempString));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setObjAlt>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <getSexComponents@plt>
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  mov    %rbp,%rdi
  lea    D8(%rip),%rdx
  mov    $0x10,%esi
  call   <snprintf@plt>
  mov    %rbp,%rdi
  call   <setStandardProcedure@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x30,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
type_infered: string

tool: D8: #:Sa%+02d*%02d#

assistant: ```c
int setObjAlt(double alt) {
  int d, m, s;
  char tempString[16];
  getSexComponents(alt, &d, &m, &s);
  snprintf(tempString, sizeof(tempString), "#:Sa%+02d*%02d#", d, m);
  return (setStandardProcedure(tempString));
}
```
user: What is the c source code of the assembly code below:

```asm
<setObjAlt>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <getSexComponents@plt>
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  mov    %rbp,%rdi
  lea    D7(%rip),%rdx
  mov    $0x10,%esi
  call   <snprintf@plt>
  mov    %rbp,%rdi
  call   <setStandardProcedure@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x30,%rsp
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string

tool: D7: #:Sa%+02d*%02d#

assistant: ```c
int setObjAlt(double alt) {
  int d, m, s;
  char tempString[16];
  getSexComponents(alt, &d, &m, &s);
  snprintf(tempString, sizeof(tempString), "#:Sa%+02d*%02d#", d, m);
  return (setStandardProcedure(tempString));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <enableAnalog@plt>
  mov    $0x7a120,%edx
  mov    $0xf4240,%esi
  mov    $0x3,%edi
  call   <initPWM@plt>
L3:
  mov    $0x5,%edi
  call   <getAnalog@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   $0xf4240,%eax,%eax
  lea    0xfff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0xc,%eax
  mov    %eax,%esi
  mov    $0x3,%edi
  call   <setDutyCyclePWM@plt>
  mov    $0x1,%edi
  call   <usleep@plt>
  jmp    L3
```

assistant: ```c
int main() {
  int value;
  enableAnalog(0);
  initPWM(3, 1 * 1000000, 500 * 1000);
  while (1) {
    value = getAnalog(5);
    setDutyCyclePWM(3, (value * 1 * 1000000) / 4096);
    usleep(1);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <enableAnalog@plt>
  mov    $0x7a120,%edx
  mov    $0xf4240,%esi
  mov    $0x3,%edi
  call   <initPWM@plt>
L4:
  mov    $0x5,%edi
  call   <getAnalog@plt>
  imul   $0xf4240,%eax,%eax
  lea    0xfff(%rax),%esi
  test   %eax,%eax
  cmovns %eax,%esi
  sar    $0xc,%esi
  mov    $0x3,%edi
  call   <setDutyCyclePWM@plt>
  mov    $0x1,%edi
  call   <usleep@plt>
  jmp    L4
```

assistant: ```c
int main() {
  int value;
  enableAnalog(0);
  initPWM(3, 1 * 1000000, 500 * 1000);
  while (1) {
    value = getAnalog(5);
    setDutyCyclePWM(3, (value * 1 * 1000000) / 4096);
    usleep(1);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  xor    %edi,%edi
  sub    $0x8,%rsp
  call   <enableAnalog@plt>
  mov    $0x7a120,%edx
  mov    $0xf4240,%esi
  mov    $0x3,%edi
  call   <initPWM@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  mov    $0x5,%edi
  call   <getAnalog@plt>
  mov    $0x3,%edi
  imul   $0xf4240,%eax,%eax
  test   %eax,%eax
  lea    0xfff(%rax),%esi
  cmovns %eax,%esi
  sar    $0xc,%esi
  call   <setDutyCyclePWM@plt>
  mov    $0x1,%edi
  call   <usleep@plt>
  jmp    L0
```

assistant: ```c
int main() {
  int value;
  enableAnalog(0);
  initPWM(3, 1 * 1000000, 500 * 1000);
  while (1) {
    value = getAnalog(5);
    setDutyCyclePWM(3, (value * 1 * 1000000) / 4096);
    usleep(1);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<welcome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D6
type_guessed: string
name: unpack_data
arguments: tag: D22
type_guessed: string
name: unpack_data
arguments: tag: D4
type_guessed: string

tool: D6: enter you name : 

tool: D22: %100s

tool: D4: Welcome %s!


assistant: ```c
void welcome() {
  char name[100];
  printf("enter you name : ");
  scanf("%100s", name);
  printf("Welcome %s!\n", name);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<welcome>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x70,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
type: string
name: unpack_data
arguments: data_label: D9
type: string
name: unpack_data
arguments: data_label: D8
type: string

tool: D6: enter you name : 

tool: D9: %100s

tool: D8: Welcome %s!


assistant: ```c
void welcome() {
  char name[100];
  printf("enter you name : ");
  scanf("%100s", name);
  printf("Welcome %s!\n", name);
}
```
user: What is the c source code of the assembly code below:

```asm
<welcome>:
  endbr64
  push   %rbp
  lea    D26(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x70,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D26
data_type: string
name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D1
data_type: string

tool: D26: enter you name : 

tool: D13: %100s

tool: D1: Welcome %s!


assistant: ```c
void welcome() {
  char name[100];
  printf("enter you name : ");
  scanf("%100s", name);
  printf("Welcome %s!\n", name);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<welcome>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x70,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
data_type: string
name: disassemble_data
arguments: label: D17
data_type: string
name: disassemble_data
arguments: label: D2
data_type: string

tool: D1: enter you name : 

tool: D17: %100s

tool: D2: Welcome %s!


assistant: ```c
void welcome() {
  char name[100];
  printf("enter you name : ");
  scanf("%100s", name);
  printf("Welcome %s!\n", name);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cfttree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  je     L3
  mov    -0x1c(%rbp),%eax
  and    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L36
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftmdl1@plt>
  jmp    L1
L36:
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftmdl2@plt>
  jmp    L1
L3:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L30
L87:
  shll   $0x2,-0x4(%rbp)
  sarl   $0x2,-0xc(%rbp)
L30:
  mov    -0xc(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  je     L87
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L73
  jmp    L97
L14:
  mov    -0x18(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftmdl1@plt>
  sarl   $0x2,-0x4(%rbp)
L97:
  cmpl   $0x80,-0x4(%rbp)
  jg     L14
  jmp    L1
L89:
  mov    -0x18(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftmdl2@plt>
  sarl   $0x2,-0x4(%rbp)
L73:
  cmpl   $0x80,-0x4(%rbp)
  jg     L89
L1:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int cfttree(int n, int j, int k, float *a) {
  void cftmdl1(int n, float *a);
  void cftmdl2(int n, float *a);
  int i, isplt, m;
  if ((k & 3) != 0) {
    isplt = k & 1;
    if (isplt != 0) {
      cftmdl1(n, &a[j - n]);
    } else {
      cftmdl2(n, &a[j - n]);
    }
  } else {
    m = n;
    for (i = k; (i & 3) == 0; i >>= 2) {
      m <<= 2;
    }
    isplt = i & 1;
    if (isplt != 0) {
      while (m > 128) {
        cftmdl1(m, &a[j - m]);
        m >>= 2;
      }
    } else {
      while (m > 128) {
        cftmdl2(m, &a[j - m]);
        m >>= 2;
      }
    }
  }
  return isplt;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cfttree>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%r12d
  mov    %rcx,%r13
  test   $0x3,%dl
  je     L104
  and    $0x1,%edx
  mov    %edx,%ebp
  je     L34
  sub    %edi,%r12d
  movslq %r12d,%r12
  lea    (%rcx,%r12,4),%rsi
  call   <cftmdl1@plt>
L32:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  sub    %edi,%r12d
  movslq %r12d,%r12
  lea    (%rcx,%r12,4),%rsi
  call   <cftmdl2@plt>
  jmp    L32
L104:
  shl    $0x2,%ebx
  sar    $0x2,%edx
  test   $0x3,%dl
  je     L104
  and    $0x1,%edx
  mov    %edx,%ebp
  jne    L85
  cmp    $0x80,%ebx
  jle    L32
L97:
  mov    %r12d,%eax
  sub    %ebx,%eax
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  mov    %ebx,%edi
  call   <cftmdl2@plt>
  sar    $0x2,%ebx
  cmp    $0x80,%ebx
  jg     L97
  jmp    L32
L85:
  cmp    $0x80,%ebx
  jle    L32
L3:
  mov    %r12d,%eax
  sub    %ebx,%eax
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  mov    %ebx,%edi
  call   <cftmdl1@plt>
  sar    $0x2,%ebx
  cmp    $0x80,%ebx
  jg     L3
  jmp    L32
```

assistant: ```c
int cfttree(int n, int j, int k, float *a) {
  void cftmdl1(int n, float *a);
  void cftmdl2(int n, float *a);
  int i, isplt, m;
  if ((k & 3) != 0) {
    isplt = k & 1;
    if (isplt != 0) {
      cftmdl1(n, &a[j - n]);
    } else {
      cftmdl2(n, &a[j - n]);
    }
  } else {
    m = n;
    for (i = k; (i & 3) == 0; i >>= 2) {
      m <<= 2;
    }
    isplt = i & 1;
    if (isplt != 0) {
      while (m > 128) {
        cftmdl1(m, &a[j - m]);
        m >>= 2;
      }
    } else {
      while (m > 128) {
        cftmdl2(m, &a[j - m]);
        m >>= 2;
      }
    }
  }
  return isplt;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cfttree>:
  endbr64
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  test   $0x3,%dl
  je     L10
  sub    %edi,%ebx
  and    $0x1,%edx
  movslq %ebx,%rbx
  mov    %edx,%r12d
  lea    (%rcx,%rbx,4),%rsi
  je     L23
  call   <cftmdl1@plt>
L45:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  sar    $0x2,%edx
  shl    $0x2,%ebp
  test   $0x3,%dl
  je     L10
  and    $0x1,%edx
  mov    %edx,%r12d
  je     L59
  cmp    $0x80,%ebp
  jle    L45
  nopl   0x0(%rax,%rax,1)
L60:
  mov    %ebx,%eax
  mov    %ebp,%edi
  sub    %ebp,%eax
  sar    $0x2,%ebp
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  call   <cftmdl1@plt>
  cmp    $0x80,%ebp
  jg     L60
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %ebx,%eax
  mov    %ebp,%edi
  sub    %ebp,%eax
  sar    $0x2,%ebp
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  call   <cftmdl2@plt>
L59:
  cmp    $0x80,%ebp
  jg     L24
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  call   <cftmdl2@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cfttree(int n, int j, int k, float *a) {
  void cftmdl1(int n, float *a);
  void cftmdl2(int n, float *a);
  int i, isplt, m;
  if ((k & 3) != 0) {
    isplt = k & 1;
    if (isplt != 0) {
      cftmdl1(n, &a[j - n]);
    } else {
      cftmdl2(n, &a[j - n]);
    }
  } else {
    m = n;
    for (i = k; (i & 3) == 0; i >>= 2) {
      m <<= 2;
    }
    isplt = i & 1;
    if (isplt != 0) {
      while (m > 128) {
        cftmdl1(m, &a[j - m]);
        m >>= 2;
      }
    } else {
      while (m > 128) {
        cftmdl2(m, &a[j - m]);
        m >>= 2;
      }
    }
  }
  return isplt;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cfttree>:
  endbr64
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  test   $0x3,%dl
  je     L41
  sub    %edi,%ebx
  and    $0x1,%edx
  movslq %ebx,%rbx
  mov    %edx,%r12d
  lea    (%rcx,%rbx,4),%rsi
  je     L4
  call   <cftmdl1@plt>
L32:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  sar    $0x2,%edx
  shl    $0x2,%ebp
  test   $0x3,%dl
  je     L41
  and    $0x1,%edx
  mov    %edx,%r12d
  je     L2
  cmp    $0x80,%ebp
  jle    L32
  nopl   0x0(%rax,%rax,1)
L61:
  mov    %ebx,%eax
  mov    %ebp,%edi
  sub    %ebp,%eax
  sar    $0x2,%ebp
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  call   <cftmdl1@plt>
  cmp    $0x80,%ebp
  jg     L61
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %ebx,%eax
  mov    %ebp,%edi
  sub    %ebp,%eax
  sar    $0x2,%ebp
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  call   <cftmdl2@plt>
L2:
  cmp    $0x80,%ebp
  jg     L18
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  call   <cftmdl2@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cfttree(int n, int j, int k, float *a) {
  void cftmdl1(int n, float *a);
  void cftmdl2(int n, float *a);
  int i, isplt, m;
  if ((k & 3) != 0) {
    isplt = k & 1;
    if (isplt != 0) {
      cftmdl1(n, &a[j - n]);
    } else {
      cftmdl2(n, &a[j - n]);
    }
  } else {
    m = n;
    for (i = k; (i & 3) == 0; i >>= 2) {
      m <<= 2;
    }
    isplt = i & 1;
    if (isplt != 0) {
      while (m > 128) {
        cftmdl1(m, &a[j - m]);
        m >>= 2;
      }
    } else {
      while (m > 128) {
        cftmdl2(m, &a[j - m]);
        m >>= 2;
      }
    }
  }
  return isplt;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fm_entry_2nd_stage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x4005b000,-0x8(%rbp)
  mov    $0x4005b000,%eax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  je     L13
  mov    -0x8(%rbp),%rax
  call   *%rax
L13:
  jmp    L13
```

assistant: ```c
void fm_entry_2nd_stage(void) {
  void (*second_stage_entry)(void) = (void *)(0x4005b000);
  if (0xffffffff != *(uint32_t *)(0x4005b000)) {
    second_stage_entry();
  }
  while ((1)) {
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<fm_entry_2nd_stage>:
  endbr64
  cmpl   $0xffffffff,0x4005b000
  je     L5
  sub    $0x8,%rsp
  mov    $0x4005b000,%eax
  call   *%rax
L10:
  jmp    L10
L5:
  jmp    L5
```

assistant: ```c
void fm_entry_2nd_stage(void) {
  void (*second_stage_entry)(void) = (void *)(0x4005b000);
  if (0xffffffff != *(uint32_t *)(0x4005b000)) {
    second_stage_entry();
  }
  while ((1)) {
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fm_entry_2nd_stage>:
  endbr64
  cmpl   $0xffffffff,0x4005b000
  je     L15
  sub    $0x8,%rsp
  mov    $0x4005b000,%eax
  call   *%rax
L24:
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L15:
  jmp    L15
```

assistant: ```c
void fm_entry_2nd_stage(void) {
  void (*second_stage_entry)(void) = (void *)(0x4005b000);
  if (0xffffffff != *(uint32_t *)(0x4005b000)) {
    second_stage_entry();
  }
  while ((1)) {
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fm_entry_2nd_stage>:
  endbr64
  cmpl   $0xffffffff,0x4005b000
  je     L1
  sub    $0x8,%rsp
  mov    $0x4005b000,%eax
  call   *%rax
L26:
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L1:
  jmp    L1
```

assistant: ```c
void fm_entry_2nd_stage(void) {
  void (*second_stage_entry)(void) = (void *)(0x4005b000);
  if (0xffffffff != *(uint32_t *)(0x4005b000)) {
    second_stage_entry();
  }
  while ((1)) {
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_setblock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    D14(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <fcntl@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L2
  mov    $0x0,%eax
  jmp    L27
L2:
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  not    %eax
  and    %eax,-0x4(%rbp)
  mov    D21(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <fcntl@plt>
  test   %rax,%rax
  jns    L31
  mov    $0x0,%eax
  jmp    L27
L31:
  mov    $0x1,%eax
L27:
  leave
  ret
```

assistant: ```c
int _setblock(int fd) {
  int flags;
  if ((flags = fcntl(fd, F_GETFL, 0)) < 0)
    return 0;
  flags &= ~O_NONBLOCK;
  if (fcntl(fd, F_SETFL, flags) < 0)
    return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_setblock>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%edx
  mov    D25(%rip),%rax
  mov    (%rax),%esi
  call   <fcntl@plt>
  mov    %rax,%rcx
  mov    $0x0,%eax
  test   %ecx,%ecx
  js     L9
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  not    %edx
  and    %ecx,%edx
  mov    D29(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <fcntl@plt>
  not    %rax
  shr    $0x3f,%rax
L9:
  pop    %rbx
  ret
```

assistant: ```c
int _setblock(int fd) {
  int flags;
  if ((flags = fcntl(fd, F_GETFL, 0)) < 0)
    return 0;
  flags &= ~O_NONBLOCK;
  if (fcntl(fd, F_SETFL, flags) < 0)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_setblock>:
  endbr64
  mov    D23(%rip),%rax
  push   %rbp
  xor    %edx,%edx
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <fcntl@plt>
  mov    %rax,%rcx
  xor    %eax,%eax
  test   %ecx,%ecx
  js     L3
  mov    D13(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  not    %edx
  mov    (%rax),%esi
  and    %ecx,%edx
  call   <fcntl@plt>
  not    %rax
  shr    $0x3f,%rax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int _setblock(int fd) {
  int flags;
  if ((flags = fcntl(fd, F_GETFL, 0)) < 0)
    return 0;
  flags &= ~O_NONBLOCK;
  if (fcntl(fd, F_SETFL, flags) < 0)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_setblock>:
  endbr64
  mov    D25(%rip),%rax
  push   %rbp
  xor    %edx,%edx
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <fcntl@plt>
  mov    %rax,%rcx
  xor    %eax,%eax
  test   %ecx,%ecx
  js     L9
  mov    D24(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%edx
  mov    D21(%rip),%rax
  not    %edx
  mov    (%rax),%esi
  and    %ecx,%edx
  call   <fcntl@plt>
  not    %rax
  shr    $0x3f,%rax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int _setblock(int fd) {
  int flags;
  if ((flags = fcntl(fd, F_GETFL, 0)) < 0)
    return 0;
  flags &= ~O_NONBLOCK;
  if (fcntl(fd, F_SETFL, flags) < 0)
    return 0;
  return 1;
}
```