ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rdi
  mov    %rsp,%rbp
  push   %rbx
  lea    -0x120(%rbp),%rbx
  mov    %rbx,%rsi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rsp,%rsi
  movslq %eax,%rdx
  lea    0xf(%rdx),%rax
  mov    %rdx,%rdi
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L89
L61:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L61
L89:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L36
L10:
  mov    %rsp,%r9
  test   %edi,%edi
  jle    L93
  lea    0x1(%rsp),%rax
  lea    -0x1(%rdi),%r8d
  mov    %rsp,%rcx
  lea    -0x1(%rbx,%rdx,1),%rdx
  add    %rax,%r8
  jmp    L84
  nopl   0x0(%rax,%rax,1)
L40:
  add    $0x1,%rax
L84:
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  mov    %sil,(%rcx)
  mov    %rax,%rcx
  cmp    %r8,%rax
  jne    L40
  mov    %edi,%ecx
  xor    %eax,%eax
  mov    $0x1,%edx
  xor    %esi,%esi
  nopl   0x0(%rax)
L77:
  movzbl (%r9,%rax,1),%edi
  cmp    %dil,(%rbx,%rax,1)
  cmovne %esi,%edx
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L77
  test   %edx,%edx
  jne    L93
  lea    D15(%rip),%rdi
  call   <printf@plt>
L107:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L56
  mov    -0x8(%rbp),%rbx
  xor    %eax,%eax
  leave
  ret
L93:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  jmp    L107
L36:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L10
L56:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D12
datatype: string
name: retrieve_data
arguments: tag: D15
datatype: string
name: retrieve_data
arguments: tag: D19
datatype: string

tool: D12: %s

tool: D15: NAO

tool: D19: SIM

assistant: ```c
int main() {
  char seq[255];
  int i, j, tam;
  int pali = 1;
  scanf("%s", seq);
  tam = strlen(seq);
  char seq2[tam];
  for (i = 0; i < tam; i++) {
    seq2[i] = seq[(tam - i) - 1];
  }
  for (i = 0; i < tam; i++) {
    if (seq[i] != seq2[i]) {
      pali = 0;
    }
  }
  if (pali == 0) {
    printf("NAO");
  } else {
    printf("SIM");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rdi
  mov    %rsp,%rbp
  push   %rbx
  lea    -0x120(%rbp),%rbx
  mov    %rbx,%rsi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rsp,%rsi
  movslq %eax,%rdx
  lea    0xf(%rdx),%rax
  mov    %rdx,%rdi
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L4
L84:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L84
L4:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L85
L8:
  mov    %rsp,%r9
  test   %edi,%edi
  jle    L76
  lea    0x1(%rsp),%rax
  lea    -0x1(%rdi),%r8d
  mov    %rsp,%rcx
  lea    -0x1(%rbx,%rdx,1),%rdx
  add    %rax,%r8
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L75:
  add    $0x1,%rax
L14:
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  mov    %sil,(%rcx)
  mov    %rax,%rcx
  cmp    %r8,%rax
  jne    L75
  mov    %edi,%ecx
  xor    %eax,%eax
  mov    $0x1,%edx
  xor    %esi,%esi
  nopl   0x0(%rax)
L97:
  movzbl (%rbx,%rax,1),%edi
  cmp    %dil,(%r9,%rax,1)
  cmovne %esi,%edx
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L97
  test   %edx,%edx
  jne    L76
  lea    D25(%rip),%rdi
  call   <printf@plt>
L99:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    -0x8(%rbp),%rbx
  xor    %eax,%eax
  leave
  ret
L76:
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L99
L85:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L8
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D25
data_type: string
name: parse_data
arguments: data_tag: D22
data_type: string

tool: D19: %s

tool: D25: NAO

tool: D22: SIM

assistant: ```c
int main() {
  char seq[255];
  int i, j, tam;
  int pali = 1;
  scanf("%s", seq);
  tam = strlen(seq);
  char seq2[tam];
  for (i = 0; i < tam; i++) {
    seq2[i] = seq[(tam - i) - 1];
  }
  for (i = 0; i < tam; i++) {
    if (seq[i] != seq2[i]) {
      pali = 0;
    }
  }
  if (pali == 0) {
    printf("NAO");
  } else {
    printf("SIM");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<doit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0x0,%eax
  call   <foo2@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void doit() {
  foo();
  foo2();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<doit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0x0,%eax
  call   <foo2@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void doit() {
  foo();
  foo2();
}
```
user: Translate the below assembly code into C programming language:

```asm
<doit>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <foo2@plt>
```

assistant: ```c
void doit() {
  foo();
  foo2();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<doit>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <foo2@plt>
```

assistant: ```c
void doit() {
  foo();
  foo2();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<transposeMatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  mov    -0x24(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x8(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x18(%rbp)
  jmp    L11
L24:
  movl   $0x0,-0x14(%rbp)
  jmp    L30
L28:
  mov    -0x18(%rbp),%ecx
  movslq %ecx,%rsi
  movslq %eax,%rcx
  imul   %rsi,%rcx
  lea    0x0(,%rcx,4),%rsi
  mov    -0x30(%rbp),%rcx
  add    %rcx,%rsi
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rdi
  movslq %edx,%rcx
  imul   %rdi,%rcx
  lea    0x0(,%rcx,4),%rdi
  mov    -0x38(%rbp),%rcx
  add    %rcx,%rdi
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rcx
  mov    (%rsi,%rcx,4),%esi
  mov    -0x18(%rbp),%ecx
  movslq %ecx,%rcx
  mov    %esi,(%rdi,%rcx,4)
  addl   $0x1,-0x14(%rbp)
L30:
  mov    -0x14(%rbp),%ecx
  cmp    -0x28(%rbp),%ecx
  jl     L28
  addl   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%ecx
  cmp    -0x24(%rbp),%ecx
  jl     L24
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int transposeMatrix(int nRows, int nCols, int matrix_ij[nRows][nCols], int matrix_ji[nCols][nRows]) {
  int i, j;
  for (i = 0; i < nRows; ++i)
    for (j = 0; j < nCols; ++j) {
      matrix_ji[j][i] = matrix_ij[i][j];
    }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<transposeMatrix>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jle    L49
  push   %rbx
  mov    %esi,%r10d
  movslq %esi,%r11
  movslq %edi,%rdi
  mov    %rcx,%r8
  shl    $0x2,%r11
  lea    -0x1(%rsi),%r9d
  lea    0x4(%rdx,%r9,4),%rsi
  lea    -0x1(%rax),%eax
  lea    0x4(%rcx,%rax,4),%rbx
  shl    $0x2,%rdi
  not    %r9
  shl    $0x2,%r9
  jmp    L52
L13:
  mov    (%rax),%ecx
  mov    %ecx,(%rdx)
  add    $0x4,%rax
  add    %rdi,%rdx
  cmp    %rsi,%rax
  jne    L13
L3:
  add    $0x4,%r8
  add    %r11,%rsi
  cmp    %rbx,%r8
  je     L0
L52:
  lea    (%r9,%rsi,1),%rax
  mov    %r8,%rdx
  test   %r10d,%r10d
  jg     L13
  jmp    L3
L0:
  mov    $0x0,%eax
  pop    %rbx
  ret
L49:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int transposeMatrix(int nRows, int nCols, int matrix_ij[nRows][nCols], int matrix_ji[nCols][nRows]) {
  int i, j;
  for (i = 0; i < nRows; ++i)
    for (j = 0; j < nCols; ++j) {
      matrix_ji[j][i] = matrix_ij[i][j];
    }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<transposeMatrix>:
  endbr64
  movslq %edi,%rdi
  test   %edi,%edi
  jle    L34
  movslq %esi,%r10
  mov    %rdi,%rax
  mov    %rcx,%r8
  mov    $0xfffffffffffffffc,%r11
  mov    %r10,%r9
  sub    $0x1,%eax
  push   %rbx
  shl    $0x2,%r10
  lea    -0x1(%r9),%ecx
  lea    0x4(%r8,%rax,4),%rbx
  shl    $0x2,%rdi
  shl    $0x2,%rcx
  lea    0x4(%rdx,%rcx,1),%rsi
  sub    %rcx,%r11
  nop
L27:
  lea    (%r11,%rsi,1),%rax
  mov    %r8,%rdx
  test   %r9d,%r9d
  jle    L21
  nopl   0x0(%rax)
L3:
  mov    (%rax),%ecx
  add    $0x4,%rax
  mov    %ecx,(%rdx)
  add    %rdi,%rdx
  cmp    %rsi,%rax
  jne    L3
L21:
  add    $0x4,%r8
  add    %r10,%rsi
  cmp    %rbx,%r8
  jne    L27
  xor    %eax,%eax
  pop    %rbx
  ret
L34:
  xor    %eax,%eax
  ret
```

assistant: ```c
int transposeMatrix(int nRows, int nCols, int matrix_ij[nRows][nCols], int matrix_ji[nCols][nRows]) {
  int i, j;
  for (i = 0; i < nRows; ++i)
    for (j = 0; j < nCols; ++j) {
      matrix_ji[j][i] = matrix_ij[i][j];
    }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<transposeMatrix>:
  endbr64
  movslq %edi,%rdi
  movslq %esi,%r9
  mov    %rdi,%rax
  test   %edi,%edi
  jle    L11
  test   %esi,%esi
  jle    L11
  shl    $0x2,%r9
  mov    %rcx,%r8
  lea    -0x1(%rsi),%ecx
  sub    $0x1,%eax
  shl    $0x2,%rcx
  mov    $0xfffffffffffffffc,%r10
  lea    0x4(%r8,%rax,4),%r11
  shl    $0x2,%rdi
  lea    0x4(%rdx,%rcx,1),%rsi
  sub    %rcx,%r10
  xchg   %ax,%ax
L14:
  lea    (%r10,%rsi,1),%rax
  mov    %r8,%rdx
  nopw   0x0(%rax,%rax,1)
L38:
  mov    (%rax),%ecx
  add    $0x4,%rax
  mov    %ecx,(%rdx)
  add    %rdi,%rdx
  cmp    %rsi,%rax
  jne    L38
  add    $0x4,%r8
  add    %r9,%rsi
  cmp    %r11,%r8
  jne    L14
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
int transposeMatrix(int nRows, int nCols, int matrix_ij[nRows][nCols], int matrix_ji[nCols][nRows]) {
  int i, j;
  for (i = 0; i < nRows; ++i)
    for (j = 0; j < nCols; ++j) {
      matrix_ji[j][i] = matrix_ij[i][j];
    }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x61,-0x11(%rbp)
  movl   $0x20,-0x10(%rbp)
  movss  D20(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D20
datatype: f32
name: retrieve_data
arguments: tag: D33
datatype: string
name: retrieve_data
arguments: tag: D7
datatype: string
name: retrieve_data
arguments: tag: D4
datatype: string

tool: D20: 2.3

tool: D33: Address of Character is %p


tool: D7: Address of Integer   is %p


tool: D4: Address of Float     is %p


assistant: ```c
int main() {
  char c = 'a';
  int i = 32;
  float f = 2.3;
  printf("Address of Character is %p\n", &c);
  printf("Address of Integer   is %p\n", &i);
  printf("Address of Float     is %p\n", &f);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x61,0xf(%rsp)
  movl   $0x20,0x10(%rsp)
  movl   $0x40133333,0x14(%rsp)
  lea    0xf(%rsp),%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: string
name: disassemble_data
arguments: tag: D28
type_guessed: string
name: disassemble_data
arguments: tag: D16
type_guessed: string

tool: D19: Address of Character is %p


tool: D28: Address of Integer   is %p


tool: D16: Address of Float     is %p


assistant: ```c
int main() {
  char c = 'a';
  int i = 32;
  float f = 2.3;
  printf("Address of Character is %p\n", &c);
  printf("Address of Integer   is %p\n", &i);
  printf("Address of Float     is %p\n", &f);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D4(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rsi
  movb   $0x61,0xf(%rsp)
  movl   $0x20,0x10(%rsp)
  movl   $0x40133333,0x14(%rsp)
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D4
type: string
name: read_data
arguments: label: D5
type: string
name: read_data
arguments: label: D25
type: string

tool: D4: Address of Character is %p


tool: D5: Address of Integer   is %p


tool: D25: Address of Float     is %p


assistant: ```c
int main() {
  char c = 'a';
  int i = 32;
  float f = 2.3;
  printf("Address of Character is %p\n", &c);
  printf("Address of Integer   is %p\n", &i);
  printf("Address of Float     is %p\n", &f);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D18(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rsi
  movb   $0x61,0xf(%rsp)
  movl   $0x20,0x10(%rsp)
  movl   $0x40133333,0x14(%rsp)
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type: string
name: extract_symbolic_data
arguments: tag: D15
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string

tool: D18: Address of Character is %p


tool: D15: Address of Integer   is %p


tool: D7: Address of Float     is %p


assistant: ```c
int main() {
  char c = 'a';
  int i = 32;
  float f = 2.3;
  printf("Address of Character is %p\n", &c);
  printf("Address of Integer   is %p\n", &i);
  printf("Address of Float     is %p\n", &f);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<substring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L34:
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L34
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L20:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L20
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *substring(char *string, int position, int length) {
  char *pointer = malloc(length + 1);
  int c;
  for (c = 0; c < position - 1; c++) {
    string++;
  }
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<substring>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %edx,%ebp
  lea    0x1(%rdx),%edi
  call   <malloc@plt>
  cmp    $0x1,%r12d
  jle    L29
  lea    -0x1(%r12),%ecx
  mov    $0x0,%edx
L36:
  add    $0x1,%edx
  cmp    %ecx,%edx
  jne    L36
  lea    -0x2(%r12),%edx
  lea    0x1(%rbx,%rdx,1),%rbx
L29:
  test   %ebp,%ebp
  jle    L18
  mov    %ebp,%esi
  mov    $0x0,%edx
L24:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L24
L9:
  movslq %ebp,%rbp
  movb   $0x0,(%rax,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    $0x0,%ebp
  jmp    L9
```

assistant: ```c
char *substring(char *string, int position, int length) {
  char *pointer = malloc(length + 1);
  int c;
  for (c = 0; c < position - 1; c++) {
    string++;
  }
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<substring>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %edx,%ebx
  call   <malloc@plt>
  cmp    $0x1,%r12d
  jle    L24
  lea    -0x2(%r12),%edx
  lea    0x1(%rbp,%rdx,1),%rbp
L24:
  test   %ebx,%ebx
  jle    L4
  movslq %ebx,%rdx
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L16:
  movzbl 0x0(%rbp,%rcx,1),%esi
  mov    %sil,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L16
  add    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L4:
  mov    %rax,%rdx
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *substring(char *string, int position, int length) {
  char *pointer = malloc(length + 1);
  int c;
  for (c = 0; c < position - 1; c++) {
    string++;
  }
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<substring>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %edx,%ebx
  call   <malloc@plt>
  mov    %rax,%r8
  cmp    $0x1,%r12d
  jle    L133
  lea    -0x2(%r12),%eax
  lea    0x1(%rbp,%rax,1),%rbp
L133:
  test   %ebx,%ebx
  jle    L62
  lea    0x1(%rbp),%rcx
  mov    %r8,%rdx
  lea    -0x1(%rbx),%eax
  mov    %ebx,%esi
  sub    %rcx,%rdx
  cmp    $0xe,%rdx
  jbe    L98
  cmp    $0x6,%eax
  jbe    L98
  cmp    $0xe,%eax
  jbe    L11
  mov    %ebx,%edx
  xor    %eax,%eax
  shr    $0x4,%edx
  shl    $0x4,%rdx
  nopl   0x0(%rax)
L88:
  movdqu 0x0(%rbp,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L88
  mov    %ebx,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%ecx
  lea    0x0(%rbp,%rcx,1),%rdx
  cmp    %eax,%ebx
  je     L16
  mov    %ebx,%esi
  sub    %eax,%esi
  lea    -0x1(%rsi),%edi
  cmp    $0x6,%edi
  jbe    L82
L153:
  mov    0x0(%rbp,%rcx,1),%rdi
  mov    %rdi,(%r8,%rcx,1)
  mov    %esi,%ecx
  and    $0xfffffff8,%ecx
  mov    %ecx,%edi
  add    %ecx,%eax
  add    %rdi,%rdx
  cmp    %ecx,%esi
  je     L16
L82:
  movzbl (%rdx),%esi
  movslq %eax,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L16
  movzbl 0x1(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x2(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L16
  movzbl 0x2(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x3(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L16
  movzbl 0x3(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x4(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L16
  movzbl 0x4(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  lea    0x5(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L16
  movzbl 0x5(%rdx),%esi
  movslq %ecx,%rcx
  add    $0x6,%eax
  mov    %sil,(%r8,%rcx,1)
  cmp    %eax,%ebx
  jle    L16
  movzbl 0x6(%rdx),%edx
  cltq
  mov    %dl,(%r8,%rax,1)
L16:
  movslq %ebx,%rbx
  mov    %r8,%rax
  add    %r8,%rbx
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L98:
  mov    %ebx,%ecx
  xor    %eax,%eax
  nopl   0x0(%rax)
L6:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L6
  jmp    L16
  nopl   0x0(%rax)
L62:
  mov    %r8,%rbx
  mov    %r8,%rax
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  mov    %rbp,%rdx
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L153
```

assistant: ```c
char *substring(char *string, int position, int length) {
  char *pointer = malloc(length + 1);
  int c;
  for (c = 0; c < position - 1; c++) {
    string++;
  }
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rvr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L14:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x1c(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x8(%rbp)
  jle    L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rvr(int *pnc, int k) {
  int i = 0, t;
  for (; i <= k / 2; ++i) {
    t = pnc[i];
    pnc[i] = pnc[k - i];
    pnc[k - i] = t;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rvr>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0xffffffff,%esi
  jl     L7
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  mov    $0x0,%eax
L17:
  mov    (%rdi,%rax,4),%ecx
  mov    (%rdx),%esi
  mov    %esi,(%rdi,%rax,4)
  mov    %ecx,(%rdx)
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%r8d
  jge    L17
L7:
  ret
```

assistant: ```c
void rvr(int *pnc, int k) {
  int i = 0, t;
  for (; i <= k / 2; ++i) {
    t = pnc[i];
    pnc[i] = pnc[k - i];
    pnc[k - i] = t;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rvr>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0xffffffff,%esi
  jl     L16
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    (%rdi,%rsi,4),%rdx
  nop
L13:
  mov    (%rdx),%esi
  mov    (%rdi,%rax,4),%ecx
  sub    $0x4,%rdx
  mov    %esi,(%rdi,%rax,4)
  add    $0x1,%rax
  mov    %ecx,0x4(%rdx)
  cmp    %eax,%r8d
  jge    L13
L16:
  ret
```

assistant: ```c
void rvr(int *pnc, int k) {
  int i = 0, t;
  for (; i <= k / 2; ++i) {
    t = pnc[i];
    pnc[i] = pnc[k - i];
    pnc[k - i] = t;
  }
}
```