ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1f,%edx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x11,%eax
  lea    0x50(%rsp),%rdi
  mov    %rbp,%rcx
  xchg   %ax,%ax
L14:
  movzwl %dx,%esi
  shr    $0x10,%edx
  add    $0x4,%rcx
  imul   $0x9069,%esi,%esi
  add    %esi,%edx
  movzwl %ax,%esi
  shr    $0x10,%eax
  imul   $0x4650,%esi,%esi
  add    %esi,%eax
  mov    %edx,%esi
  shl    $0x10,%esi
  add    %eax,%esi
  mov    %esi,-0x4(%rcx)
  cmp    %rdi,%rcx
  jne    L14
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <sort@plt>
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <myprint@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x60,%rsp
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  int buf[20];
  int *ptr = buf;
  unsigned int i, m_w, m_z;
  m_w = 17;
  m_z = 31;
  for (i = 0; i < 20; i++, ptr++) {
    m_z = 36969 * (m_z & 65535) + (m_z >> 16);
    m_w = 18000 * (m_w & 65535) + (m_w >> 16);
    *ptr = (m_z << 16) + m_w;
  }
  sort(buf, 20);
  myprint(buf, 20);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1f,%edx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x11,%eax
  lea    0x50(%rsp),%rdi
  mov    %rbp,%rcx
  xchg   %ax,%ax
L12:
  movzwl %dx,%esi
  shr    $0x10,%edx
  add    $0x4,%rcx
  imul   $0x9069,%esi,%esi
  add    %esi,%edx
  movzwl %ax,%esi
  shr    $0x10,%eax
  imul   $0x4650,%esi,%esi
  add    %esi,%eax
  mov    %edx,%esi
  shl    $0x10,%esi
  add    %eax,%esi
  mov    %esi,-0x4(%rcx)
  cmp    %rdi,%rcx
  jne    L12
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <sort@plt>
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <myprint@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x60,%rsp
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  int buf[20];
  int *ptr = buf;
  unsigned int i, m_w, m_z;
  m_w = 17;
  m_z = 31;
  for (i = 0; i < 20; i++, ptr++) {
    m_z = 36969 * (m_z & 65535) + (m_z >> 16);
    m_w = 18000 * (m_w & 65535) + (m_w >> 16);
    *ptr = (m_z << 16) + m_w;
  }
  sort(buf, 20);
  myprint(buf, 20);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <initLibrary@plt>
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L4
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
L4:
  mov    $0x0,%eax
  call   <println@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D8: OK

assistant: ```c
int main(int argc, int *argv) {
  int *i;
  initLibrary();
  i = malloc(2);
  *i = 0;
  ++*i;
  if (*i == 1)
    print("OK");
  println();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <initLibrary@plt>
  mov    $0x2,%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  lea    D7(%rip),%rdi
  call   <print@plt>
  mov    $0x0,%eax
  call   <println@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type: string

tool: D7: OK

assistant: ```c
int main(int argc, int *argv) {
  int *i;
  initLibrary();
  i = malloc(2);
  *i = 0;
  ++*i;
  if (*i == 1)
    print("OK");
  println();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initLibrary@plt>
  mov    $0x2,%edi
  call   <malloc@plt>
  lea    D8(%rip),%rdi
  movl   $0x1,(%rax)
  call   <print@plt>
  xor    %eax,%eax
  call   <println@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D8: OK

assistant: ```c
int main(int argc, int *argv) {
  int *i;
  initLibrary();
  i = malloc(2);
  *i = 0;
  ++*i;
  if (*i == 1)
    print("OK");
  println();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initLibrary@plt>
  mov    $0x2,%edi
  call   <malloc@plt>
  lea    D3(%rip),%rdi
  movl   $0x1,(%rax)
  call   <print@plt>
  xor    %eax,%eax
  call   <println@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D3: OK

assistant: ```c
int main(int argc, int *argv) {
  int *i;
  initLibrary();
  i = malloc(2);
  *i = 0;
  ++*i;
  if (*i == 1)
    print("OK");
  println();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<drv_D4D_send_extra>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %ecx,%eax
  mov    %r8d,%esi
  mov    %r9d,%ecx
  mov    %dl,-0x10(%rbp)
  mov    %al,-0x14(%rbp)
  mov    %esi,%eax
  mov    %al,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x1c(%rbp)
  movzbl -0x1c(%rbp),%r8d
  movzbl -0x18(%rbp),%edi
  movzbl -0x14(%rbp),%ecx
  movzbl -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <drv_D4D_send_extra_nowait@plt>
  mov    $0x0,%eax
  call   <drv_D4D_receive_ACK@plt>
  nop
  leave
  ret
```

assistant: ```c
void drv_D4D_send_extra(const char *data, const unsigned int len, unsigned char pos1, unsigned char pos2, unsigned char pos3, unsigned char pos4) {
  drv_D4D_send_extra_nowait(data, len, pos1, pos2, pos3, pos4);
  drv_D4D_receive_ACK();
}
```
user: Translate the below assembly code into C programming language:

```asm
<drv_D4D_send_extra>:
  endbr64
  sub    $0x8,%rsp
  movzbl %cl,%ecx
  movzbl %dl,%edx
  movzbl %r9b,%r9d
  movzbl %r8b,%r8d
  call   <drv_D4D_send_extra_nowait@plt>
  mov    $0x0,%eax
  call   <drv_D4D_receive_ACK@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void drv_D4D_send_extra(const char *data, const unsigned int len, unsigned char pos1, unsigned char pos2, unsigned char pos3, unsigned char pos4) {
  drv_D4D_send_extra_nowait(data, len, pos1, pos2, pos3, pos4);
  drv_D4D_receive_ACK();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<drv_D4D_send_extra>:
  endbr64
  sub    $0x8,%rsp
  movzbl %cl,%ecx
  movzbl %dl,%edx
  movzbl %r9b,%r9d
  movzbl %r8b,%r8d
  call   <drv_D4D_send_extra_nowait@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <drv_D4D_receive_ACK@plt>
```

assistant: ```c
void drv_D4D_send_extra(const char *data, const unsigned int len, unsigned char pos1, unsigned char pos2, unsigned char pos3, unsigned char pos4) {
  drv_D4D_send_extra_nowait(data, len, pos1, pos2, pos3, pos4);
  drv_D4D_receive_ACK();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<drv_D4D_send_extra>:
  endbr64
  sub    $0x8,%rsp
  movzbl %cl,%ecx
  movzbl %dl,%edx
  movzbl %r9b,%r9d
  movzbl %r8b,%r8d
  call   <drv_D4D_send_extra_nowait@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <drv_D4D_receive_ACK@plt>
```

assistant: ```c
void drv_D4D_send_extra(const char *data, const unsigned int len, unsigned char pos1, unsigned char pos2, unsigned char pos3, unsigned char pos4) {
  drv_D4D_send_extra_nowait(data, len, pos1, pos2, pos3, pos4);
  drv_D4D_receive_ACK();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<varint_get>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
L30:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jae    L33
  mov    $0x0,%eax
  jmp    L11
L33:
  shll   $0x7,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  and    $0x7f,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  ja     L4
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x30(%rbp),%rdx
  mov    %ecx,(%rdx)
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  and    $0x80,%eax
  test   %eax,%eax
  jne    L30
L4:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0x1,%eax
L11:
  pop    %rbp
  ret
```

assistant: ```c
int varint_get(uint32_t *v, const uint8_t *banner, unsigned length, unsigned *j) {
  uint32_t accum = 0;
  unsigned bytes = 0;
  do {
    if (*j > length)
      return 0;
    accum <<= 7;
    accum += banner[*j] & 0x7f;
    bytes++;
  } while (bytes < 4 && banner[(*j)++] & 0x80);
  *v = accum;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<varint_get>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  mov    $0x4,%r8d
  mov    $0x0,%esi
L5:
  mov    (%rcx),%eax
  cmp    %edx,%eax
  ja     L10
  shl    $0x7,%esi
  mov    %eax,%edi
  lea    (%r11,%rdi,4),%r9
  mov    (%r9),%edi
  and    $0x7f,%edi
  add    %edi,%esi
  sub    $0x1,%r8d
  je     L17
  add    $0x1,%eax
  mov    %eax,(%rcx)
  testb  $0x80,(%r9)
  jne    L5
L17:
  mov    %esi,(%r10)
  mov    $0x1,%eax
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int varint_get(uint32_t *v, const uint8_t *banner, unsigned length, unsigned *j) {
  uint32_t accum = 0;
  unsigned bytes = 0;
  do {
    if (*j > length)
      return 0;
    accum <<= 7;
    accum += banner[*j] & 0x7f;
    bytes++;
  } while (bytes < 4 && banner[(*j)++] & 0x80);
  *v = accum;
  return 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
<varint_get>:
  endbr64
  mov    (%rcx),%eax
  mov    %rdi,%r10
  xor    %edi,%edi
  lea    0x3(%rax),%r11d
L10:
  cmp    %eax,%edx
  jb     L22
  mov    %eax,%r8d
  shl    $0x7,%edi
  lea    (%rsi,%r8,4),%r9
  mov    (%r9),%r8d
  and    $0x7f,%r8d
  add    %r8d,%edi
  cmp    %r11d,%eax
  je     L24
  add    $0x1,%eax
  mov    %eax,(%rcx)
  testb  $0x80,(%r9)
  jne    L10
L24:
  mov    %edi,(%r10)
  mov    $0x1,%eax
  ret
L22:
  xor    %eax,%eax
  ret
```

assistant: ```c
int varint_get(uint32_t *v, const uint8_t *banner, unsigned length, unsigned *j) {
  uint32_t accum = 0;
  unsigned bytes = 0;
  do {
    if (*j > length)
      return 0;
    accum <<= 7;
    accum += banner[*j] & 0x7f;
    bytes++;
  } while (bytes < 4 && banner[(*j)++] & 0x80);
  *v = accum;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<varint_get>:
  endbr64
  mov    %rdi,%r9
  mov    %rsi,%rdi
  mov    %edx,%esi
  mov    (%rcx),%edx
  cmp    %esi,%edx
  ja     L49
  mov    %edx,%eax
  lea    0x1(%rdx),%r8d
  lea    (%rdi,%rax,4),%r10
  mov    (%r10),%eax
  mov    %r8d,(%rcx)
  and    $0x7f,%eax
  testb  $0x80,(%r10)
  je     L17
  cmp    %r8d,%esi
  jb     L49
  lea    (%rdi,%r8,4),%r10
  shl    $0x7,%eax
  mov    (%r10),%r8d
  and    $0x7f,%r8d
  add    %r8d,%eax
  lea    0x2(%rdx),%r8d
  mov    %r8d,(%rcx)
  testb  $0x80,(%r10)
  je     L17
  cmp    %r8d,%esi
  jb     L49
  lea    (%rdi,%r8,4),%r10
  shl    $0x7,%eax
  add    $0x3,%edx
  mov    (%r10),%r8d
  mov    %edx,(%rcx)
  and    $0x7f,%r8d
  add    %r8d,%eax
  testb  $0x80,(%r10)
  je     L17
  cmp    %edx,%esi
  jb     L49
  mov    (%rdi,%rdx,4),%edx
  shl    $0x7,%eax
  and    $0x7f,%edx
  add    %edx,%eax
L17:
  mov    %eax,(%r9)
  mov    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L49:
  xor    %eax,%eax
  ret
```

assistant: ```c
int varint_get(uint32_t *v, const uint8_t *banner, unsigned length, unsigned *j) {
  uint32_t accum = 0;
  unsigned bytes = 0;
  do {
    if (*j > length)
      return 0;
    accum <<= 7;
    accum += banner[*j] & 0x7f;
    bytes++;
  } while (bytes < 4 && banner[(*j)++] & 0x80);
  *v = accum;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <maxdiag1@plt>
  mov    %eax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <maxdiag2@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  call   <maxbottom@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <maxleft@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  cmp    %eax,%edx
  cmovl  %eax,%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  cmovl  %eax,%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D8: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int diag1, diag2, bottom, left;
  diag1 = maxdiag1();
  diag2 = maxdiag2();
  bottom = maxbottom();
  left = maxleft();
  printf("%d\n", ((diag2) > (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left))))) ? (diag2) : (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left)))))));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <maxdiag1@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <maxdiag2@plt>
  mov    %eax,%r12d
  mov    $0x0,%eax
  call   <maxbottom@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <maxleft@plt>
  cmp    %r12d,%ebx
  cmovl  %r12d,%ebx
  cmp    %ebp,%ebx
  cmovl  %ebp,%ebx
  cmp    %eax,%ebx
  cmovge %ebx,%eax
  mov    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D2
data_type: string

tool: D2: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int diag1, diag2, bottom, left;
  diag1 = maxdiag1();
  diag2 = maxdiag2();
  bottom = maxbottom();
  left = maxleft();
  printf("%d\n", ((diag2) > (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left))))) ? (diag2) : (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left)))))));
  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   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <maxdiag1@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <maxdiag2@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <maxbottom@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <maxleft@plt>
  cmp    %r12d,%ebx
  lea    D4(%rip),%rdi
  cmovl  %r12d,%ebx
  cmp    %ebp,%ebx
  cmovl  %ebp,%ebx
  cmp    %eax,%ebx
  cmovge %ebx,%eax
  mov    %eax,%esi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int diag1, diag2, bottom, left;
  diag1 = maxdiag1();
  diag2 = maxdiag2();
  bottom = maxbottom();
  left = maxleft();
  printf("%d\n", ((diag2) > (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left))))) ? (diag2) : (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left)))))));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <maxdiag1@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <maxdiag2@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <maxbottom@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <maxleft@plt>
  cmp    %r12d,%ebx
  lea    D7(%rip),%rdi
  cmovl  %r12d,%ebx
  cmp    %ebp,%ebx
  cmovl  %ebp,%ebx
  cmp    %eax,%ebx
  cmovge %ebx,%eax
  mov    %eax,%esi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
data_type: string

tool: D7: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int diag1, diag2, bottom, left;
  diag1 = maxdiag1();
  diag2 = maxdiag2();
  bottom = maxbottom();
  left = maxleft();
  printf("%d\n", ((diag2) > (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left))))) ? (diag2) : (((diag1) > (((bottom) > (left) ? (bottom) : (left))) ? (diag1) : (((bottom) > (left) ? (bottom) : (left)))))));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isPalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    %rax
  mov    %eax,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x8(%rbp),%rax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  movl   $0x20,-0x18(%rbp)
  movl   $0x2c,-0x14(%rbp)
  movl   $0x1,-0x1c(%rbp)
  jmp    L126
L155:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x18(%rbp)
  je     L76
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L54
L76:
  mov    $0x1,%eax
  jmp    L20
L54:
  mov    $0x0,%eax
L20:
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x18(%rbp)
  je     L135
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L93
L135:
  mov    $0x1,%eax
  jmp    L58
L93:
  mov    $0x0,%eax
L58:
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L70
  addl   $0x1,-0x24(%rbp)
  subl   $0x1,-0x20(%rbp)
  jmp    L16
L70:
  cmpl   $0x0,-0x10(%rbp)
  jne    L85
  cmpl   $0x0,-0xc(%rbp)
  je     L124
L85:
  cmpl   $0x0,-0x10(%rbp)
  je     L59
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
L59:
  cmpl   $0x0,-0xc(%rbp)
  je     L16
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jmp    L16
L124:
  movl   $0x0,-0x1c(%rbp)
L16:
  subl   $0x1,-0x28(%rbp)
L126:
  cmpl   $0x0,-0x28(%rbp)
  jle    L29
  cmpl   $0x0,-0x1c(%rbp)
  jne    L155
L29:
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
bool isPalindrome(char *str) {
  size_t len = strlen(str);
  int half = len / 2;
  int start = 0;
  int end = len - 1;
  int space = 32;
  int comma = 44;
  bool isPal = true;
  bool startSpace;
  bool endSpace;
  while (half > 0 && isPal) {
    startSpace = str[start] == space || str[start] == comma;
    endSpace = str[end] == space || str[end] == comma;
    if (str[start] == str[end]) {
      start++;
      end--;
    } else if (startSpace || endSpace) {
      startSpace &&start++;
      endSpace &&end--;
    } else {
      isPal = false;
    }
    half--;
  }
  return isPal;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isPalindrome>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rax,%rdx
  shr    %rdx
  sub    $0x1,%eax
  test   %edx,%edx
  jle    L91
  mov    %edx,%edi
  mov    $0x0,%r8d
  jmp    L67
L45:
  add    $0x1,%r8d
  sub    $0x1,%eax
L31:
  sub    $0x1,%edi
  je     L82
L67:
  movslq %r8d,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  movslq %eax,%rdx
  movzbl (%rbx,%rdx,1),%edx
  cmp    %dl,%cl
  je     L45
  cmp    $0x2c,%cl
  sete   %sil
  cmp    $0x20,%cl
  sete   %r9b
  cmp    $0x2c,%dl
  sete   %cl
  cmp    $0x20,%dl
  sete   %dl
  or     %edx,%ecx
  or     %r9b,%sil
  jne    L24
  test   %cl,%cl
  jne    L24
  mov    $0x0,%eax
L35:
  pop    %rbx
  ret
L24:
  test   %sil,%sil
  je     L48
  add    $0x1,%r8d
  test   %cl,%cl
  je     L31
L48:
  sub    $0x1,%eax
  jmp    L31
L82:
  mov    $0x1,%eax
  jmp    L35
L91:
  mov    $0x1,%eax
  jmp    L35
```

assistant: ```c
bool isPalindrome(char *str) {
  size_t len = strlen(str);
  int half = len / 2;
  int start = 0;
  int end = len - 1;
  int space = 32;
  int comma = 44;
  bool isPal = true;
  bool startSpace;
  bool endSpace;
  while (half > 0 && isPal) {
    startSpace = str[start] == space || str[start] == comma;
    endSpace = str[end] == space || str[end] == comma;
    if (str[start] == str[end]) {
      start++;
      end--;
    } else if (startSpace || endSpace) {
      startSpace &&start++;
      endSpace &&end--;
    } else {
      isPal = false;
    }
    half--;
  }
  return isPal;
}
```
user: What is the c source code of the assembly code below?

```asm
<isPalindrome>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rax,%rdx
  sub    $0x1,%eax
  shr    %rdx
  test   %edx,%edx
  jle    L59
  mov    %edx,%r8d
  xor    %edi,%edi
  nop
L56:
  movslq %edi,%rdx
  movzbl (%rbx,%rdx,1),%edx
  cmp    $0x2c,%dl
  sete   %cl
  cmp    $0x20,%dl
  sete   %sil
  or     %ecx,%esi
  movslq %eax,%rcx
  movzbl (%rbx,%rcx,1),%ecx
  cmp    $0x20,%cl
  je     L55
  cmp    $0x2c,%cl
  je     L55
  cmp    %cl,%dl
  je     L53
  test   %sil,%sil
  jne    L26
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L55:
  cmp    %cl,%dl
  je     L53
  cmp    $0x1,%sil
  sbb    $0xffffffff,%edi
  sub    $0x1,%eax
L14:
  sub    $0x1,%r8d
  jne    L56
L59:
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L53:
  add    $0x1,%edi
  sub    $0x1,%eax
  jmp    L14
L26:
  add    $0x1,%edi
  jmp    L14
```

assistant: ```c
bool isPalindrome(char *str) {
  size_t len = strlen(str);
  int half = len / 2;
  int start = 0;
  int end = len - 1;
  int space = 32;
  int comma = 44;
  bool isPal = true;
  bool startSpace;
  bool endSpace;
  while (half > 0 && isPal) {
    startSpace = str[start] == space || str[start] == comma;
    endSpace = str[end] == space || str[end] == comma;
    if (str[start] == str[end]) {
      start++;
      end--;
    } else if (startSpace || endSpace) {
      startSpace &&start++;
      endSpace &&end--;
    } else {
      isPal = false;
    }
    half--;
  }
  return isPal;
}
```