ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<proc_inc>:
  endbr64
  mov    %rdi,%rax
  cmp    $0x1,%r8d
  je     L26
  mov    (%rdi),%edi
  sub    $0x1,%edx
  cmp    %edx,%edi
  je     L17
  add    $0x1,%edi
  mov    %edi,(%rax)
  ret
  nopl   0x0(%rax)
L17:
  movl   $0x0,(%rax)
  mov    (%rsi),%eax
  sub    $0x1,%ecx
  lea    0x1(%rax),%edx
  cmp    %ecx,%eax
  mov    $0x0,%eax
  cmovne %edx,%eax
  mov    %eax,(%rsi)
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  mov    (%rsi),%edi
  sub    $0x1,%ecx
  cmp    %ecx,%edi
  je     L4
  add    $0x1,%edi
  mov    %edi,(%rsi)
  ret
  nop
L4:
  movl   $0x0,(%rsi)
  mov    (%rax),%ecx
  sub    $0x1,%edx
  cmp    %edx,%ecx
  lea    0x1(%rcx),%esi
  mov    $0x0,%edx
  cmovne %esi,%edx
  mov    %edx,(%rax)
  ret
```

assistant: ```c
void proc_inc(myrow, mycol, nprow, npcol, major) int *myrow, *mycol, nprow, npcol, major;
{
  if (major == 1) {
    if (*mycol == npcol - 1) {
      *mycol = 0;
      if (*myrow == nprow - 1) {
        *myrow = 0;
      } else {
        *myrow = *myrow + 1;
      }
    } else {
      *mycol = *mycol + 1;
    }
  } else {
    if (*myrow == nprow - 1) {
      *myrow = 0;
      if (*mycol == npcol - 1) {
        *mycol = 0;
      } else {
        *mycol = *mycol + 1;
      }
    } else {
      *myrow = *myrow + 1;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<proc_inc>:
  endbr64
  mov    %rdi,%rax
  cmp    $0x1,%r8d
  je     L25
  mov    (%rdi),%edi
  sub    $0x1,%edx
  cmp    %edx,%edi
  je     L22
  add    $0x1,%edi
  mov    %edi,(%rax)
  ret
  nopl   0x0(%rax)
L22:
  movl   $0x0,(%rax)
  mov    (%rsi),%eax
  sub    $0x1,%ecx
  lea    0x1(%rax),%edx
  cmp    %ecx,%eax
  mov    $0x0,%eax
  cmovne %edx,%eax
  mov    %eax,(%rsi)
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  mov    (%rsi),%edi
  sub    $0x1,%ecx
  cmp    %ecx,%edi
  je     L4
  add    $0x1,%edi
  mov    %edi,(%rsi)
  ret
  nop
L4:
  movl   $0x0,(%rsi)
  mov    (%rax),%ecx
  sub    $0x1,%edx
  cmp    %edx,%ecx
  lea    0x1(%rcx),%esi
  mov    $0x0,%edx
  cmovne %esi,%edx
  mov    %edx,(%rax)
  ret
```

assistant: ```c
void proc_inc(myrow, mycol, nprow, npcol, major) int *myrow, *mycol, nprow, npcol, major;
{
  if (major == 1) {
    if (*mycol == npcol - 1) {
      *mycol = 0;
      if (*myrow == nprow - 1) {
        *myrow = 0;
      } else {
        *myrow = *myrow + 1;
      }
    } else {
      *mycol = *mycol + 1;
    }
  } else {
    if (*myrow == nprow - 1) {
      *myrow = 0;
      if (*mycol == npcol - 1) {
        *mycol = 0;
      } else {
        *mycol = *mycol + 1;
      }
    } else {
      *myrow = *myrow + 1;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_XcursorNextPath>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L4
  mov    $0x0,%eax
  jmp    L3
L4:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
L3:
  leave
  ret
```

assistant: ```c
const char *_XcursorNextPath(const char *path) {
  char *colon = strchr(path, ':');
  if (!colon)
    return NULL;
  return colon + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_XcursorNextPath>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3a,%esi
  call   <strchr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *_XcursorNextPath(const char *path) {
  char *colon = strchr(path, ':');
  if (!colon)
    return NULL;
  return colon + 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_XcursorNextPath>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3a,%esi
  call   <strchr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *_XcursorNextPath(const char *path) {
  char *colon = strchr(path, ':');
  if (!colon)
    return NULL;
  return colon + 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_XcursorNextPath>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3a,%esi
  call   <strchr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *_XcursorNextPath(const char *path) {
  char *colon = strchr(path, ':');
  if (!colon)
    return NULL;
  return colon + 1;
}
```
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    $0x410,%rsp
  mov    %edi,-0x404(%rbp)
  mov    %rsi,-0x410(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3f4(%rbp)
  cmpl   $0x2,-0x404(%rbp)
  je     L24
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L48
L32:
  mov    -0x410(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  lea    -0x3f0(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L24
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x3f4(%rbp)
L24:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getline_x@plt>
  test   %rax,%rax
  jg     L32
L48:
  mov    -0x3f4(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
type: string
name: parse_data
arguments: data_label: D16
type: string

tool: D5: Usage: find pattern

tool: D16: %s

assistant: ```c
int main(int argc, char *argv[]) {
  char line[1000];
  int found = 0;
  if (argc != 2)
    printf("Usage: find pattern\n");
  else
    while (getline_x(line, 1000) > 0)
      if (strstr(line, argv[1]) != NULL) {
        printf("%s", line);
        found++;
      }
  return found;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r12d
  lea    D12(%rip),%r13
  cmp    $0x2,%edi
  jne    L13
L2:
  mov    %rsp,%rbx
L3:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getline_x@plt>
  test   %rax,%rax
  jle    L16
  mov    0x8(%rbp),%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L3
  mov    %rsp,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%r12d
  jmp    L2
L13:
  lea    D14(%rip),%rdi
  call   <puts@plt>
L16:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    %r12d,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D12
type_guessed: string
name: unpack_data
arguments: data_label: D14
type_guessed: string

tool: D12: %s

tool: D14: Usage: find pattern

assistant: ```c
int main(int argc, char *argv[]) {
  char line[1000];
  int found = 0;
  if (argc != 2)
    printf("Usage: find pattern\n");
  else
    while (getline_x(line, 1000) > 0)
      if (strstr(line, argv[1]) != NULL) {
        printf("%s", line);
        found++;
      }
  return found;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  jne    L3
  mov    %rsi,%rbp
  xor    %r12d,%r12d
  lea    D12(%rip),%r13
  mov    %rsp,%rbx
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L40:
  mov    0x8(%rbp),%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L13
L24:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getline_x@plt>
  test   %rax,%rax
  jg     L40
L14:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x3f8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  lea    D1(%rip),%rdi
  xor    %r12d,%r12d
  call   <puts@plt>
  jmp    L14
L13:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  jmp    L24
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D1
data_type: string

tool: D12: %s

tool: D1: Usage: find pattern

assistant: ```c
int main(int argc, char *argv[]) {
  char line[1000];
  int found = 0;
  if (argc != 2)
    printf("Usage: find pattern\n");
  else
    while (getline_x(line, 1000) > 0)
      if (strstr(line, argv[1]) != NULL) {
        printf("%s", line);
        found++;
      }
  return found;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  jne    L69
  mov    %rsi,%rbp
  xor    %r12d,%r12d
  lea    D1(%rip),%r13
  mov    %rsp,%rbx
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L59:
  mov    0x8(%rbp),%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L3
L52:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getline_x@plt>
  test   %rax,%rax
  jg     L59
L29:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x3f8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L69:
  lea    D15(%rip),%rdi
  xor    %r12d,%r12d
  call   <puts@plt>
  jmp    L29
L3:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  jmp    L52
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D1
type_infered: string
name: parse_data
arguments: label: D15
type_infered: string

tool: D1: %s

tool: D15: Usage: find pattern

assistant: ```c
int main(int argc, char *argv[]) {
  char line[1000];
  int found = 0;
  if (argc != 2)
    printf("Usage: find pattern\n");
  else
    while (getline_x(line, 1000) > 0)
      if (strstr(line, argv[1]) != NULL) {
        printf("%s", line);
        found++;
      }
  return found;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    %r9d,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L38
L5:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  addl   $0x1,-0x10(%rbp)
L38:
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jle    L5
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L28
L6:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  sub    -0x1c(%rbp),%eax
  add    %eax,-0x10(%rbp)
L28:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L6
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void foo(int *a, int i, int j, int k, int l, int m) {
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = 0; j <= (6 * l + 4 * k); j++)
    a[j] = 1;
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = m; j <= l; j += (k + l - m))
    a[j] = 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<foo>:
  endbr64
  lea    (%r8,%r8,2),%edx
  lea    0x0(,%rcx,4),%eax
  lea    (%rax,%rdx,2),%edx
  test   %edx,%edx
  js     L36
  mov    %rdi,%rax
  mov    %edx,%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L19:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L19
L36:
  cmp    %r9d,%r8d
  jl     L20
  movslq %r9d,%rax
  movslq %ecx,%rsi
  sub    %rax,%rsi
  movslq %r8d,%rdx
  add    %rdx,%rsi
  shl    $0x2,%rsi
  lea    (%rdi,%rax,4),%rdx
  mov    %r9d,%eax
  add    %r8d,%ecx
  sub    %r9d,%ecx
L22:
  movl   $0x1,(%rdx)
  add    %ecx,%eax
  add    %rsi,%rdx
  cmp    %eax,%r8d
  jge    L22
L20:
  ret
```

assistant: ```c
void foo(int *a, int i, int j, int k, int l, int m) {
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = 0; j <= (6 * l + 4 * k); j++)
    a[j] = 1;
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = m; j <= l; j += (k + l - m))
    a[j] = 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  lea    (%r8,%r8,2),%esi
  lea    0x0(,%rcx,4),%eax
  mov    %rdi,%rdx
  lea    (%rax,%rsi,2),%esi
  test   %esi,%esi
  js     L5
  movslq %esi,%rsi
  mov    %rdi,%rax
  lea    0x4(%rdi,%rsi,4),%rsi
  nopl   0x0(%rax)
L17:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L17
L5:
  cmp    %r9d,%r8d
  jl     L15
  add    %r8d,%ecx
  sub    %r9d,%ecx
  movslq %r9d,%r9
  movslq %ecx,%rax
  nopl   0x0(%rax,%rax,1)
L16:
  movl   $0x1,(%rdx,%r9,4)
  add    %rax,%r9
  cmp    %r9d,%r8d
  jge    L16
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
void foo(int *a, int i, int j, int k, int l, int m) {
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = 0; j <= (6 * l + 4 * k); j++)
    a[j] = 1;
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = m; j <= l; j += (k + l - m))
    a[j] = 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  mov    %ecx,%edx
  lea    (%r8,%r8,2),%ecx
  mov    %rdi,%rsi
  lea    0x0(,%rdx,4),%eax
  lea    (%rax,%rcx,2),%edi
  test   %edi,%edi
  js     L87
  lea    0x1(%rdi),%r10d
  cmp    $0x2,%edi
  jle    L39
  mov    %r10d,%ecx
  movdqa D3(%rip),%xmm0
  mov    %rsi,%rax
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rsi,%rcx,1),%rcx
  nopl   (%rax)
L45:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L45
  mov    %r10d,%eax
  and    $0xfffffffc,%eax
  and    $0x3,%r10d
  je     L87
L16:
  movslq %eax,%rcx
  movl   $0x1,(%rsi,%rcx,4)
  lea    0x0(,%rcx,4),%r10
  lea    0x1(%rax),%ecx
  cmp    %ecx,%edi
  jl     L87
  movl   $0x1,0x4(%rsi,%r10,1)
  add    $0x2,%eax
  cmp    %eax,%edi
  jl     L87
  movl   $0x1,0x8(%rsi,%r10,1)
L87:
  cmp    %r9d,%r8d
  jl     L77
  add    %r8d,%edx
  sub    %r9d,%edx
  movslq %r9d,%r9
  movslq %edx,%rax
  nopl   0x0(%rax,%rax,1)
L12:
  movl   $0x1,(%rsi,%r9,4)
  add    %rax,%r9
  cmp    %r9d,%r8d
  jge    L12
  ret
  nopl   0x0(%rax)
L77:
  ret
L39:
  xor    %eax,%eax
  jmp    L16
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_infered: byte[16]

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
void foo(int *a, int i, int j, int k, int l, int m) {
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = 0; j <= (6 * l + 4 * k); j++)
    a[j] = 1;
#pragma omp parallel for num_threads(3 * i) schedule(dynamic, i * 4)
  for (j = m; j <= l; j += (k + l - m))
    a[j] = 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %r8d,-0x1c(%rbp)
  mov    %r9d,-0x20(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L8
  cmpl   $0x0,-0x20(%rbp)
  jg     L45
L8:
  mov    0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  jmp    L58
L45:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x14(%rbp),%ecx
  mov    -0x20(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L37
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  subl   $0x1,0x10(%rbp)
  mov    0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%r9d
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%r8d
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <bt>
  add    $0x10,%rsp
  jmp    L58
L37:
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  imul   %eax,%ecx
  mov    -0x20(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L31
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%r9d
  mov    -0x1c(%rbp),%r8d
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <bt>
  add    $0x10,%rsp
  jmp    L58
L31:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%r10d
  mov    -0x20(%rbp),%r8d
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %r8d,%r9d
  mov    %r10d,%r8d
  mov    %rax,%rdi
  call   <bt>
  add    $0x10,%rsp
L58:
  leave
  ret
```

assistant: ```c
char *bt(int *ib, char *sb, int s, int t, int i, int j, int p) {
  if (i < 1 || j < 1)
    return sb + p;
  if (sb[i - 1] == sb[s + j - 1]) {
    sb[--p] = sb[i - 1];
    return bt(ib, sb, s, t, i - 1, j - 1, p);
  }
  if (ib[i * (t + 1) + j - 1] > ib[(i - 1) * (t + 1) + j])
    return bt(ib, sb, s, t, i, j - 1, p);
  return bt(ib, sb, s, t, i - 1, j, p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bt>:
  endbr64
  push   %rbx
  test   %r8d,%r8d
  jle    L43
  mov    %edx,%r10d
  mov    %r8d,%eax
  test   %r9d,%r9d
  jle    L43
  movslq %r8d,%rdx
  movzbl -0x1(%rsi,%rdx,1),%r8d
  lea    (%r9,%r10,1),%edx
  movslq %edx,%rdx
  cmp    -0x1(%rsi,%rdx,1),%r8b
  je     L53
  lea    0x1(%rcx),%ebx
  mov    %ebx,%edx
  imul   %eax,%edx
  lea    -0x1(%rax),%r8d
  lea    (%rdx,%r9,1),%r11d
  movslq %r11d,%r11
  sub    %ebx,%edx
  add    %r9d,%edx
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%ebx
  cmp    %ebx,-0x4(%rdi,%r11,4)
  jg     L33
  sub    $0x8,%rsp
  mov    0x18(%rsp),%eax
  push   %rax
  mov    %r10d,%edx
  call   <bt>
  add    $0x10,%rsp
L31:
  pop    %rbx
  ret
L43:
  movslq 0x10(%rsp),%rax
  add    %rsi,%rax
  jmp    L31
L53:
  mov    0x10(%rsp),%ebx
  lea    -0x1(%rbx),%edx
  movslq %edx,%r11
  mov    %r8b,(%rsi,%r11,1)
  sub    $0x8,%rsp
  push   %rdx
  lea    -0x1(%r9),%r9d
  lea    -0x1(%rax),%r8d
  mov    %r10d,%edx
  call   <bt>
  add    $0x10,%rsp
  jmp    L31
L33:
  sub    $0x8,%rsp
  mov    0x18(%rsp),%ebx
  push   %rbx
  lea    -0x1(%r9),%r9d
  mov    %eax,%r8d
  mov    %r10d,%edx
  call   <bt>
  add    $0x10,%rsp
  jmp    L31
```

assistant: ```c
char *bt(int *ib, char *sb, int s, int t, int i, int j, int p) {
  if (i < 1 || j < 1)
    return sb + p;
  if (sb[i - 1] == sb[s + j - 1]) {
    sb[--p] = sb[i - 1];
    return bt(ib, sb, s, t, i - 1, j - 1, p);
  }
  if (ib[i * (t + 1) + j - 1] > ib[(i - 1) * (t + 1) + j])
    return bt(ib, sb, s, t, i, j - 1, p);
  return bt(ib, sb, s, t, i - 1, j, p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bt>:
  endbr64
  push   %rbx
  mov    %rsi,%rax
  mov    0x10(%rsp),%ebx
  test   %r8d,%r8d
  jle    L80
  test   %r9d,%r9d
  jle    L80
  mov    %edx,%esi
  lea    0x1(%rcx),%r10d
  jmp    L45
  xchg   %ax,%ax
L1:
  mov    %r10d,%edx
  imul   %r8d,%edx
  lea    (%rdx,%r9,1),%ecx
  sub    %r10d,%edx
  add    %r9d,%edx
  movslq %ecx,%rcx
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,-0x4(%rdi,%rcx,4)
  jle    L15
  sub    $0x1,%r9d
L59:
  test   %r8d,%r8d
  jle    L80
L78:
  test   %r9d,%r9d
  jle    L80
L45:
  movslq %r8d,%rdx
  lea    -0x1(%r8),%r11d
  movzbl -0x1(%rax,%rdx,1),%ecx
  lea    (%rsi,%r9,1),%edx
  movslq %edx,%rdx
  cmp    -0x1(%rax,%rdx,1),%cl
  jne    L1
  sub    $0x1,%ebx
  mov    %r11d,%r8d
  sub    $0x1,%r9d
  movslq %ebx,%rdx
  mov    %cl,(%rax,%rdx,1)
  test   %r8d,%r8d
  jg     L78
L80:
  movslq %ebx,%rbx
  add    %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L15:
  mov    %r11d,%r8d
  jmp    L59
```

assistant: ```c
char *bt(int *ib, char *sb, int s, int t, int i, int j, int p) {
  if (i < 1 || j < 1)
    return sb + p;
  if (sb[i - 1] == sb[s + j - 1]) {
    sb[--p] = sb[i - 1];
    return bt(ib, sb, s, t, i - 1, j - 1, p);
  }
  if (ib[i * (t + 1) + j - 1] > ib[(i - 1) * (t + 1) + j])
    return bt(ib, sb, s, t, i, j - 1, p);
  return bt(ib, sb, s, t, i - 1, j, p);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bt>:
  endbr64
  push   %rbx
  mov    %rsi,%rax
  mov    0x10(%rsp),%ebx
  test   %r8d,%r8d
  jle    L58
  test   %r9d,%r9d
  jle    L58
  mov    %edx,%esi
  lea    0x1(%rcx),%r10d
  jmp    L62
  xchg   %ax,%ax
L86:
  mov    %r10d,%edx
  imul   %r8d,%edx
  lea    (%rdx,%r9,1),%ecx
  sub    %r10d,%edx
  add    %r9d,%edx
  movslq %ecx,%rcx
  movslq %edx,%rdx
  mov    (%rdi,%rdx,4),%edx
  cmp    %edx,-0x4(%rdi,%rcx,4)
  jle    L27
  sub    $0x1,%r9d
L2:
  test   %r8d,%r8d
  jle    L58
L51:
  test   %r9d,%r9d
  jle    L58
L62:
  movslq %r8d,%rdx
  lea    -0x1(%r8),%r11d
  movzbl -0x1(%rax,%rdx,1),%ecx
  lea    (%rsi,%r9,1),%edx
  movslq %edx,%rdx
  cmp    -0x1(%rax,%rdx,1),%cl
  jne    L86
  sub    $0x1,%ebx
  mov    %r11d,%r8d
  sub    $0x1,%r9d
  movslq %ebx,%rdx
  mov    %cl,(%rax,%rdx,1)
  test   %r8d,%r8d
  jg     L51
L58:
  movslq %ebx,%rbx
  add    %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L27:
  mov    %r11d,%r8d
  jmp    L2
```

assistant: ```c
char *bt(int *ib, char *sb, int s, int t, int i, int j, int p) {
  if (i < 1 || j < 1)
    return sb + p;
  if (sb[i - 1] == sb[s + j - 1]) {
    sb[--p] = sb[i - 1];
    return bt(ib, sb, s, t, i - 1, j - 1, p);
  }
  if (ib[i * (t + 1) + j - 1] > ib[(i - 1) * (t + 1) + j])
    return bt(ib, sb, s, t, i, j - 1, p);
  return bt(ib, sb, s, t, i - 1, j, p);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<num_active_dice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  jne    L20
  mov    -0x4(%rbp),%eax
  jmp    L6
L20:
  addl   $0x1,-0x4(%rbp)
L27:
  cmpl   $0x5,-0x4(%rbp)
  jle    L10
  mov    $0x6,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int num_active_dice(int *dice) {
  int i;
  for (i = 0; i < 6; i++) {
    if (dice[i] == -1) {
      return i;
    }
  }
  return 6;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<num_active_dice>:
  endbr64
  mov    $0x0,%eax
L18:
  mov    %eax,%edx
  cmpl   $0xffffffff,(%rdi,%rax,4)
  je     L10
  add    $0x1,%rax
  cmp    $0x6,%rax
  jne    L18
  mov    $0x6,%edx
L10:
  mov    %edx,%eax
  ret
```

assistant: ```c
int num_active_dice(int *dice) {
  int i;
  for (i = 0; i < 6; i++) {
    if (dice[i] == -1) {
      return i;
    }
  }
  return 6;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<num_active_dice>:
  endbr64
  xor    %eax,%eax
L5:
  cmpl   $0xffffffff,(%rdi,%rax,4)
  mov    %eax,%r8d
  je     L14
  add    $0x1,%rax
  cmp    $0x6,%rax
  jne    L5
  mov    $0x6,%r8d
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int num_active_dice(int *dice) {
  int i;
  for (i = 0; i < 6; i++) {
    if (dice[i] == -1) {
      return i;
    }
  }
  return 6;
}
```