ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<search>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  sub    $0x428,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  lea    0x10(%rsp),%rdx
  mov    %r12d,%esi
  mov    %r15,%rdi
  mov    %eax,%r13d
  mov    %eax,0xc(%rsp)
  call   <searchwrongchar@plt>
  sub    %r12d,%r13d
  js     L68
  lea    -0x1(%r12),%ebx
  xor    %r14d,%r14d
  nopl   (%rax)
L31:
  test   %ebx,%ebx
  js     L44
  movslq %r14d,%rcx
  movslq %ebx,%rax
  add    %rbp,%rcx
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L39:
  sub    $0x1,%rax
  test   %eax,%eax
  js     L44
L50:
  movsbq (%rcx,%rax,1),%rdx
  mov    %eax,%esi
  cmp    %dl,(%r15,%rax,1)
  je     L39
  sub    0x10(%rsp,%rdx,4),%esi
  mov    $0x1,%edi
  call   <findmax@plt>
  add    %eax,%r14d
L66:
  cmp    %r13d,%r14d
  jle    L31
L68:
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L44:
  mov    %r14d,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    (%r12,%r14,1),%eax
  cmp    0xc(%rsp),%eax
  jl     L82
  mov    $0x1,%eax
L45:
  add    %eax,%r14d
  jmp    L66
  nopl   0x0(%rax,%rax,1)
L82:
  cltq
  mov    %r12d,%edi
  movsbq 0x0(%rbp,%rax,1),%rax
  sub    0x10(%rsp,%rax,4),%edi
  mov    %edi,%eax
  jmp    L45
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type: string

tool: D4: 

Pattern Matches at shift of = %d.

assistant: ```c
void search(char *text, char *pattern) {
  int patLen = strlen(pattern);
  int txtLen = strlen(text);
  int extra[256];
  searchwrongchar(pattern, patLen, extra);
  int s = 0;
  while (s <= (txtLen - patLen)) {
    int j = patLen - 1;
    while (j >= 0 && pattern[j] == text[s + j])
      j--;
    if (j < 0) {
      printf("\n\nPattern Matches at shift of = %d.", s);
      s += (s + patLen < txtLen) ? patLen - extra[text[s + patLen]] : 1;
    } else
      s += findmax(1, j - extra[text[s + j]]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<search>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  sub    $0x428,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  lea    0x10(%rsp),%rdx
  mov    %r12d,%esi
  mov    %r15,%rdi
  mov    %eax,%r13d
  mov    %eax,0xc(%rsp)
  call   <searchwrongchar@plt>
  sub    %r12d,%r13d
  js     L23
  lea    -0x1(%r12),%ebx
  xor    %r14d,%r14d
  nopl   (%rax)
L50:
  test   %ebx,%ebx
  js     L69
  movslq %r14d,%rcx
  movslq %ebx,%rax
  add    %rbp,%rcx
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L18:
  sub    $0x1,%rax
  test   %eax,%eax
  js     L69
L33:
  movsbq (%rcx,%rax,1),%rdx
  mov    %eax,%esi
  cmp    %dl,(%r15,%rax,1)
  je     L18
  sub    0x10(%rsp,%rdx,4),%esi
  mov    $0x1,%edi
  call   <findmax@plt>
  add    %eax,%r14d
L30:
  cmp    %r13d,%r14d
  jle    L50
L23:
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L73
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L69:
  mov    %r14d,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    (%r12,%r14,1),%eax
  cmp    0xc(%rsp),%eax
  jl     L99
  mov    $0x1,%eax
L47:
  add    %eax,%r14d
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L99:
  cltq
  mov    %r12d,%edi
  movsbq 0x0(%rbp,%rax,1),%rax
  sub    0x10(%rsp,%rax,4),%edi
  mov    %edi,%eax
  jmp    L47
L73:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_infered: string

tool: D0: 

Pattern Matches at shift of = %d.

assistant: ```c
void search(char *text, char *pattern) {
  int patLen = strlen(pattern);
  int txtLen = strlen(text);
  int extra[256];
  searchwrongchar(pattern, patLen, extra);
  int s = 0;
  while (s <= (txtLen - patLen)) {
    int j = patLen - 1;
    while (j >= 0 && pattern[j] == text[s + j])
      j--;
    if (j < 0) {
      printf("\n\nPattern Matches at shift of = %d.", s);
      s += (s + patLen < txtLen) ? patLen - extra[text[s + patLen]] : 1;
    } else
      s += findmax(1, j - extra[text[s + j]]);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<run_test_suites>:
  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    %r8d,-0x20(%rbp)
  mov    %r9d,-0x2c(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L25
L6:
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%ecx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <run_test_suite@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L10
  mov    $0xffffffff,%eax
  jmp    L22
L10:
  mov    -0x4(%rbp),%eax
  or     %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  add    %eax,-0xc(%rbp)
L25:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L6
  mov    -0x8(%rbp),%eax
L22:
  leave
  ret
```

assistant: ```c
int run_test_suites(int core_idx, int cores, int argc, char *argv[], int verbose, int stderr_log) {
  int suite_idx = core_idx + 1;
  int retval = 0;
  while (suite_idx < argc - 1) {
    const int r = run_test_suite(argv[suite_idx + 1], verbose, stderr_log);
    if (-1 == r) {
      return -1;
    }
    retval |= r;
    suite_idx += cores;
  }
  return retval;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<run_test_suites>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  lea    0x1(%rdi),%ebx
  lea    -0x1(%rdx),%r12d
  cmp    %r12d,%ebx
  jge    L9
  mov    %rcx,%r13
  mov    %r8d,%r14d
  mov    %r9d,%r15d
  movslq %esi,%rax
  mov    %rax,0x8(%rsp)
  movslq %ebx,%rbx
  mov    $0x0,%ebp
L30:
  mov    0x8(%r13,%rbx,8),%rdi
  mov    %r15d,%edx
  mov    %r14d,%esi
  call   <run_test_suite@plt>
  cmp    $0xffffffff,%eax
  je     L31
  or     %eax,%ebp
  add    0x8(%rsp),%rbx
  cmp    %ebx,%r12d
  jg     L30
  mov    %ebp,%eax
L31:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  mov    $0x0,%eax
  jmp    L31
```

assistant: ```c
int run_test_suites(int core_idx, int cores, int argc, char *argv[], int verbose, int stderr_log) {
  int suite_idx = core_idx + 1;
  int retval = 0;
  while (suite_idx < argc - 1) {
    const int r = run_test_suite(argv[suite_idx + 1], verbose, stderr_log);
    if (-1 == r) {
      return -1;
    }
    retval |= r;
    suite_idx += cores;
  }
  return retval;
}
```
user: What is the c source code of the assembly code below:

```asm
<run_test_suites>:
  endbr64
  push   %r15
  lea    -0x1(%rdx),%eax
  lea    0x1(%rdi),%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %eax,0xc(%rsp)
  cmp    %eax,%r15d
  jge    L15
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %r9d,%r14d
  movslq %esi,%rbp
  movslq %r15d,%r15
  xor    %ebx,%ebx
  jmp    L4
  nopl   (%rax)
L48:
  or     %eax,%ebx
  add    %rbp,%r15
  cmp    %r15d,0xc(%rsp)
  jle    L13
L4:
  mov    0x8(%r12,%r15,8),%rdi
  mov    %r14d,%edx
  mov    %r13d,%esi
  call   <run_test_suite@plt>
  cmp    $0xffffffff,%eax
  jne    L48
L6:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x18,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  xor    %eax,%eax
  jmp    L6
```

assistant: ```c
int run_test_suites(int core_idx, int cores, int argc, char *argv[], int verbose, int stderr_log) {
  int suite_idx = core_idx + 1;
  int retval = 0;
  while (suite_idx < argc - 1) {
    const int r = run_test_suite(argv[suite_idx + 1], verbose, stderr_log);
    if (-1 == r) {
      return -1;
    }
    retval |= r;
    suite_idx += cores;
  }
  return retval;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<run_test_suites>:
  endbr64
  push   %r15
  lea    -0x1(%rdx),%eax
  lea    0x1(%rdi),%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %eax,0xc(%rsp)
  cmp    %eax,%r15d
  jge    L38
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %r9d,%r14d
  movslq %esi,%rbp
  movslq %r15d,%r15
  xor    %ebx,%ebx
  jmp    L11
  nopl   (%rax)
L18:
  or     %eax,%ebx
  add    %rbp,%r15
  cmp    %r15d,0xc(%rsp)
  jle    L36
L11:
  mov    0x8(%r12,%r15,8),%rdi
  mov    %r14d,%edx
  mov    %r13d,%esi
  call   <run_test_suite@plt>
  cmp    $0xffffffff,%eax
  jne    L18
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  add    $0x18,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L38:
  xor    %eax,%eax
  jmp    L0
```

assistant: ```c
int run_test_suites(int core_idx, int cores, int argc, char *argv[], int verbose, int stderr_log) {
  int suite_idx = core_idx + 1;
  int retval = 0;
  while (suite_idx < argc - 1) {
    const int r = run_test_suite(argv[suite_idx + 1], verbose, stderr_log);
    if (-1 == r) {
      return -1;
    }
    retval |= r;
    suite_idx += cores;
  }
  return retval;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<WhatBitOS>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x40,%eax
  pop    %rbp
  ret
```

assistant: ```c
int WhatBitOS() {
  if (sizeof(int *) == 4)
    return 32;
  else if (sizeof(int *) == 8)
    return 64;
  else
    return (sizeof(int *) * 8);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<WhatBitOS>:
  endbr64
  mov    $0x40,%eax
  ret
```

assistant: ```c
int WhatBitOS() {
  if (sizeof(int *) == 4)
    return 32;
  else if (sizeof(int *) == 8)
    return 64;
  else
    return (sizeof(int *) * 8);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<WhatBitOS>:
  endbr64
  mov    $0x40,%eax
  ret
```

assistant: ```c
int WhatBitOS() {
  if (sizeof(int *) == 4)
    return 32;
  else if (sizeof(int *) == 8)
    return 64;
  else
    return (sizeof(int *) * 8);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<WhatBitOS>:
  endbr64
  mov    $0x40,%eax
  ret
```

assistant: ```c
int WhatBitOS() {
  if (sizeof(int *) == 4)
    return 32;
  else if (sizeof(int *) == 8)
    return 64;
  else
    return (sizeof(int *) * 8);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<moveable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L58
  mov    $0x0,%eax
  jmp    L11
L58:
  mov    -0x3c(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <row@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <column@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <quadrant@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <inrow@plt>
  mov    %eax,%ebx
  mov    -0x1c(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <incolumn@plt>
  or     %eax,%ebx
  mov    -0x18(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <inquadrant@plt>
  or     %ebx,%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x1,-0x24(%rbp)
  jmp    L93
L97:
  mov    -0x24(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L70
  cmpl   $0x0,-0x28(%rbp)
  jne    L70
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  jmp    L17
L70:
  mov    -0x24(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L17
  cmpl   $0x0,-0x28(%rbp)
  jle    L17
  mov    $0x0,%eax
  jmp    L11
L17:
  addl   $0x1,-0x24(%rbp)
L93:
  cmpl   $0x9,-0x24(%rbp)
  jle    L97
  mov    -0x28(%rbp),%eax
L11:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int moveable(sudoku, a)
int sudoku[];
int a;
{
  if (sudoku[a] != 0) {
    return (0);
  }
  int b = row(a);
  int c = column(a);
  int d = quadrant(a);
  int e = inrow(sudoku, b) | incolumn(sudoku, c) | inquadrant(sudoku, d);
  int j = 0;
  int i;
  for (i = 1; i <= 9; i++) {
    if (((e & (1 << i)) == 0) && (j == 0)) {
      j = i;
    } else if (((e & (1 << i)) == 0) && (j > 0)) {
      return (0);
    }
  }
  return (j);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<moveable>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movslq %esi,%rax
  mov    (%rdi,%rax,4),%ebx
  mov    $0x0,%eax
  test   %ebx,%ebx
  je     L37
L49:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %esi,%edi
  call   <row@plt>
  mov    %eax,%r15d
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <column@plt>
  mov    %eax,%r14d
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <quadrant@plt>
  mov    %eax,%r13d
  mov    %r15d,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <inrow@plt>
  mov    %eax,%ebp
  mov    %r14d,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <incolumn@plt>
  or     %eax,%ebp
  mov    %r13d,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <inquadrant@plt>
  mov    %ebp,%esi
  or     %eax,%esi
  mov    %ebx,%eax
  mov    $0x1,%ecx
  jmp    L0
L43:
  mov    %ecx,%eax
L57:
  add    $0x1,%ecx
  cmp    $0xa,%ecx
  je     L49
L0:
  mov    %esi,%edx
  sar    %cl,%edx
  and    $0x1,%edx
  mov    %edx,%edi
  or     %eax,%edi
  je     L43
  test   %eax,%eax
  jle    L57
  test   %edx,%edx
  jne    L57
  mov    %ebx,%eax
  jmp    L49
```

assistant: ```c
int moveable(sudoku, a)
int sudoku[];
int a;
{
  if (sudoku[a] != 0) {
    return (0);
  }
  int b = row(a);
  int c = column(a);
  int d = quadrant(a);
  int e = inrow(sudoku, b) | incolumn(sudoku, c) | inquadrant(sudoku, d);
  int j = 0;
  int i;
  for (i = 1; i <= 9; i++) {
    if (((e & (1 << i)) == 0) && (j == 0)) {
      j = i;
    } else if (((e & (1 << i)) == 0) && (j > 0)) {
      return (0);
    }
  }
  return (j);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<moveable>:
  endbr64
  push   %r15
  movslq %esi,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi,%rax,4),%r12d
  test   %r12d,%r12d
  je     L31
L33:
  xor    %r12d,%r12d
L23:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L31:
  mov    %rdi,%rbp
  mov    %rax,%r13
  mov    %eax,%edi
  xor    %eax,%eax
  call   <row@plt>
  mov    %r13d,%edi
  mov    %eax,%r15d
  xor    %eax,%eax
  call   <column@plt>
  mov    %r13d,%edi
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <quadrant@plt>
  mov    %r15d,%esi
  mov    %rbp,%rdi
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <inrow@plt>
  mov    %r14d,%esi
  mov    %rbp,%rdi
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <incolumn@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  or     %eax,%ebx
  xor    %eax,%eax
  call   <inquadrant@plt>
  mov    $0x1,%ecx
  or     %ebx,%eax
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L48:
  test   %edx,%edx
  jne    L44
  test   %r12d,%r12d
  jne    L33
L44:
  add    $0x1,%ecx
  cmp    $0xa,%ecx
  je     L23
L6:
  mov    %eax,%edx
  sar    %cl,%edx
  and    $0x1,%edx
  mov    %edx,%ebx
  or     %r12d,%ebx
  jne    L48
  mov    %ecx,%r12d
  jmp    L44
```

assistant: ```c
int moveable(sudoku, a)
int sudoku[];
int a;
{
  if (sudoku[a] != 0) {
    return (0);
  }
  int b = row(a);
  int c = column(a);
  int d = quadrant(a);
  int e = inrow(sudoku, b) | incolumn(sudoku, c) | inquadrant(sudoku, d);
  int j = 0;
  int i;
  for (i = 1; i <= 9; i++) {
    if (((e & (1 << i)) == 0) && (j == 0)) {
      j = i;
    } else if (((e & (1 << i)) == 0) && (j > 0)) {
      return (0);
    }
  }
  return (j);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<BruteForceMatching>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L76
L85:
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L90:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x8(%rbp)
  jge    L98
  mov    $0xffffffff,%eax
  jmp    L41
L98:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L35
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jne    L34
  mov    -0x10(%rbp),%eax
  jmp    L41
L34:
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L90
  jmp    L10
L35:
  nop
L10:
  addl   $0x1,-0x10(%rbp)
L76:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L85
  mov    $0xffffffff,%eax
L41:
  leave
  ret
```

assistant: ```c
int BruteForceMatching(const char *mStr, const char *pStr) {
  int i, j;
  int mStrLen = strlen(mStr);
  int pStrLen = strlen(pStr);
  for (i = 0; i < mStrLen; i++) {
    for (j = 0; j < pStrLen; j++) {
      if (i + j > mStrLen) {
        return -1;
      }
      if (mStr[i + j] != pStr[j]) {
        break;
      }
      if (j + 1 == pStrLen) {
        return i;
      }
    }
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<BruteForceMatching>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <strlen@plt>
  test   %ebx,%ebx
  jle    L3
  mov    %eax,%edi
  mov    %r12,%rsi
  mov    $0x0,%eax
  lea    -0x1(%rdi),%r8d
  mov    $0x0,%r9d
L32:
  mov    %r9,%rdx
L42:
  cmp    %edx,%edi
  jle    L87
  lea    (%rax,%rdx,1),%ecx
  cmp    %ecx,%ebx
  jl     L61
  movzbl 0x0(%rbp,%rdx,1),%ecx
  cmp    %cl,(%rsi,%rdx,1)
  jne    L87
  lea    0x1(%rdx),%rcx
  cmp    %r8,%rdx
  je     L47
  mov    %rcx,%rdx
  jmp    L42
L87:
  add    $0x1,%eax
  add    $0x1,%rsi
  cmp    %eax,%ebx
  jne    L32
  mov    $0xffffffff,%eax
  jmp    L47
L3:
  mov    $0xffffffff,%eax
  jmp    L47
L61:
  mov    $0xffffffff,%eax
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int BruteForceMatching(const char *mStr, const char *pStr) {
  int i, j;
  int mStrLen = strlen(mStr);
  int pStrLen = strlen(pStr);
  for (i = 0; i < mStrLen; i++) {
    for (j = 0; j < pStrLen; j++) {
      if (i + j > mStrLen) {
        return -1;
      }
      if (mStr[i + j] != pStr[j]) {
        break;
      }
      if (j + 1 == pStrLen) {
        return i;
      }
    }
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<BruteForceMatching>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  test   %ebp,%ebp
  jle    L21
  mov    %eax,%r8d
  mov    %rbx,%rdi
  xor    %eax,%eax
  lea    -0x1(%r8),%esi
L9:
  xor    %edx,%edx
  jmp    L23
  nopl   0x0(%rax)
L26:
  lea    (%rax,%rdx,1),%ecx
  cmp    %ecx,%ebp
  jl     L21
  movzbl (%r12,%rdx,1),%ebx
  cmp    %bl,(%rdi,%rdx,1)
  jne    L40
  lea    0x1(%rdx),%rcx
  cmp    %rsi,%rdx
  je     L4
  mov    %rcx,%rdx
L23:
  cmp    %edx,%r8d
  jg     L26
L40:
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    %eax,%ebp
  jne    L9
  nopl   (%rax)
L21:
  mov    $0xffffffff,%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int BruteForceMatching(const char *mStr, const char *pStr) {
  int i, j;
  int mStrLen = strlen(mStr);
  int pStrLen = strlen(pStr);
  for (i = 0; i < mStrLen; i++) {
    for (j = 0; j < pStrLen; j++) {
      if (i + j > mStrLen) {
        return -1;
      }
      if (mStr[i + j] != pStr[j]) {
        break;
      }
      if (j + 1 == pStrLen) {
        return i;
      }
    }
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BruteForceMatching>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  test   %ebp,%ebp
  jle    L17
  mov    %eax,%r8d
  mov    %rbx,%rdi
  xor    %eax,%eax
  lea    -0x1(%r8),%esi
L33:
  xor    %edx,%edx
  jmp    L20
  nopl   0x0(%rax)
L48:
  lea    (%rax,%rdx,1),%ecx
  cmp    %ecx,%ebp
  jl     L17
  movzbl (%r12,%rdx,1),%ebx
  cmp    %bl,(%rdi,%rdx,1)
  jne    L3
  lea    0x1(%rdx),%rcx
  cmp    %rsi,%rdx
  je     L21
  mov    %rcx,%rdx
L20:
  cmp    %edx,%r8d
  jg     L48
L3:
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    %eax,%ebp
  jne    L33
  nopl   (%rax)
L17:
  mov    $0xffffffff,%eax
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int BruteForceMatching(const char *mStr, const char *pStr) {
  int i, j;
  int mStrLen = strlen(mStr);
  int pStrLen = strlen(pStr);
  for (i = 0; i < mStrLen; i++) {
    for (j = 0; j < pStrLen; j++) {
      if (i + j > mStrLen) {
        return -1;
      }
      if (mStr[i + j] != pStr[j]) {
        break;
      }
      if (j + 1 == pStrLen) {
        return i;
      }
    }
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L13:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L13
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
data_type: string
name: parse_data
arguments: data_label: D1
data_type: string

tool: D0: %s[%llu]=

tool: D1: %02x

assistant: ```c
void print(unsigned char *name, unsigned char *var, unsigned long long len) {
  int i;
  printf("%s[%llu]=", name, len);
  for (i = 0; i < len; ++i)
    printf("%02x", var[i]);
}
```