ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<split>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,%eax
  mov    %al,-0x2c(%rbp)
  movq   $0x0,-0x18(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L12
  movsbl -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <count_tokens@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L18
L29:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x2c(%rbp)
  jne    L7
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
L7:
  addq   $0x1,-0x10(%rbp)
L18:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
L12:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **split(const char *str, char delim) {
  char **tokens = NULL;
  if (str != NULL) {
    int count = count_tokens(str, delim);
    int str_len = strlen(str);
    char *cpy = malloc(str_len + 1);
    strcpy(cpy, str);
    tokens = malloc(sizeof(char *) * (count + 1));
    char **tkn_ptr = tokens;
    *tkn_ptr++ = cpy;
    while (*cpy != '\0') {
      if (*cpy == delim) {
        *cpy++ = '\0';
        *tkn_ptr++ = cpy;
      }
      cpy++;
    }
    *tkn_ptr = NULL;
  }
  return tokens;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<split>:
  endbr64
  mov    %rdi,%rax
  test   %rdi,%rdi
  je     L30
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  movsbl %sil,%esi
  call   <count_tokens@plt>
  mov    %eax,%r13d
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    0x8(,%r13,8),%edi
  call   <malloc@plt>
  lea    0x8(%rax),%rsi
  mov    %rbx,(%rax)
  movzbl (%rbx),%ecx
  test   %cl,%cl
  jne    L25
L14:
  movq   $0x0,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  lea    0x1(%rdx),%rbx
  movzbl 0x1(%rdx),%ecx
  test   %cl,%cl
  je     L14
L25:
  mov    %rbx,%rdx
  cmp    %cl,%r12b
  jne    L21
  lea    0x1(%rbx),%rdx
  movb   $0x0,(%rbx)
  mov    %rdx,(%rsi)
  lea    0x8(%rsi),%rsi
  jmp    L21
L30:
  ret
```

assistant: ```c
char **split(const char *str, char delim) {
  char **tokens = NULL;
  if (str != NULL) {
    int count = count_tokens(str, delim);
    int str_len = strlen(str);
    char *cpy = malloc(str_len + 1);
    strcpy(cpy, str);
    tokens = malloc(sizeof(char *) * (count + 1));
    char **tkn_ptr = tokens;
    *tkn_ptr++ = cpy;
    while (*cpy != '\0') {
      if (*cpy == delim) {
        *cpy++ = '\0';
        *tkn_ptr++ = cpy;
      }
      cpy++;
    }
    *tkn_ptr = NULL;
  }
  return tokens;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<split>:
  endbr64
  test   %rdi,%rdi
  je     L13
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  movsbl %sil,%esi
  push   %rbx
  sub    $0x8,%rsp
  call   <count_tokens@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <strcpy@plt>
  lea    0x8(,%r13,8),%edi
  call   <malloc@plt>
  mov    %rbx,(%rax)
  movzbl (%rbx),%ecx
  lea    0x8(%rax),%rsi
  test   %cl,%cl
  je     L5
  nopl   0x0(%rax)
L14:
  mov    %rbx,%rdx
  cmp    %cl,%bpl
  jne    L37
  add    $0x1,%rdx
  movb   $0x0,(%rbx)
  add    $0x8,%rsi
  mov    %rdx,-0x8(%rsi)
L37:
  movzbl 0x1(%rdx),%ecx
  lea    0x1(%rdx),%rbx
  test   %cl,%cl
  jne    L14
L5:
  movq   $0x0,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
char **split(const char *str, char delim) {
  char **tokens = NULL;
  if (str != NULL) {
    int count = count_tokens(str, delim);
    int str_len = strlen(str);
    char *cpy = malloc(str_len + 1);
    strcpy(cpy, str);
    tokens = malloc(sizeof(char *) * (count + 1));
    char **tkn_ptr = tokens;
    *tkn_ptr++ = cpy;
    while (*cpy != '\0') {
      if (*cpy == delim) {
        *cpy++ = '\0';
        *tkn_ptr++ = cpy;
      }
      cpy++;
    }
    *tkn_ptr = NULL;
  }
  return tokens;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<split>:
  endbr64
  test   %rdi,%rdi
  je     L27
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  movsbl %sil,%esi
  push   %rbx
  sub    $0x8,%rsp
  call   <count_tokens@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <strcpy@plt>
  lea    0x8(,%r13,8),%edi
  call   <malloc@plt>
  mov    %rbx,(%rax)
  movzbl (%rbx),%edx
  lea    0x8(%rax),%rsi
  test   %dl,%dl
  jne    L57
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L31:
  movzbl 0x1(%rbx),%edx
  lea    0x1(%rbx),%rcx
  test   %dl,%dl
  je     L39
L11:
  mov    %rcx,%rbx
L57:
  cmp    %dl,%bpl
  jne    L31
  lea    0x1(%rbx),%rdx
  movb   $0x0,(%rbx)
  lea    0x8(%rsi),%rdi
  mov    %rdx,(%rsi)
  movzbl 0x2(%rbx),%edx
  lea    0x2(%rbx),%rcx
  mov    %rdi,%rsi
  test   %dl,%dl
  jne    L11
L39:
  movq   $0x0,(%rsi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
char **split(const char *str, char delim) {
  char **tokens = NULL;
  if (str != NULL) {
    int count = count_tokens(str, delim);
    int str_len = strlen(str);
    char *cpy = malloc(str_len + 1);
    strcpy(cpy, str);
    tokens = malloc(sizeof(char *) * (count + 1));
    char **tkn_ptr = tokens;
    *tkn_ptr++ = cpy;
    while (*cpy != '\0') {
      if (*cpy == delim) {
        *cpy++ = '\0';
        *tkn_ptr++ = cpy;
      }
      cpy++;
    }
    *tkn_ptr = NULL;
  }
  return tokens;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isValidPattern>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L9
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L22
L9:
  mov    $0x1,%eax
  jmp    L18
L22:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
bool isValidPattern(uint8_t firstBytes[], uint8_t patternOne[], uint8_t patternTwo[]) {
  if ((memcmp(patternOne, firstBytes, 4) == 0) || (memcmp(patternTwo, firstBytes, 4) == 0))
    return true;
  return false;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isValidPattern>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%rbp
  mov    $0x4,%edx
  mov    %rbx,%rsi
  call   <memcmp@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L11
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    $0x4,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L17
```

assistant: ```c
bool isValidPattern(uint8_t firstBytes[], uint8_t patternOne[], uint8_t patternTwo[]) {
  if ((memcmp(patternOne, firstBytes, 4) == 0) || (memcmp(patternTwo, firstBytes, 4) == 0))
    return true;
  return false;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isValidPattern>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  mov    $0x4,%edx
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <memcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L2
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L2:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
bool isValidPattern(uint8_t firstBytes[], uint8_t patternOne[], uint8_t patternTwo[]) {
  if ((memcmp(patternOne, firstBytes, 4) == 0) || (memcmp(patternTwo, firstBytes, 4) == 0))
    return true;
  return false;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isValidPattern>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  mov    $0x4,%edx
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <memcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L0
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L0:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
bool isValidPattern(uint8_t firstBytes[], uint8_t patternOne[], uint8_t patternTwo[]) {
  if ((memcmp(patternOne, firstBytes, 4) == 0) || (memcmp(patternTwo, firstBytes, 4) == 0))
    return true;
  return false;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mbc_to_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xc3,-0x4(%rbp)
  je     L1
  cmpl   $0xc3,-0x4(%rbp)
  jg     L7
  cmpl   $0xf,-0x4(%rbp)
  jg     L123
  cmpl   $0x0,-0x4(%rbp)
  js     L7
  cmpl   $0xf,-0x4(%rbp)
  ja     L7
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D43(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D43(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
L123:
  cmpl   $0xc1,-0x4(%rbp)
  je     L98
  jmp    L7
  lea    D21(%rip),%rax
  jmp    L20
  lea    D107(%rip),%rax
  jmp    L20
  lea    D3(%rip),%rax
  jmp    L20
  lea    D24(%rip),%rax
  jmp    L20
  lea    D99(%rip),%rax
  jmp    L20
  lea    D86(%rip),%rax
  jmp    L20
L98:
  lea    D62(%rip),%rax
  jmp    L20
L1:
  lea    D37(%rip),%rax
  jmp    L20
L7:
  lea    D36(%rip),%rax
L20:
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D21
type: string
name: parse_data
arguments: data_tag: D107
type: string
name: parse_data
arguments: data_tag: D3
type: string
name: parse_data
arguments: data_tag: D24
type: string
name: parse_data
arguments: data_tag: D99
type: string
name: parse_data
arguments: data_tag: D86
type: string
name: parse_data
arguments: data_tag: D62
type: string
name: parse_data
arguments: data_tag: D37
type: string
name: parse_data
arguments: data_tag: D36
type: string

tool: D21: NONE

tool: D107: MBC1

tool: D3: MBC2

tool: D24: MBC3

tool: D99: MBC5

tool: D86: RUMBL

tool: D62: HUC1

tool: D37: HUC3

tool: D36: UNKN

assistant: ```c
const char *mbc_to_string(int mbc_type) {
  switch (mbc_type) {
  case 0:
    return "NONE";
  case 1:
    return "MBC1";
  case 2:
    return "MBC2";
  case 3:
    return "MBC3";
  case 5:
    return "MBC5";
  case 15:
    return "RUMBL";
  case 0xC1:
    return "HUC1";
  case 0xC3:
    return "HUC3";
  default:
    return "UNKN";
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mbc_to_string>:
  endbr64
  cmp    $0xf,%edi
  jg     L5
  lea    D69(%rip),%rax
  test   %edi,%edi
  js     L23
  cmp    $0xf,%edi
  ja     L29
  mov    %edi,%edi
  lea    D116(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D16(%rip),%rax
  ret
L29:
  lea    D69(%rip),%rax
  ret
L5:
  lea    D114(%rip),%rax
  cmp    $0xc1,%edi
  je     L23
  cmp    $0xc3,%edi
  lea    D69(%rip),%rax
  lea    D0(%rip),%rdx
  cmove  %rdx,%rax
  ret
  lea    D18(%rip),%rax
  ret
  lea    D115(%rip),%rax
  ret
  lea    D9(%rip),%rax
  ret
  lea    D106(%rip),%rax
  ret
  lea    D105(%rip),%rax
L23:
  ret
```

tool_calls: name: read_data
arguments: data_label: D69
type_infered: string
name: read_data
arguments: data_label: D16
type_infered: string
name: read_data
arguments: data_label: D114
type_infered: string
name: read_data
arguments: data_label: D0
type_infered: string
name: read_data
arguments: data_label: D18
type_infered: string
name: read_data
arguments: data_label: D115
type_infered: string
name: read_data
arguments: data_label: D9
type_infered: string
name: read_data
arguments: data_label: D106
type_infered: string
name: read_data
arguments: data_label: D105
type_infered: string

tool: D69: UNKN

tool: D16: MBC1

tool: D114: HUC1

tool: D0: HUC3

tool: D18: MBC2

tool: D115: MBC3

tool: D9: MBC5

tool: D106: RUMBL

tool: D105: NONE

assistant: ```c
const char *mbc_to_string(int mbc_type) {
  switch (mbc_type) {
  case 0:
    return "NONE";
  case 1:
    return "MBC1";
  case 2:
    return "MBC2";
  case 3:
    return "MBC3";
  case 5:
    return "MBC5";
  case 15:
    return "RUMBL";
  case 0xC1:
    return "HUC1";
  case 0xC3:
    return "HUC3";
  default:
    return "UNKN";
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<backtrace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L49
L29:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <getreturnaddr@plt>
  mov    %rax,(%rbx)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L14
  addl   $0x1,-0x14(%rbp)
L49:
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <getframeaddr@plt>
  test   %rax,%rax
  je     L44
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L29
  jmp    L44
L14:
  nop
L44:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int backtrace(void **buffer, int size) {
  int i;
  for (i = 1; getframeaddr(i + 1) != NULL && i != size + 1; i++) {
    buffer[i - 1] = getreturnaddr(i);
    if (buffer[i - 1] == NULL)
      break;
  }
  return i - 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
<backtrace>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%r12d
  add    $0x2,%r12
  mov    $0x2,%ebx
L4:
  mov    %ebx,%ebp
  lea    -0x1(%rbx),%r14d
  mov    %ebx,%edi
  call   <getframeaddr@plt>
  test   %rax,%rax
  je     L19
  cmp    %r12,%rbx
  je     L19
  mov    %r14d,%edi
  call   <getreturnaddr@plt>
  mov    %rax,-0x10(%r13,%rbx,8)
  add    $0x1,%rbx
  test   %rax,%rax
  jne    L4
L19:
  lea    -0x2(%rbp),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int backtrace(void **buffer, int size) {
  int i;
  for (i = 1; getframeaddr(i + 1) != NULL && i != size + 1; i++) {
    buffer[i - 1] = getreturnaddr(i);
    if (buffer[i - 1] == NULL)
      break;
  }
  return i - 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<backtrace>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  add    $0x2,%r12
  push   %rbx
  mov    $0x2,%ebx
  jmp    L1
  nopl   (%rax)
L29:
  cmp    %r12,%rbx
  je     L11
  mov    %r14d,%edi
  call   <getreturnaddr@plt>
  mov    %rax,-0x10(%r13,%rbx,8)
  add    $0x1,%rbx
  test   %rax,%rax
  je     L11
L1:
  mov    %ebx,%edi
  mov    %ebx,%ebp
  lea    -0x1(%rbx),%r14d
  call   <getframeaddr@plt>
  test   %rax,%rax
  jne    L29
L11:
  pop    %rbx
  lea    -0x2(%rbp),%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int backtrace(void **buffer, int size) {
  int i;
  for (i = 1; getframeaddr(i + 1) != NULL && i != size + 1; i++) {
    buffer[i - 1] = getreturnaddr(i);
    if (buffer[i - 1] == NULL)
      break;
  }
  return i - 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<backtrace>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  add    $0x2,%r12
  push   %rbx
  mov    $0x2,%ebx
  jmp    L21
  nopl   (%rax)
L1:
  cmp    %r12,%rbx
  je     L0
  mov    %r14d,%edi
  call   <getreturnaddr@plt>
  mov    %rax,-0x10(%r13,%rbx,8)
  add    $0x1,%rbx
  test   %rax,%rax
  je     L0
L21:
  mov    %ebx,%edi
  mov    %ebx,%ebp
  lea    -0x1(%rbx),%r14d
  call   <getframeaddr@plt>
  test   %rax,%rax
  jne    L1
L0:
  pop    %rbx
  lea    -0x2(%rbp),%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int backtrace(void **buffer, int size) {
  int i;
  for (i = 1; getframeaddr(i + 1) != NULL && i != size + 1; i++) {
    buffer[i - 1] = getreturnaddr(i);
    if (buffer[i - 1] == NULL)
      break;
  }
  return i - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printDiff>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x20(%rsp)
  mov    %rsi,0x28(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %ecx,%ebx
  mov    %ecx,0x1c(%rsp)
  mov    %r8d,%r14d
  movss  %xmm0,0xc(%rsp)
  cvtss2sd %xmm0,%xmm0
  mov    %r8d,%esi
  lea    D18(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L91
  movl   $0x0,0x18(%rsp)
  mov    $0x0,%r12d
  movl   $0x0,0x10(%rsp)
  jmp    L40
L10:
  mov    0x10(%rsp),%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L93
L12:
  add    $0x1,%r12d
L56:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  je     L55
L86:
  mov    %ebx,0x8(%rsp)
  movss  0x0(%r13,%rbx,4),%xmm0
  subss  0x0(%rbp,%rbx,4),%xmm0
  call   <fabs@plt>
  movaps %xmm0,%xmm2
  comiss 0xc(%rsp),%xmm0
  jbe    L56
  cmp    %r12d,%r14d
  jle    L12
  pxor   %xmm0,%xmm0
  cvtss2sd 0x0(%r13,%rbx,4),%xmm0
  cvtss2sd %xmm2,%xmm2
  pxor   %xmm1,%xmm1
  cvtss2sd 0x0(%rbp,%rbx,4),%xmm1
  mov    0x10(%rsp),%edx
  mov    0x8(%rsp),%esi
  lea    D20(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L12
L55:
  addl   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%edx
  add    %edx,0x18(%rsp)
  cmp    %eax,0x1c(%rsp)
  je     L104
L40:
  cmp    %r12d,%r14d
  jg     L10
L93:
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L55
  mov    %eax,%r15d
  movslq 0x18(%rsp),%rbp
  shl    $0x2,%rbp
  mov    0x20(%rsp),%rax
  lea    (%rax,%rbp,1),%r13
  add    0x28(%rsp),%rbp
  mov    $0x0,%ebx
  jmp    L86
L91:
  mov    $0x0,%r12d
L104:
  mov    %r12d,%esi
  lea    D31(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: label: D18
type_guessed: string
name: disassemble_data
arguments: label: D9
type_guessed: string
name: disassemble_data
arguments: label: D20
type_guessed: string
name: disassemble_data
arguments: label: D31
type_guessed: string

tool: D18: Listing first %d Differences > %.6f...


tool: D9: 
  Row %d:


tool: D20:     Loc(%d,%d)	CPU=%.5f	GPU=%.5f	Diff=%.6f


tool: D31:  
  Total Errors = %d



assistant: ```c
void printDiff(float *data1, float *data2, int width, int height, int listLength, float listTol) {
  printf("Listing first %d Differences > %.6f...\n", listLength, listTol);
  int i, j, k;
  int error_count = 0;
  for (j = 0; j < height; j++) {
    if (error_count < listLength) {
      printf("\n  Row %d:\n", j);
    }
    for (i = 0; i < width; i++) {
      k = j * width + i;
      float diff = fabs(data1[k] - data2[k]);
      if (diff > listTol) {
        if (error_count < listLength) {
          printf("    Loc(%d,%d)\tCPU=%.5f\tGPU=%.5f\tDiff=%.6f\n", i, j, data1[k], data2[k], diff);
        }
        error_count++;
      }
    }
  }
  printf(" \n  Total Errors = %d\n\n", error_count);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<printDiff>:
  endbr64
  push   %r15
  mov    $0x1,%eax
  push   %r14
  mov    %edx,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x38,%rsp
  mov    %rdi,0x18(%rsp)
  lea    D7(%rip),%rdi
  mov    %rsi,0x20(%rsp)
  mov    %r8d,%esi
  movss  %xmm0,0x4(%rsp)
  cvtss2sd %xmm0,%xmm0
  mov    %edx,0x10(%rsp)
  mov    %ecx,0x28(%rsp)
  mov    %r8d,0x8(%rsp)
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L56
  lea    -0x1(%r14),%eax
  movl   $0x0,0x14(%rsp)
  xor    %ebp,%ebp
  movl   $0x0,0xc(%rsp)
  mov    %eax,0x2c(%rsp)
  xchg   %ax,%ax
L41:
  cmp    %ebp,0x8(%rsp)
  jg     L97
L85:
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  jle    L44
  movslq 0x14(%rsp),%rbx
  mov    0x18(%rsp),%rax
  xor    %r15d,%r15d
  mov    0x2c(%rsp),%r14d
  shl    $0x2,%rbx
  lea    (%rax,%rbx,1),%r12
  add    0x20(%rsp),%rbx
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L31:
  add    $0x1,%ebp
L81:
  lea    0x1(%r15),%rax
  cmp    %r14,%r15
  je     L44
  mov    %rax,%r15
L23:
  movss  (%r12,%r15,4),%xmm0
  subss  (%rbx,%r15,4),%xmm0
  call   <fabs@plt>
  comiss 0x4(%rsp),%xmm0
  jbe    L81
  cmp    %ebp,0x8(%rsp)
  jle    L31
  mov    0xc(%rsp),%edx
  pxor   %xmm3,%xmm3
  pxor   %xmm2,%xmm2
  pxor   %xmm1,%xmm1
  cvtss2sd (%r12,%r15,4),%xmm3
  cvtss2sd %xmm0,%xmm2
  mov    %r15d,%esi
  movapd %xmm3,%xmm0
  lea    D0(%rip),%rdi
  mov    $0x3,%eax
  cvtss2sd (%rbx,%r15,4),%xmm1
  call   <printf@plt>
  jmp    L31
  nopw   0x0(%rax,%rax,1)
L44:
  addl   $0x1,0xc(%rsp)
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%eax
  add    %edx,0x14(%rsp)
  cmp    %eax,0x28(%rsp)
  jne    L41
L71:
  add    $0x38,%rsp
  mov    %ebp,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <printf@plt>
  nopl   0x0(%rax)
L97:
  mov    0xc(%rsp),%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L85
L56:
  xor    %ebp,%ebp
  jmp    L71
```

tool_calls: name: read_data
arguments: data_tag: D7
data_type: string
name: read_data
arguments: data_tag: D0
data_type: string
name: read_data
arguments: data_tag: D19
data_type: string
name: read_data
arguments: data_tag: D11
data_type: string

tool: D7: Listing first %d Differences > %.6f...


tool: D0:     Loc(%d,%d)	CPU=%.5f	GPU=%.5f	Diff=%.6f


tool: D19:  
  Total Errors = %d



tool: D11: 
  Row %d:


assistant: ```c
void printDiff(float *data1, float *data2, int width, int height, int listLength, float listTol) {
  printf("Listing first %d Differences > %.6f...\n", listLength, listTol);
  int i, j, k;
  int error_count = 0;
  for (j = 0; j < height; j++) {
    if (error_count < listLength) {
      printf("\n  Row %d:\n", j);
    }
    for (i = 0; i < width; i++) {
      k = j * width + i;
      float diff = fabs(data1[k] - data2[k]);
      if (diff > listTol) {
        if (error_count < listLength) {
          printf("    Loc(%d,%d)\tCPU=%.5f\tGPU=%.5f\tDiff=%.6f\n", i, j, data1[k], data2[k], diff);
        }
        error_count++;
      }
    }
  }
  printf(" \n  Total Errors = %d\n\n", error_count);
}
```