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

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x40000000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <arraywrite@plt>
  movsbl (%rbx),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: tag: D4
data_type: string

tool: D4: r = %d


assistant: ```c
int main(int argc, char **argv) {
  char *arr = (char *)malloc(1024 * 1024 * 1024);
  arraywrite(arr);
  printf("r = %d\n", arr[0]);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x40000000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <arraywrite@plt>
  movsbl (%rbx),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_guessed: string

tool: D7: r = %d


assistant: ```c
int main(int argc, char **argv) {
  char *arr = (char *)malloc(1024 * 1024 * 1024);
  arraywrite(arr);
  printf("r = %d\n", arr[0]);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x40000000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <arraywrite@plt>
  movsbl (%rbx),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
data_type: string

tool: D4: r = %d


assistant: ```c
int main(int argc, char **argv) {
  char *arr = (char *)malloc(1024 * 1024 * 1024);
  arraywrite(arr);
  printf("r = %d\n", arr[0]);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<energymon_strencpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L38
  cmpq   $0x0,-0x20(%rbp)
  jne    L53
L38:
  mov    $0x0,%eax
  jmp    L54
L53:
  movq   $0x0,-0x8(%rbp)
  jmp    L26
L30:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x8(%rbp)
L26:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L34
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L30
L34:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L44
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L39
L44:
  cmpq   $0x0,-0x28(%rbp)
  je     L39
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L39:
  mov    -0x18(%rbp),%rax
L54:
  pop    %rbp
  ret
```

assistant: ```c
char *energymon_strencpy(char *dest, const char *src, size_t n) {
  if (dest == NULL || src == NULL) {
    return NULL;
  }
  size_t i;
  for (i = 0; i < n && src[i] != '\0'; i++) {
    dest[i] = src[i];
  }
  if (i < n) {
    dest[i] = '\0';
  } else if (n > 0) {
    dest[n - 1] = '\0';
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<energymon_strencpy>:
  endbr64
  test   %rdi,%rdi
  je     L12
  test   %rsi,%rsi
  je     L12
  mov    $0x0,%ecx
  test   %rdx,%rdx
  je     L48
L56:
  movzbl (%rsi,%rcx,1),%r8d
  test   %r8b,%r8b
  je     L28
  mov    %r8b,(%rdi,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L56
L6:
  movb   $0x0,-0x1(%rdi,%rdx,1)
  mov    %rdi,%rax
  ret
L48:
  mov    %rdi,%rax
  ret
L12:
  mov    $0x0,%eax
  ret
L28:
  cmp    %rcx,%rdx
  jbe    L6
  movb   $0x0,(%rdi,%rcx,1)
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *energymon_strencpy(char *dest, const char *src, size_t n) {
  if (dest == NULL || src == NULL) {
    return NULL;
  }
  size_t i;
  for (i = 0; i < n && src[i] != '\0'; i++) {
    dest[i] = src[i];
  }
  if (i < n) {
    dest[i] = '\0';
  } else if (n > 0) {
    dest[n - 1] = '\0';
  }
  return dest;
}
```
user: Translate the below assembly code into C programming language:

```asm
<energymon_strencpy>:
  endbr64
  test   %rdi,%rdi
  je     L24
  test   %rsi,%rsi
  je     L24
  test   %rdx,%rdx
  je     L15
  mov    %rdi,%rcx
  lea    (%rdx,%rdi,1),%rax
  jmp    L16
  nopl   0x0(%rax)
L14:
  mov    %r8b,(%rcx)
  add    $0x1,%rcx
  add    $0x1,%rsi
  cmp    %rax,%rcx
  je     L39
L16:
  movzbl (%rsi),%r8d
  test   %r8b,%r8b
  jne    L14
  movb   $0x0,(%rcx)
  mov    %rdi,%rax
  ret
L39:
  movb   $0x0,-0x1(%rdi,%rdx,1)
L15:
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *energymon_strencpy(char *dest, const char *src, size_t n) {
  if (dest == NULL || src == NULL) {
    return NULL;
  }
  size_t i;
  for (i = 0; i < n && src[i] != '\0'; i++) {
    dest[i] = src[i];
  }
  if (i < n) {
    dest[i] = '\0';
  } else if (n > 0) {
    dest[n - 1] = '\0';
  }
  return dest;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<energymon_strencpy>:
  endbr64
  test   %rdi,%rdi
  je     L33
  test   %rsi,%rsi
  je     L33
  test   %rdx,%rdx
  je     L24
  mov    %rdi,%rcx
  lea    (%rdx,%rdi,1),%rax
  jmp    L2
  nopl   0x0(%rax)
L3:
  mov    %r8b,(%rcx)
  add    $0x1,%rcx
  add    $0x1,%rsi
  cmp    %rax,%rcx
  je     L15
L2:
  movzbl (%rsi),%r8d
  test   %r8b,%r8b
  jne    L3
  movb   $0x0,(%rcx)
  mov    %rdi,%rax
  ret
L15:
  movb   $0x0,-0x1(%rdi,%rdx,1)
L24:
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L33:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *energymon_strencpy(char *dest, const char *src, size_t n) {
  if (dest == NULL || src == NULL) {
    return NULL;
  }
  size_t i;
  for (i = 0; i < n && src[i] != '\0'; i++) {
    dest[i] = src[i];
  }
  if (i < n) {
    dest[i] = '\0';
  } else if (n > 0) {
    dest[n - 1] = '\0';
  }
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<inserisci>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%r12)
  jle    L29
  mov    %r13,%rbp
  mov    $0x0,%ebx
  mov    %rsp,%r14
L55:
  mov    0x0(%rbp),%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L78
  add    $0x1,%ebx
  add    $0x8,%rbp
  cmp    %ebx,(%r12)
  jg     L55
  jmp    L3
L78:
  mov    (%r12),%edx
  cmp    %ebx,%edx
  jle    L3
  movslq %edx,%rcx
  shl    $0x3,%rcx
  lea    0x0(%r13,%rcx,1),%rax
  lea    -0x8(%r13,%rcx,1),%rcx
  sub    $0x1,%edx
  sub    %ebx,%edx
  shl    $0x3,%rdx
  sub    %rdx,%rcx
L22:
  mov    -0x8(%rax),%rdx
  mov    %rdx,(%rax)
  sub    $0x8,%rax
  cmp    %rcx,%rax
  jne    L22
L3:
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movslq %ebx,%rbx
  mov    %rax,0x0(%r13,%rbx,8)
  mov    %rbp,%rsi
  call   <strcpy@plt>
  mov    (%r12),%eax
  add    $0x1,%eax
  mov    %eax,(%r12)
  test   %eax,%eax
  jle    L4
  mov    $0x0,%ebx
  lea    D13(%rip),%rbp
L19:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,(%r12)
  jg     L19
L4:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L29:
  mov    $0x0,%ebx
  jmp    L3
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D4
type_infered: string
name: read_data
arguments: label: D7
type_infered: string
name: read_data
arguments: label: D13
type_infered: string

tool: D4: Parola da inserire: 

tool: D7: %s

tool: D13: [%d] %s


assistant: ```c
void inserisci(char **p, int *n) {
  char inp[25 + 1];
  int i, j;
  printf("Parola da inserire: ");
  scanf("%s", inp);
  for (i = 0; i < *n && strcmp(inp, p[i]) > 0; i++)
    ;
  for (j = *n; j > i; j--)
    p[j] = p[j - 1];
  p[i] = (char *)malloc((strlen(inp) + 1) * sizeof(char));
  strcpy(p[i], inp);
  *n = *n + 1;
  for (i = 0; i < *n; i++)
    printf("[%d] %s\n", i, p[i]);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inserisci>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  lea    D4(%rip),%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %r13,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <printf@plt>
  mov    %r14,%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    (%r12),%eax
  test   %eax,%eax
  jg     L33
  jmp    L80
  nopl   0x0(%rax,%rax,1)
L26:
  add    $0x1,%ebp
  add    $0x8,%rbx
  cmp    %ebp,(%r12)
  jle    L47
L33:
  mov    (%rbx),%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L26
  movslq (%r12),%rax
  cmp    %ebp,%eax
  jle    L80
  mov    %eax,%edx
  shl    $0x3,%rax
  sub    %ebp,%edx
  sub    $0x1,%edx
  mov    %rdx,%rdi
  lea    0x8(,%rdx,8),%rdx
  neg    %rdi
  shl    $0x3,%rdi
  lea    -0x8(%rax,%rdi,1),%rsi
  add    %rax,%rdi
  add    %r13,%rsi
  add    %r13,%rdi
  call   <memmove@plt>
L80:
  mov    %r14,%rdi
  lea    D7(%rip),%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r14,%rsi
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  xor    %ebx,%ebx
  call   <strcpy@plt>
  mov    (%r12),%eax
  add    $0x1,%eax
  mov    %eax,(%r12)
  test   %eax,%eax
  jle    L63
  nopw   0x0(%rax,%rax,1)
L40:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,(%r12)
  jg     L40
L63:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  movslq %ebp,%rbp
  lea    0x0(%r13,%rbp,8),%rbx
  jmp    L80
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type: string
name: read_data
arguments: data_tag: D20
type: string
name: read_data
arguments: data_tag: D7
type: string

tool: D4: Parola da inserire: 

tool: D20: %s

tool: D7: [%d] %s


assistant: ```c
void inserisci(char **p, int *n) {
  char inp[25 + 1];
  int i, j;
  printf("Parola da inserire: ");
  scanf("%s", inp);
  for (i = 0; i < *n && strcmp(inp, p[i]) > 0; i++)
    ;
  for (j = *n; j > i; j--)
    p[j] = p[j - 1];
  p[i] = (char *)malloc((strlen(inp) + 1) * sizeof(char));
  strcpy(p[i], inp);
  *n = *n + 1;
  for (i = 0; i < *n; i++)
    printf("[%d] %s\n", i, p[i]);
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inserisci>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  lea    D3(%rip),%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %r13,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <printf@plt>
  mov    %r14,%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  mov    (%r12),%eax
  test   %eax,%eax
  jg     L37
  jmp    L63
  nopl   0x0(%rax,%rax,1)
L38:
  add    $0x1,%ebp
  add    $0x8,%rbx
  cmp    %ebp,(%r12)
  jle    L82
L37:
  mov    (%rbx),%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L38
  movslq (%r12),%rax
  cmp    %ebp,%eax
  jle    L63
  mov    %eax,%edx
  shl    $0x3,%rax
  sub    %ebp,%edx
  sub    $0x1,%edx
  mov    %rdx,%rdi
  lea    0x8(,%rdx,8),%rdx
  neg    %rdi
  shl    $0x3,%rdi
  lea    -0x8(%rax,%rdi,1),%rsi
  add    %rax,%rdi
  add    %r13,%rsi
  add    %r13,%rdi
  call   <memmove@plt>
L63:
  mov    %r14,%rdi
  lea    D25(%rip),%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r14,%rsi
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  xor    %ebx,%ebx
  call   <strcpy@plt>
  mov    (%r12),%eax
  add    $0x1,%eax
  mov    %eax,(%r12)
  test   %eax,%eax
  jle    L19
  nopw   0x0(%rax,%rax,1)
L54:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,(%r12)
  jg     L54
L19:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L82:
  movslq %ebp,%rbp
  lea    0x0(%r13,%rbp,8),%rbx
  jmp    L63
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_infered: string
name: disassemble_data
arguments: data_label: D5
type_infered: string
name: disassemble_data
arguments: data_label: D25
type_infered: string

tool: D3: Parola da inserire: 

tool: D5: %s

tool: D25: [%d] %s


assistant: ```c
void inserisci(char **p, int *n) {
  char inp[25 + 1];
  int i, j;
  printf("Parola da inserire: ");
  scanf("%s", inp);
  for (i = 0; i < *n && strcmp(inp, p[i]) > 0; i++)
    ;
  for (j = *n; j > i; j--)
    p[j] = p[j - 1];
  p[i] = (char *)malloc((strlen(inp) + 1) * sizeof(char));
  strcpy(p[i], inp);
  *n = *n + 1;
  for (i = 0; i < *n; i++)
    printf("[%d] %s\n", i, p[i]);
  printf("\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<read_mysql_tiny>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edx
  lea    D1(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <logger@plt>
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
data_type: string

tool: D1: TINY INT: %d


assistant: ```c
int read_mysql_tiny(const char *buf) {
  logger(0, "TINY INT: %d\n", *buf);
  return *buf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<read_mysql_tiny>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edx
  lea    D5(%rip),%rsi
  mov    $0x0,%edi
  call   <logger@plt>
  movsbl (%rbx),%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type_guessed: string

tool: D5: TINY INT: %d


assistant: ```c
int read_mysql_tiny(const char *buf) {
  logger(0, "TINY INT: %d\n", *buf);
  return *buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_mysql_tiny>:
  endbr64
  push   %rbx
  movsbl (%rdi),%edx
  mov    %rdi,%rbx
  lea    D6(%rip),%rsi
  xor    %edi,%edi
  call   <logger@plt>
  movsbl (%rbx),%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type: string

tool: D6: TINY INT: %d


assistant: ```c
int read_mysql_tiny(const char *buf) {
  logger(0, "TINY INT: %d\n", *buf);
  return *buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_mysql_tiny>:
  endbr64
  push   %rbx
  movsbl (%rdi),%edx
  mov    %rdi,%rbx
  lea    D6(%rip),%rsi
  xor    %edi,%edi
  call   <logger@plt>
  movsbl (%rbx),%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
datatype: string

tool: D6: TINY INT: %d


assistant: ```c
int read_mysql_tiny(const char *buf) {
  logger(0, "TINY INT: %d\n", *buf);
  return *buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bb_print_timestamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x58(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x64(%rbp)
  lea    -0x64(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x60(%rbp)
  mov    -0x60(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdx,%rcx
  lea    D0(%rip),%rdx
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <strftime@plt>
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    -0x50(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D14
type_infered: string

tool: D0: %T

tool: D14: %s.%03u

assistant: ```c
void bb_print_timestamp(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  time_t timer = (time_t)tv.tv_sec;
  struct tm *tm_info = localtime(&timer);
  char buf[64];
  strftime(buf, sizeof(buf), "%T", tm_info);
  printf("%s.%03u", buf, (unsigned)(tv.tv_usec / 1000));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bb_print_timestamp>:
  endbr64
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  mov    %rax,%rcx
  lea    0x10(%rsp),%rbx
  lea    D5(%rip),%rdx
  mov    $0x40,%esi
  mov    %rbx,%rdi
  call   <strftime@plt>
  mov    0x8(%rsp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %rbx,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x60,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
data_type: string
name: unpack_data
arguments: data_tag: D17
data_type: string

tool: D5: %T

tool: D17: %s.%03u

assistant: ```c
void bb_print_timestamp(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  time_t timer = (time_t)tv.tv_sec;
  struct tm *tm_info = localtime(&timer);
  char buf[64];
  strftime(buf, sizeof(buf), "%T", tm_info);
  printf("%s.%03u", buf, (unsigned)(tv.tv_usec / 1000));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bb_print_timestamp>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <gettimeofday@plt>
  mov    0xc(%rsp),%eax
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  lea    D1(%rip),%rdx
  mov    $0x40,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <strftime@plt>
  movslq 0x8(%rsp),%rdx
  mov    %rbp,%rsi
  lea    D10(%rip),%rdi
  mov    %rdx,%rax
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%eax
  sar    $0x26,%rdx
  sub    %eax,%edx
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x60,%rsp
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
type_infered: string
name: retrieve_data
arguments: tag: D10
type_infered: string

tool: D1: %T

tool: D10: %s.%03u

assistant: ```c
void bb_print_timestamp(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  time_t timer = (time_t)tv.tv_sec;
  struct tm *tm_info = localtime(&timer);
  char buf[64];
  strftime(buf, sizeof(buf), "%T", tm_info);
  printf("%s.%03u", buf, (unsigned)(tv.tv_usec / 1000));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bb_print_timestamp>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <gettimeofday@plt>
  mov    0xc(%rsp),%eax
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  lea    D0(%rip),%rdx
  mov    $0x40,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <strftime@plt>
  movslq 0x8(%rsp),%rdx
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  mov    %rdx,%rax
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%eax
  sar    $0x26,%rdx
  sub    %eax,%edx
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x60,%rsp
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
datatype: string
name: extract_symbolic_data
arguments: data_tag: D16
datatype: string

tool: D0: %T

tool: D16: %s.%03u

assistant: ```c
void bb_print_timestamp(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  time_t timer = (time_t)tv.tv_sec;
  struct tm *tm_info = localtime(&timer);
  char buf[64];
  strftime(buf, sizeof(buf), "%T", tm_info);
  printf("%s.%03u", buf, (unsigned)(tv.tv_usec / 1000));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rotr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int help;
  help = *c;
  *c = *b;
  *b = *a;
  *a = help;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rotr>:
  endbr64
  mov    (%rdx),%eax
  mov    (%rsi),%ecx
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int help;
  help = *c;
  *c = *b;
  *b = *a;
  *a = help;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rotr>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int help;
  help = *c;
  *c = *b;
  *b = *a;
  *a = help;
}
```
user: What is the c source code of the assembly code below?

```asm
<rotr>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int help;
  help = *c;
  *c = *b;
  *b = *a;
  *a = help;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_rx_queue_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  lea    D1(%rip),%rax
  mov    %rax,%rdx
  mov    $0x14,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    D8(%rip),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D1: MProc_Client_%u_RX

assistant: ```c
const char *get_rx_queue_name(unsigned id) {
  static char buffer[sizeof("MProc_Client_%u_RX") + 2];
  snprintf(buffer, sizeof(buffer) - 1, "MProc_Client_%u_RX", id);
  return buffer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_rx_queue_name>:
  endbr64
  push   %rbx
  mov    %edi,%ecx
  lea    D0(%rip),%rdx
  mov    $0x14,%esi
  lea    D12(%rip),%rbx
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

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

tool: D0: MProc_Client_%u_RX

assistant: ```c
const char *get_rx_queue_name(unsigned id) {
  static char buffer[sizeof("MProc_Client_%u_RX") + 2];
  snprintf(buffer, sizeof(buffer) - 1, "MProc_Client_%u_RX", id);
  return buffer;
}
```