ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<ShiftByAmount>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strLen@plt>
  test   %eax,%eax
  jle    L10
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rsi
  nopl   0x0(%rax,%rax,1)
L4:
  movsbl (%rdi),%eax
  add    $0x1,%rdi
  lea    -0x61(%rbp,%rax,1),%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x4ec4ec4f,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x23,%rax
  sub    %ecx,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%edx
  add    $0x61,%edx
  mov    %dl,-0x1(%rdi)
  cmp    %rdi,%rsi
  jne    L4
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ShiftByAmount(char *str, int amount) {
  int i, len = strLen(str);
  for (i = 0; i < len; i++)
    str[i] = (str[i] - 'a' + amount) % 26 + 'a';
}
```
user: What is the c source code of the assembly code below:

```asm
<sysfsparser_thermal_kernel_support>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <chdir@plt>
  test   %rax,%rax
  jns    L13
  mov    $0x0,%eax
  jmp    L3
L13:
  mov    $0x1,%eax
L3:
  pop    %rbp
  ret
```

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

tool: D0: /sys/class/thermal

assistant: ```c
signed char sysfsparser_thermal_kernel_support() {
  if (chdir("/sys/class"
            "/thermal") < 0)
    return 0;
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sysfsparser_thermal_kernel_support>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <chdir@plt>
  not    %rax
  shr    $0x3f,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_guessed: string

tool: D1: /sys/class/thermal

assistant: ```c
signed char sysfsparser_thermal_kernel_support() {
  if (chdir("/sys/class"
            "/thermal") < 0)
    return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<sysfsparser_thermal_kernel_support>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <chdir@plt>
  add    $0x8,%rsp
  not    %rax
  shr    $0x3f,%rax
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: string

tool: D3: /sys/class/thermal

assistant: ```c
signed char sysfsparser_thermal_kernel_support() {
  if (chdir("/sys/class"
            "/thermal") < 0)
    return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sysfsparser_thermal_kernel_support>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <chdir@plt>
  add    $0x8,%rsp
  not    %rax
  shr    $0x3f,%rax
  ret
```

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

tool: D7: /sys/class/thermal

assistant: ```c
signed char sysfsparser_thermal_kernel_support() {
  if (chdir("/sys/class"
            "/thermal") < 0)
    return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %rdi,-0x418(%rbp)
  mov    %esi,-0x41c(%rbp)
  mov    %rdx,-0x428(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x428(%rbp),%rax
  mov    %rax,-0x400(%rbp)
  mov    -0x428(%rbp),%rax
  add    $0x1388,%rax
  mov    %rax,-0x3f8(%rbp)
  movl   $0x0,-0x408(%rbp)
  jmp    L31
L24:
  mov    -0x408(%rbp),%eax
  cmp    -0x41c(%rbp),%eax
  jge    L22
  mov    -0x404(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x400(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x3f8(%rbp)
  jae    L42
L22:
  mov    $0xffffffff,%eax
  jmp    L47
L42:
  mov    -0x404(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movb   $0x0,-0x3f0(%rbp,%rax,1)
  lea    -0x3f0(%rbp),%rdx
  mov    -0x400(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x408(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x408(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x418(%rbp),%rax
  add    %rax,%rdx
  mov    -0x400(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x404(%rbp),%eax
  cltq
  add    %rax,-0x400(%rbp)
L31:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getline1@plt>
  mov    %eax,-0x404(%rbp)
  cmpl   $0x0,-0x404(%rbp)
  jg     L24
  mov    -0x408(%rbp),%eax
L47:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

assistant: ```c
int readlines(char *lineptr[], int maxlines, char *buf) {
  int len, nlines;
  char *p, line[1000], *bufstop;
  p = buf;
  bufstop = buf + 5000;
  nlines = 0;
  while ((len = getline1(line, 1000)) > 0)
    if (nlines >= maxlines || p + len > bufstop)
      return -1;
    else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
      p += len;
    }
  return nlines;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x408,%rsp
  mov    %rdi,%r14
  mov    %esi,0xc(%rsp)
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x1388(%rdx),%r15
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r13
  jmp    L51
L30:
  sub    $0x1,%eax
  cltq
  movb   $0x0,0x10(%rsp,%rax,1)
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,(%r14,%rbx,8)
  add    $0x1,%rbx
  mov    %rbp,%r12
L51:
  mov    %ebx,%ebp
  mov    $0x3e8,%esi
  mov    %r13,%rdi
  call   <getline1@plt>
  test   %eax,%eax
  jle    L54
  cmp    %ebx,0xc(%rsp)
  jle    L18
  movslq %eax,%rbp
  add    %r12,%rbp
  cmp    %r15,%rbp
  jbe    L30
  mov    $0xffffffff,%ebp
  jmp    L54
L18:
  mov    $0xffffffff,%ebp
L54:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    %ebp,%eax
  add    $0x408,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readlines(char *lineptr[], int maxlines, char *buf) {
  int len, nlines;
  char *p, line[1000], *bufstop;
  p = buf;
  bufstop = buf + 5000;
  nlines = 0;
  while ((len = getline1(line, 1000)) > 0)
    if (nlines >= maxlines || p + len > bufstop)
      return -1;
    else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
      p += len;
    }
  return nlines;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readlines>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x1388(%rdx),%rax
  lea    0x10(%rsp),%rbx
  mov    %rax,0x8(%rsp)
  jmp    L42
  cs nopw 0x0(%rax,%rax,1)
L36:
  cmp    %r15d,%ebp
  jle    L53
  movslq %eax,%r13
  add    %r14,%r13
  cmp    0x8(%rsp),%r13
  ja     L53
  sub    $0x1,%eax
  mov    %r14,%rdi
  mov    %rbx,%rsi
  cltq
  movb   $0x0,0x10(%rsp,%rax,1)
  call   <strcpy@plt>
  mov    %r14,(%r12,%r15,8)
  add    $0x1,%r15
  mov    %r13,%r14
L42:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  mov    %r15d,%r13d
  call   <getline1@plt>
  test   %eax,%eax
  jg     L36
L29:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x408,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L53:
  mov    $0xffffffff,%r13d
  jmp    L29
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readlines(char *lineptr[], int maxlines, char *buf) {
  int len, nlines;
  char *p, line[1000], *bufstop;
  p = buf;
  bufstop = buf + 5000;
  nlines = 0;
  while ((len = getline1(line, 1000)) > 0)
    if (nlines >= maxlines || p + len > bufstop)
      return -1;
    else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
      p += len;
    }
  return nlines;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<readlines>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x1388(%rdx),%rax
  lea    0x10(%rsp),%rbx
  mov    %rax,0x8(%rsp)
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L5:
  cmp    %r15d,%ebp
  jle    L3
  movslq %eax,%r13
  add    %r14,%r13
  cmp    0x8(%rsp),%r13
  ja     L3
  sub    $0x1,%eax
  mov    %r14,%rdi
  mov    %rbx,%rsi
  cltq
  movb   $0x0,0x10(%rsp,%rax,1)
  call   <strcpy@plt>
  mov    %r14,(%r12,%r15,8)
  add    $0x1,%r15
  mov    %r13,%r14
L15:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  mov    %r15d,%r13d
  call   <getline1@plt>
  test   %eax,%eax
  jg     L5
L10:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x408,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L3:
  mov    $0xffffffff,%r13d
  jmp    L10
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readlines(char *lineptr[], int maxlines, char *buf) {
  int len, nlines;
  char *p, line[1000], *bufstop;
  p = buf;
  bufstop = buf + 5000;
  nlines = 0;
  while ((len = getline1(line, 1000)) > 0)
    if (nlines >= maxlines || p + len > bufstop)
      return -1;
    else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
      p += len;
    }
  return nlines;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xcalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fixup_null_alloc@plt>
  mov    %rax,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *xcalloc(size_t n, size_t s) {
  void *p;
  p = calloc(n, s);
  if (p == 0)
    p = fixup_null_alloc(n);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xcalloc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <calloc@plt>
  test   %rax,%rax
  je     L8
L6:
  pop    %rbx
  ret
L8:
  mov    %rbx,%rdi
  call   <fixup_null_alloc@plt>
  jmp    L6
```

assistant: ```c
void *xcalloc(size_t n, size_t s) {
  void *p;
  p = calloc(n, s);
  if (p == 0)
    p = fixup_null_alloc(n);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xcalloc>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <calloc@plt>
  test   %rax,%rax
  je     L3
  pop    %rbp
  ret
L3:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <fixup_null_alloc@plt>
```

assistant: ```c
void *xcalloc(size_t n, size_t s) {
  void *p;
  p = calloc(n, s);
  if (p == 0)
    p = fixup_null_alloc(n);
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xcalloc>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <calloc@plt>
  test   %rax,%rax
  je     L2
  pop    %rbp
  ret
L2:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <fixup_null_alloc@plt>
```

assistant: ```c
void *xcalloc(size_t n, size_t s) {
  void *p;
  p = calloc(n, s);
  if (p == 0)
    p = fixup_null_alloc(n);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<updwtmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    %edx,%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L12
  mov    D5(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <lockf@plt>
  test   %rax,%rax
  jne    L12
  mov    -0x20(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    D31(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <lockf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
L12:
  nop
  leave
  ret
```

assistant: ```c
extern void updwtmp(const char *wtmp_file, const struct utmp *lutmp) {
  int fd;
  fd = open(wtmp_file, O_APPEND | O_WRONLY, 0);
  if (fd >= 0) {
    if (lockf(fd, F_LOCK, 0) == 0) {
      write(fd, (const char *)lutmp, sizeof(struct utmp));
      lockf(fd, F_ULOCK, 0);
      close(fd);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<updwtmp>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    D35(%rip),%rax
  mov    (%rax),%esi
  mov    D27(%rip),%rax
  or     (%rax),%esi
  mov    $0x0,%edx
  call   <open@plt>
  test   %eax,%eax
  jns    L19
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  mov    %eax,%ebx
  mov    $0x0,%edx
  mov    D23(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <lockf@plt>
  test   %rax,%rax
  jne    L8
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    $0x0,%edx
  mov    D34(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <lockf@plt>
  mov    %ebx,%edi
  call   <close@plt>
  jmp    L8
```

assistant: ```c
extern void updwtmp(const char *wtmp_file, const struct utmp *lutmp) {
  int fd;
  fd = open(wtmp_file, O_APPEND | O_WRONLY, 0);
  if (fd >= 0) {
    if (lockf(fd, F_LOCK, 0) == 0) {
      write(fd, (const char *)lutmp, sizeof(struct utmp));
      lockf(fd, F_ULOCK, 0);
      close(fd);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<updwtmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %edx,%edx
  push   %rbp
  sub    $0x8,%rsp
  mov    D27(%rip),%rax
  mov    (%rax),%esi
  mov    D39(%rip),%rax
  or     (%rax),%esi
  call   <open@plt>
  test   %eax,%eax
  jns    L15
L14:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %eax,%ebp
  mov    D28(%rip),%rax
  xor    %edx,%edx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <lockf@plt>
  test   %rax,%rax
  jne    L14
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x4,%edx
  call   <write@plt>
  mov    D37(%rip),%rax
  mov    %ebp,%edi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <lockf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10a0 <close@plt>
```

assistant: ```c
extern void updwtmp(const char *wtmp_file, const struct utmp *lutmp) {
  int fd;
  fd = open(wtmp_file, O_APPEND | O_WRONLY, 0);
  if (fd >= 0) {
    if (lockf(fd, F_LOCK, 0) == 0) {
      write(fd, (const char *)lutmp, sizeof(struct utmp));
      lockf(fd, F_ULOCK, 0);
      close(fd);
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<updwtmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %edx,%edx
  push   %rbp
  sub    $0x8,%rsp
  mov    D29(%rip),%rax
  mov    (%rax),%esi
  mov    D3(%rip),%rax
  or     (%rax),%esi
  call   <open@plt>
  test   %eax,%eax
  jns    L10
L5:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %eax,%ebp
  mov    D19(%rip),%rax
  xor    %edx,%edx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <lockf@plt>
  test   %rax,%rax
  jne    L5
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x4,%edx
  call   <write@plt>
  mov    D14(%rip),%rax
  mov    %ebp,%edi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <lockf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10a0 <close@plt>
```

assistant: ```c
extern void updwtmp(const char *wtmp_file, const struct utmp *lutmp) {
  int fd;
  fd = open(wtmp_file, O_APPEND | O_WRONLY, 0);
  if (fd >= 0) {
    if (lockf(fd, F_LOCK, 0) == 0) {
      write(fd, (const char *)lutmp, sizeof(struct utmp));
      lockf(fd, F_ULOCK, 0);
      close(fd);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pmc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L13
L17:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <pc@plt>
L13:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void pmc(const char *s) {
  while (*s)
    pc(*s++);
}
```
user: Translate the below assembly code into C programming language:

```asm
<pmc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L7
L4:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <pc@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L4
L7:
  pop    %rbx
  ret
```

assistant: ```c
void pmc(const char *s) {
  while (*s)
    pc(*s++);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pmc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L0
L8:
  call   <pc@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L8
L0:
  pop    %rbx
  ret
```

assistant: ```c
void pmc(const char *s) {
  while (*s)
    pc(*s++);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<pmc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L7
L2:
  call   <pc@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L2
L7:
  pop    %rbx
  ret
```

assistant: ```c
void pmc(const char *s) {
  while (*s)
    pc(*s++);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<htoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L84
L115:
  cmpb   $0x78,-0x5(%rbp)
  jne    L50
  jmp    L84
L50:
  cmpb   $0x2f,-0x5(%rbp)
  jle    L15
  cmpb   $0x39,-0x5(%rbp)
  jg     L15
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movsbl -0x5(%rbp),%eax
  sub    $0x30,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L84
L15:
  cmpb   $0x60,-0x5(%rbp)
  jle    L44
  cmpb   $0x66,-0x5(%rbp)
  jg     L44
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movsbl -0x5(%rbp),%eax
  sub    $0x57,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L84
L44:
  cmpb   $0x40,-0x5(%rbp)
  jle    L73
  cmpb   $0x46,-0x5(%rbp)
  jg     L73
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movsbl -0x5(%rbp),%eax
  sub    $0x37,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
L84:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  cmpb   $0x0,-0x5(%rbp)
  jne    L115
L73:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int htoi(s)
char *s;
{
  int i = 0;
  char c;
  while ((c = *s++) != '\0') {
    if (c == 'x')
      continue;
    if ('0' <= c && c <= '9')
      i = (i * 16) + (c - '0');
    else if ('a' <= c && c <= 'f')
      i = (i * 16) + (c - 'a' + 10);
    else if ('A' <= c && c <= 'F')
      i = (i * 16) + (c - 'A' + 10);
    else
      break;
  }
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<htoi>:
  endbr64
  lea    0x1(%rdi),%rcx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L27
  mov    $0x0,%eax
  jmp    L47
L84:
  lea    -0x61(%rdx),%esi
  cmp    $0x5,%sil
  ja     L88
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x57(%rax,%rdx,1),%eax
L22:
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  je     L0
L47:
  cmp    $0x78,%dl
  je     L22
  mov    %edx,%edi
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%sil
  ja     L84
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x30(%rax,%rdx,1),%eax
  jmp    L22
L88:
  sub    $0x41,%edi
  cmp    $0x5,%dil
  ja     L38
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x37(%rax,%rdx,1),%eax
  jmp    L22
L0:
  ret
L27:
  mov    $0x0,%eax
L38:
  ret
```

assistant: ```c
int htoi(s)
char *s;
{
  int i = 0;
  char c;
  while ((c = *s++) != '\0') {
    if (c == 'x')
      continue;
    if ('0' <= c && c <= '9')
      i = (i * 16) + (c - '0');
    else if ('a' <= c && c <= 'f')
      i = (i * 16) + (c - 'a' + 10);
    else if ('A' <= c && c <= 'F')
      i = (i * 16) + (c - 'A' + 10);
    else
      break;
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<htoi>:
  endbr64
  movzbl (%rdi),%edx
  lea    0x1(%rdi),%rcx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L44
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L30:
  shl    $0x4,%eax
  movsbl %sil,%esi
  add    %esi,%eax
L62:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  je     L8
L44:
  cmp    $0x78,%dl
  je     L62
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%sil
  jbe    L30
  lea    -0x61(%rdx),%esi
  cmp    $0x5,%sil
  ja     L86
  sub    $0x57,%edx
  shl    $0x4,%eax
  add    $0x1,%rcx
  movsbl %dl,%edx
  add    %edx,%eax
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L44
L8:
  ret
  nopl   0x0(%rax,%rax,1)
L86:
  lea    -0x41(%rdx),%esi
  cmp    $0x5,%sil
  ja     L28
  sub    $0x37,%edx
  shl    $0x4,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  jmp    L62
  cs nopw 0x0(%rax,%rax,1)
L28:
  ret
```

assistant: ```c
int htoi(s)
char *s;
{
  int i = 0;
  char c;
  while ((c = *s++) != '\0') {
    if (c == 'x')
      continue;
    if ('0' <= c && c <= '9')
      i = (i * 16) + (c - '0');
    else if ('a' <= c && c <= 'f')
      i = (i * 16) + (c - 'a' + 10);
    else if ('A' <= c && c <= 'F')
      i = (i * 16) + (c - 'A' + 10);
    else
      break;
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<htoi>:
  endbr64
  movzbl (%rdi),%edx
  lea    0x1(%rdi),%rcx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L86
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L75:
  shl    $0x4,%eax
  movsbl %sil,%esi
  add    %esi,%eax
L32:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  je     L14
L86:
  cmp    $0x78,%dl
  je     L32
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%sil
  jbe    L75
  lea    -0x61(%rdx),%esi
  cmp    $0x5,%sil
  ja     L17
  sub    $0x57,%edx
  shl    $0x4,%eax
  add    $0x1,%rcx
  movsbl %dl,%edx
  add    %edx,%eax
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L86
L14:
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  lea    -0x41(%rdx),%esi
  cmp    $0x5,%sil
  ja     L40
  sub    $0x37,%edx
  shl    $0x4,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  jmp    L32
  cs nopw 0x0(%rax,%rax,1)
L40:
  ret
```

assistant: ```c
int htoi(s)
char *s;
{
  int i = 0;
  char c;
  while ((c = *s++) != '\0') {
    if (c == 'x')
      continue;
    if ('0' <= c && c <= '9')
      i = (i * 16) + (c - '0');
    else if ('a' <= c && c <= 'f')
      i = (i * 16) + (c - 'a' + 10);
    else if ('A' <= c && c <= 'F')
      i = (i * 16) + (c - 'A' + 10);
    else
      break;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x7a120,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <countPrimes@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_guessed: string

tool: D2: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 500000;
  printf("%d\n", countPrimes(n));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7a120,%edi
  call   <countPrimes@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type_guessed: string

tool: D6: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 500000;
  printf("%d\n", countPrimes(n));
  return 0;
}
```