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

```asm
<logger_line_log_level>:
  endbr64
  test   %edi,%edi
  jle    L42
  lea    -0x1(%rdi),%eax
  push   %r12
  lea    0x8(%rsi,%rax,8),%r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L46
L35:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L0
L46:
  mov    (%rbx),%rdi
  mov    $0x3,%edx
  mov    %rbp,%rsi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L35
  mov    (%rbx),%rax
  movsbl 0x3(%rax),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L35
  mov    (%rbx),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  movsbl 0x3(%rax),%eax
  sub    $0x30,%eax
  ret
  xchg   %ax,%ax
L0:
  pop    %rbx
  mov    $0x9,%eax
  pop    %rbp
  pop    %r12
  ret
L42:
  mov    $0x9,%eax
  ret
```

tool_calls: name: parse_data
arguments: label: D6
data_type: string

tool: D6: log

assistant: ```c
int logger_line_log_level(int tags_count, const char **tags) {
  int i;
  for (i = 0; i < tags_count; i++) {
    if (strncmp(tags[i], "log", 3) == 0) {
      if (isdigit(tags[i][3])) {
        return (tags[i][3] - '0');
      }
    }
  }
  return 9;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<eeprom_update_dword>:
  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   <eeprom_read_dword@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  je     L1
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <eeprom_write_dword@plt>
L1:
  nop
  leave
  ret
```

assistant: ```c
void eeprom_update_dword(uint32_t *addr, uint32_t value) {
  uint32_t orig = eeprom_read_dword(addr);
  if (orig != value) {
    eeprom_write_dword(addr, value);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<eeprom_update_dword>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <eeprom_read_dword@plt>
  cmp    %rbp,%rax
  jne    L6
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <eeprom_write_dword@plt>
  jmp    L14
```

assistant: ```c
void eeprom_update_dword(uint32_t *addr, uint32_t value) {
  uint32_t orig = eeprom_read_dword(addr);
  if (orig != value) {
    eeprom_write_dword(addr, value);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<eeprom_update_dword>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <eeprom_read_dword@plt>
  cmp    %r12,%rax
  jne    L3
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <eeprom_write_dword@plt>
```

assistant: ```c
void eeprom_update_dword(uint32_t *addr, uint32_t value) {
  uint32_t orig = eeprom_read_dword(addr);
  if (orig != value) {
    eeprom_write_dword(addr, value);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<eeprom_update_dword>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <eeprom_read_dword@plt>
  cmp    %r12,%rax
  jne    L0
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <eeprom_write_dword@plt>
```

assistant: ```c
void eeprom_update_dword(uint32_t *addr, uint32_t value) {
  uint32_t orig = eeprom_read_dword(addr);
  if (orig != value) {
    eeprom_write_dword(addr, value);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<string_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L18
L13:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L18:
  leave
  ret
```

assistant: ```c
char *string_copy(char *s) {
  char *new;
  if (s == NULL)
    return NULL;
  new = (char *)malloc(strlen(s) + 1);
  strcpy(new, s);
  return new;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_copy>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L10
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L15:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    %rdi,%rbp
  jmp    L15
```

assistant: ```c
char *string_copy(char *s) {
  char *new;
  if (s == NULL)
    return NULL;
  new = (char *)malloc(strlen(s) + 1);
  strcpy(new, s);
  return new;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_copy>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L3
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string_copy(char *s) {
  char *new;
  if (s == NULL)
    return NULL;
  new = (char *)malloc(strlen(s) + 1);
  strcpy(new, s);
  return new;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<string_copy>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L9
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string_copy(char *s) {
  char *new;
  if (s == NULL)
    return NULL;
  new = (char *)malloc(strlen(s) + 1);
  strcpy(new, s);
  return new;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<UpnpContentDirectory_GetSearchCapabilities>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  lea    D11(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <UpnpResponse_ContentDirectory_GetSearchCapabilities@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D18
type_guessed: string
name: parse_data
arguments: tag: D11
type_guessed: string

tool: D18: Invoke: UpnpContentDirectory_GetSearchCapabilities();


tool: D11: Sample String

assistant: ```c
void UpnpContentDirectory_GetSearchCapabilities(void *upnptoken) {
  printf("Invoke: UpnpContentDirectory_GetSearchCapabilities();\r\n");
  UpnpResponse_ContentDirectory_GetSearchCapabilities(upnptoken, "Sample String");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<UpnpContentDirectory_GetSearchCapabilities>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <UpnpResponse_ContentDirectory_GetSearchCapabilities@plt>
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type: string
name: unpack_data
arguments: label: D0
type: string

tool: D7: Invoke: UpnpContentDirectory_GetSearchCapabilities();


tool: D0: Sample String

assistant: ```c
void UpnpContentDirectory_GetSearchCapabilities(void *upnptoken) {
  printf("Invoke: UpnpContentDirectory_GetSearchCapabilities();\r\n");
  UpnpResponse_ContentDirectory_GetSearchCapabilities(upnptoken, "Sample String");
}
```
user: Translate the below assembly code into C programming language:

```asm
<UpnpContentDirectory_GetSearchCapabilities>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    D0(%rip),%rsi
  pop    %rbp
  jmp    1060 <UpnpResponse_ContentDirectory_GetSearchCapabilities@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string

tool: D12: Invoke: UpnpContentDirectory_GetSearchCapabilities();


tool: D0: Sample String

assistant: ```c
void UpnpContentDirectory_GetSearchCapabilities(void *upnptoken) {
  printf("Invoke: UpnpContentDirectory_GetSearchCapabilities();\r\n");
  UpnpResponse_ContentDirectory_GetSearchCapabilities(upnptoken, "Sample String");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<UpnpContentDirectory_GetSearchCapabilities>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    D2(%rip),%rsi
  pop    %rbp
  jmp    1060 <UpnpResponse_ContentDirectory_GetSearchCapabilities@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
type_infered: string
name: retrieve_data
arguments: tag: D2
type_infered: string

tool: D15: Invoke: UpnpContentDirectory_GetSearchCapabilities();


tool: D2: Sample String

assistant: ```c
void UpnpContentDirectory_GetSearchCapabilities(void *upnptoken) {
  printf("Invoke: UpnpContentDirectory_GetSearchCapabilities();\r\n");
  UpnpResponse_ContentDirectory_GetSearchCapabilities(upnptoken, "Sample String");
}
```
user: Translate the below assembly code into C programming language:

```asm
<put_x>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  cltd
  shr    $0x1c,%edx
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  mov    %eax,%edi
  call   <toHexChar@plt>
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x1c(%rbp),%eax
  lea    0xf(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x4,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L1
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *put_x(char *p, uint64_t v) {
  char *base = p;
  do {
    *p++ = toHexChar(v % 16);
  } while ((v /= 16) != 0);
  reverse(base, p);
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<put_x>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %rdi,%rbp
L1:
  mov    %ebx,%eax
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rbx,%rax,1),%edi
  and    $0xf,%edi
  sub    %eax,%edi
  call   <toHexChar@plt>
  add    $0x1,%rbp
  mov    %al,-0x1(%rbp)
  lea    0xf(%rbx),%eax
  test   %ebx,%ebx
  cmovns %ebx,%eax
  sar    $0x4,%eax
  mov    %eax,%ebx
  jne    L1
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <reverse@plt>
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *put_x(char *p, uint64_t v) {
  char *base = p;
  do {
    *p++ = toHexChar(v % 16);
  } while ((v /= 16) != 0);
  reverse(base, p);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<put_x>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
L5:
  mov    %ebx,%eax
  add    $0x1,%r12
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rbx,%rax,1),%edi
  and    $0xf,%edi
  sub    %eax,%edi
  call   <toHexChar@plt>
  test   %ebx,%ebx
  mov    %al,-0x1(%r12)
  lea    0xf(%rbx),%eax
  cmovs  %eax,%ebx
  sar    $0x4,%ebx
  jne    L5
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <reverse@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *put_x(char *p, uint64_t v) {
  char *base = p;
  do {
    *p++ = toHexChar(v % 16);
  } while ((v /= 16) != 0);
  reverse(base, p);
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<put_x>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
L3:
  mov    %ebx,%eax
  add    $0x1,%r12
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rbx,%rax,1),%edi
  and    $0xf,%edi
  sub    %eax,%edi
  call   <toHexChar@plt>
  test   %ebx,%ebx
  mov    %al,-0x1(%r12)
  lea    0xf(%rbx),%eax
  cmovs  %eax,%ebx
  sar    $0x4,%ebx
  jne    L3
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <reverse@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *put_x(char *p, uint64_t v) {
  char *base = p;
  do {
    *p++ = toHexChar(v % 16);
  } while ((v /= 16) != 0);
  reverse(base, p);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isOddNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L16
  mov    $0x0,%eax
  jmp    L17
L16:
  mov    $0x1,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int isOddNumber(int value) {
  if (value % 2 == 0) {
    return 0;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isOddNumber>:
  endbr64
  mov    %edi,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int isOddNumber(int value) {
  if (value % 2 == 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
<isOddNumber>:
  endbr64
  mov    %edi,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int isOddNumber(int value) {
  if (value % 2 == 0) {
    return 0;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isOddNumber>:
  endbr64
  mov    %edi,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int isOddNumber(int value) {
  if (value % 2 == 0) {
    return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<removeDuplicates>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L8:
  cmpl   $0x1,-0xc(%rbp)
  jg     L15
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  jmp    L24
L15:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L24
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L24:
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%eax
  leave
  ret
```

assistant: ```c
int removeDuplicates(int *nums, int numsSize) {
  int *ans = (int *)malloc(sizeof(int) * numsSize);
  int dest = 0;
  for (int i = 0; i < numsSize; i++) {
    if (i < 2) {
      *(ans + dest++) = nums[i];
    } else {
      if (*(ans + dest - 2) != nums[i]) {
        *(ans + dest++) = nums[i];
      }
    }
  }
  memcpy(nums, ans, sizeof(int) * dest);
  return dest;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<removeDuplicates>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  test   %ebx,%ebx
  jle    L43
  mov    %ebx,%edi
  mov    $0x0,%edx
  mov    $0x0,%ebx
  jmp    L16
L12:
  mov    0x0(%rbp,%rdx,4),%ecx
  movslq %ebx,%rax
  mov    %ecx,(%rsi,%rax,4)
  lea    0x1(%rbx),%ebx
L20:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L42
L16:
  cmp    $0x1,%edx
  jle    L12
  movslq %ebx,%rax
  mov    0x0(%rbp,%rdx,4),%ecx
  cmp    %ecx,-0x8(%rsi,%rax,4)
  je     L20
  add    $0x1,%ebx
  mov    %ecx,(%rsi,%rax,4)
  jmp    L20
L43:
  mov    $0x0,%ebx
L42:
  lea    0x0(,%rbx,4),%edx
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int removeDuplicates(int *nums, int numsSize) {
  int *ans = (int *)malloc(sizeof(int) * numsSize);
  int dest = 0;
  for (int i = 0; i < numsSize; i++) {
    if (i < 2) {
      *(ans + dest++) = nums[i];
    } else {
      if (*(ans + dest - 2) != nums[i]) {
        *(ans + dest++) = nums[i];
      }
    }
  }
  memcpy(nums, ans, sizeof(int) * dest);
  return dest;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<removeDuplicates>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  test   %ebx,%ebx
  jle    L30
  xor    %edx,%edx
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L25:
  movslq %r12d,%rcx
  mov    0x0(%rbp,%rdx,4),%edi
  lea    0x0(,%rcx,4),%rax
  cmp    $0x1,%rdx
  jbe    L24
  cmp    %edi,-0x8(%rsi,%rax,1)
  je     L34
L24:
  mov    %edi,(%rsi,%rcx,4)
  add    $0x1,%r12d
L34:
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jne    L25
  lea    0x0(,%r12,4),%edx
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  xor    %r12d,%r12d
  mov    %rbp,%rdi
  xor    %edx,%edx
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int removeDuplicates(int *nums, int numsSize) {
  int *ans = (int *)malloc(sizeof(int) * numsSize);
  int dest = 0;
  for (int i = 0; i < numsSize; i++) {
    if (i < 2) {
      *(ans + dest++) = nums[i];
    } else {
      if (*(ans + dest - 2) != nums[i]) {
        *(ans + dest++) = nums[i];
      }
    }
  }
  memcpy(nums, ans, sizeof(int) * dest);
  return dest;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<removeDuplicates>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%rsi,4),%edi
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  mov    %rax,%rsi
  test   %ebx,%ebx
  jle    L26
  lea    -0x1(%rbx),%ecx
  mov    $0x1,%edx
  mov    %rbp,%r8
  test   %ecx,%ecx
  cmovle %ecx,%edx
  xor    %edi,%edi
  sub    %rax,%r8
L22:
  lea    0x1(%rdi),%r12d
  mov    (%r8,%rax,1),%edi
  add    $0x4,%rax
  mov    %edi,-0x4(%rax)
  mov    $0x1,%edi
  cmp    %r12d,%edx
  jge    L22
  cmp    %r12d,%ebx
  jle    L5
  movslq %r12d,%rdx
  sub    %r12d,%ecx
  add    %rdx,%rcx
  lea    0x0(%rbp,%rdx,4),%rax
  lea    0x4(%rbp,%rcx,4),%rdi
  jmp    L39
  nopl   0x0(%rax)
L21:
  movslq %r12d,%rdx
L39:
  mov    (%rax),%ecx
  cmp    %ecx,-0x8(%rsi,%rdx,4)
  je     L42
  mov    %ecx,(%rsi,%rdx,4)
  add    $0x1,%r12d
L42:
  add    $0x4,%rax
  cmp    %rax,%rdi
  jne    L21
L5:
  lea    0x0(,%r12,4),%edx
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  xor    %r12d,%r12d
  mov    %rbp,%rdi
  xor    %edx,%edx
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int removeDuplicates(int *nums, int numsSize) {
  int *ans = (int *)malloc(sizeof(int) * numsSize);
  int dest = 0;
  for (int i = 0; i < numsSize; i++) {
    if (i < 2) {
      *(ans + dest++) = nums[i];
    } else {
      if (*(ans + dest - 2) != nums[i]) {
        *(ans + dest++) = nums[i];
      }
    }
  }
  memcpy(nums, ans, sizeof(int) * dest);
  return dest;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<iyogrt_set_debug_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <yogrt_set_debug@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int iyogrt_set_debug_(int *val) {
  yogrt_set_debug(*val);
  return *val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<iyogrt_set_debug_>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%edi
  call   <yogrt_set_debug@plt>
  mov    (%rbx),%eax
  pop    %rbx
  ret
```

assistant: ```c
int iyogrt_set_debug_(int *val) {
  yogrt_set_debug(*val);
  return *val;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<iyogrt_set_debug_>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%edi
  call   <yogrt_set_debug@plt>
  mov    (%rbx),%eax
  pop    %rbx
  ret
```

assistant: ```c
int iyogrt_set_debug_(int *val) {
  yogrt_set_debug(*val);
  return *val;
}
```
user: What is the c source code of the assembly code below:

```asm
<iyogrt_set_debug_>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%edi
  call   <yogrt_set_debug@plt>
  mov    (%rbx),%eax
  pop    %rbx
  ret
```

assistant: ```c
int iyogrt_set_debug_(int *val) {
  yogrt_set_debug(*val);
  return *val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,%eax
  mov    %al,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L21:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L17
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x24(%rbp)
  jne    L21
L17:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *my_strcpy(char *dest, const char *src, char c) {
  int i;
  i = 0;
  while (src[i] && (src[i] != c)) {
    dest[i] = src[i];
    i++;
  }
  dest[i] = '\0';
  return (dest);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<my_strcpy>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%edi
  movzbl (%rsi),%ecx
  cmp    %dl,%cl
  je     L30
  test   %cl,%cl
  je     L30
  mov    $0x0,%edx
L41:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rsi,%rdx,1),%ecx
  cmp    %dil,%cl
  je     L40
  test   %cl,%cl
  jne    L41
L40:
  movb   $0x0,(%rax,%rdx,1)
  ret
L30:
  mov    $0x0,%edx
  jmp    L40
```

assistant: ```c
char *my_strcpy(char *dest, const char *src, char c) {
  int i;
  i = 0;
  while (src[i] && (src[i] != c)) {
    dest[i] = src[i];
    i++;
  }
  dest[i] = '\0';
  return (dest);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strcpy>:
  endbr64
  movzbl (%rsi),%ecx
  mov    %rdi,%rax
  mov    %edx,%edi
  cmp    %dl,%cl
  je     L12
  test   %cl,%cl
  je     L12
  xor    %edx,%edx
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L15:
  test   %cl,%cl
  je     L6
L38:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rsi,%rdx,1),%ecx
  cmp    %dil,%cl
  jne    L15
L6:
  add    %rax,%rdx
  movb   $0x0,(%rdx)
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rax,%rdx
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
char *my_strcpy(char *dest, const char *src, char c) {
  int i;
  i = 0;
  while (src[i] && (src[i] != c)) {
    dest[i] = src[i];
    i++;
  }
  dest[i] = '\0';
  return (dest);
}
```