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

```asm
<xor_bufs_and_store>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltd
  shr    $0x1c,%edx
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x34(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L45
  mov    -0x34(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  jmp    L33
L45:
  mov    $0x0,%eax
L33:
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  shr    $0x3,%rax
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L52
L26:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  xor    %rcx,%rdx
  mov    %rdx,(%rax)
  addl   $0x1,-0x20(%rbp)
L52:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L26
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L41
L38:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%ecx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x30(%rbp),%rax
  add    %rsi,%rax
  xor    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x20(%rbp)
L41:
  mov    -0x20(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L38
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void xor_bufs_and_store(char *buf1, char *buf2, int blocksize) {
  int residual_bytes = (blocksize % 16);
  int fast_blocksize = blocksize > residual_bytes ? (blocksize - residual_bytes) : 0;
  int fast_int_blocksize = fast_blocksize / sizeof(unsigned long);
  int i;
  unsigned long *_buf1 = (unsigned long *)buf1;
  unsigned long *_buf2 = (unsigned long *)buf2;
  for (i = 0; i < fast_int_blocksize; i++) {
    _buf2[i] = _buf1[i] ^ _buf2[i];
  }
  for (i = fast_blocksize; i < blocksize; i++) {
    buf2[i] ^= buf1[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xor_bufs_and_store>:
  endbr64
  mov    %rsi,%rcx
  mov    %edx,%r8d
  mov    %edx,%eax
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rdx,%rax,1),%edx
  and    $0xf,%edx
  sub    %eax,%edx
  mov    $0x0,%eax
  cmp    %edx,%r8d
  jle    L10
  mov    %r8d,%eax
  sub    %edx,%eax
  movslq %eax,%rdx
  shr    $0x3,%rdx
  test   %edx,%edx
  jle    L10
  lea    -0x1(%rdx),%r9d
  mov    $0x0,%edx
L14:
  mov    (%rdi,%rdx,8),%rsi
  xor    %rsi,(%rcx,%rdx,8)
  mov    %rdx,%rsi
  add    $0x1,%rdx
  cmp    %r9,%rsi
  jne    L14
L10:
  cmp    %r8d,%eax
  jge    L7
  cltq
L27:
  movzbl (%rdi,%rax,1),%edx
  xor    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L27
L7:
  ret
```

assistant: ```c
void xor_bufs_and_store(char *buf1, char *buf2, int blocksize) {
  int residual_bytes = (blocksize % 16);
  int fast_blocksize = blocksize > residual_bytes ? (blocksize - residual_bytes) : 0;
  int fast_int_blocksize = fast_blocksize / sizeof(unsigned long);
  int i;
  unsigned long *_buf1 = (unsigned long *)buf1;
  unsigned long *_buf2 = (unsigned long *)buf2;
  for (i = 0; i < fast_int_blocksize; i++) {
    _buf2[i] = _buf1[i] ^ _buf2[i];
  }
  for (i = fast_blocksize; i < blocksize; i++) {
    buf2[i] ^= buf1[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<xor_bufs_and_store>:
  endbr64
  mov    %edx,%eax
  mov    %edx,%r8d
  mov    %rsi,%rcx
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rdx,%rax,1),%edx
  and    $0xf,%edx
  sub    %eax,%edx
  xor    %eax,%eax
  cmp    %edx,%r8d
  jle    L5
  mov    %r8d,%eax
  sub    %edx,%eax
  movslq %eax,%rdx
  shr    $0x3,%rdx
  test   %edx,%edx
  jle    L5
  lea    -0x1(%rdx),%r9d
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L22:
  mov    (%rdi,%rdx,8),%rsi
  xor    %rsi,(%rcx,%rdx,8)
  mov    %rdx,%rsi
  add    $0x1,%rdx
  cmp    %rsi,%r9
  jne    L22
L5:
  cmp    %r8d,%eax
  jge    L43
  cltq
  nopl   0x0(%rax,%rax,1)
L41:
  movzbl (%rdi,%rax,1),%edx
  xor    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L41
L43:
  ret
```

assistant: ```c
void xor_bufs_and_store(char *buf1, char *buf2, int blocksize) {
  int residual_bytes = (blocksize % 16);
  int fast_blocksize = blocksize > residual_bytes ? (blocksize - residual_bytes) : 0;
  int fast_int_blocksize = fast_blocksize / sizeof(unsigned long);
  int i;
  unsigned long *_buf1 = (unsigned long *)buf1;
  unsigned long *_buf2 = (unsigned long *)buf2;
  for (i = 0; i < fast_int_blocksize; i++) {
    _buf2[i] = _buf1[i] ^ _buf2[i];
  }
  for (i = fast_blocksize; i < blocksize; i++) {
    buf2[i] ^= buf1[i];
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fun_call_align_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  movsbl -0x8(%rbp),%edx
  movsbl -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0x4(%rbp),%edx
  movsbl -0x8(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D6
type_infered: string

tool: D6: a = %d, b = %d


assistant: ```c
int fun_call_align_2(char a, char b) {
  printf("a = %d, b = %d\n", a, b);
  return a + b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fun_call_align_2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl %sil,%ebx
  movsbl %dil,%ebp
  mov    %ebx,%edx
  mov    %ebp,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D9: a = %d, b = %d


assistant: ```c
int fun_call_align_2(char a, char b) {
  printf("a = %d, b = %d\n", a, b);
  return a + b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fun_call_align_2>:
  endbr64
  push   %rbp
  movsbl %dil,%ebp
  lea    D1(%rip),%rdi
  push   %rbx
  movsbl %sil,%ebx
  mov    %ebp,%esi
  mov    %ebx,%edx
  sub    $0x8,%rsp
  call   <printf@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
type: string

tool: D1: a = %d, b = %d


assistant: ```c
int fun_call_align_2(char a, char b) {
  printf("a = %d, b = %d\n", a, b);
  return a + b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fun_call_align_2>:
  endbr64
  push   %rbp
  movsbl %dil,%ebp
  lea    D9(%rip),%rdi
  push   %rbx
  movsbl %sil,%ebx
  mov    %ebp,%esi
  mov    %ebx,%edx
  sub    $0x8,%rsp
  call   <printf@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type: string

tool: D9: a = %d, b = %d


assistant: ```c
int fun_call_align_2(char a, char b) {
  printf("a = %d, b = %d\n", a, b);
  return a + b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  xor    $0x29be1cdb,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  or     $0x762cc408,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  char copy13;
  char copy14;
  char copy16;
  {
    state[0UL] = input[0UL] ^ 700325083U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] != local1) {
        if (state[0UL] == local1) {
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
          copy12 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy13;
          copy14 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy14;
        }
      } else {
        copy16 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy16;
        copy16 = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy16;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] | 1982645256U;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  xor    $0x29be1cdb,%eax
  or     $0x762cc408,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  char copy13;
  char copy14;
  char copy16;
  {
    state[0UL] = input[0UL] ^ 700325083U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] != local1) {
        if (state[0UL] == local1) {
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
          copy12 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy13;
          copy14 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy14;
        }
      } else {
        copy16 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy16;
        copy16 = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy16;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] | 1982645256U;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  xor    $0x29be1cdb,%eax
  or     $0x762cc408,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  char copy13;
  char copy14;
  char copy16;
  {
    state[0UL] = input[0UL] ^ 700325083U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] != local1) {
        if (state[0UL] == local1) {
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
          copy12 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy13;
          copy14 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy14;
        }
      } else {
        copy16 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy16;
        copy16 = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy16;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] | 1982645256U;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  xor    $0x29be1cdb,%eax
  or     $0x762cc408,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  char copy13;
  char copy14;
  char copy16;
  {
    state[0UL] = input[0UL] ^ 700325083U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] != local1) {
        if (state[0UL] == local1) {
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
          copy12 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy13;
          copy14 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy14;
        }
      } else {
        copy16 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy16;
        copy16 = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy16;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] | 1982645256U;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<regexR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L12
  mov    -0x8(%rbp),%rax
  jmp    L18
L12:
  lea    D2(%rip),%rax
L18:
  pop    %rbp
  ret
```

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

tool: D2: .*

assistant: ```c
char *regexR(char *regex) {
  if (regex != NULL) {
    return regex;
  }
  return ".*";
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<regexR>:
  endbr64
  test   %rdi,%rdi
  lea    D1(%rip),%rax
  cmovne %rdi,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
data_type: string

tool: D1: .*

assistant: ```c
char *regexR(char *regex) {
  if (regex != NULL) {
    return regex;
  }
  return ".*";
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<regexR>:
  endbr64
  test   %rdi,%rdi
  lea    D2(%rip),%rax
  cmovne %rdi,%rax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: .*

assistant: ```c
char *regexR(char *regex) {
  if (regex != NULL) {
    return regex;
  }
  return ".*";
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<regexR>:
  endbr64
  test   %rdi,%rdi
  lea    D6(%rip),%rax
  cmovne %rdi,%rax
  ret
```

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

tool: D6: .*

assistant: ```c
char *regexR(char *regex) {
  if (regex != NULL) {
    return regex;
  }
  return ".*";
}
```
user: Translate the below assembly code into C programming language:

```asm
<inf_loop_cb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
L8:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L8
```

assistant: ```c
void inf_loop_cb(void *env) {
  (void)env;
  for (;;) {
    sleep(1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<inf_loop_cb>:
  endbr64
  sub    $0x8,%rsp
L2:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L2
```

assistant: ```c
void inf_loop_cb(void *env) {
  (void)env;
  for (;;) {
    sleep(1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inf_loop_cb>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L4
```

assistant: ```c
void inf_loop_cb(void *env) {
  (void)env;
  for (;;) {
    sleep(1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<inf_loop_cb>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L3
```

assistant: ```c
void inf_loop_cb(void *env) {
  (void)env;
  for (;;) {
    sleep(1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ansimove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    $0x0,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    $0x5b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ansiparm@plt>
  mov    $0x3b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ansiparm@plt>
  mov    $0x48,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int ansimove(row, col)
int row, col;
{
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ansiparm(row + 1);
  ttputc(';');
  ansiparm(col + 1);
  ttputc('H');
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ansimove>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    $0x0,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    $0x5b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  lea    0x1(%rbp),%edi
  mov    $0x0,%eax
  call   <ansiparm@plt>
  mov    $0x3b,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  lea    0x1(%rbx),%edi
  mov    $0x0,%eax
  call   <ansiparm@plt>
  mov    $0x48,%edi
  mov    $0x0,%eax
  call   <ttputc@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int ansimove(row, col)
int row, col;
{
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ansiparm(row + 1);
  ttputc(';');
  ansiparm(col + 1);
  ttputc('H');
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ansimove>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <ttflush@plt>
  mov    $0x1b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  mov    $0x5b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  lea    0x1(%rbp),%edi
  xor    %eax,%eax
  call   <ansiparm@plt>
  mov    $0x3b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  lea    0x1(%rbx),%edi
  xor    %eax,%eax
  call   <ansiparm@plt>
  mov    $0x48,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int ansimove(row, col)
int row, col;
{
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ansiparm(row + 1);
  ttputc(';');
  ansiparm(col + 1);
  ttputc('H');
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ansimove>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <ttflush@plt>
  mov    $0x1b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  mov    $0x5b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  lea    0x1(%rbp),%edi
  xor    %eax,%eax
  call   <ansiparm@plt>
  mov    $0x3b,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  lea    0x1(%rbx),%edi
  xor    %eax,%eax
  call   <ansiparm@plt>
  mov    $0x48,%edi
  xor    %eax,%eax
  call   <ttputc@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int ansimove(row, col)
int row, col;
{
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ansiparm(row + 1);
  ttputc(';');
  ansiparm(col + 1);
  ttputc('H');
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0xc(%rbp)
  mov    -0x18(%rbp),%rsi
  lea    -0xc(%rbp),%rdx
  mov    (%rdx),%eax
  mov    $0x0,%ecx
  lock cmpxchg %ecx,(%rsi)
  mov    %eax,%ecx
  sete   %al
  test   %al,%al
  jne    L0
  mov    %ecx,(%rdx)
L0:
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int foo(int *a) {
  int x = 3;
  return __atomic_compare_exchange_n(a, &x, 0, 1, 2, 2);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo>:
  endbr64
  mov    $0x3,%eax
  mov    $0x0,%edx
  lock cmpxchg %edx,(%rdi)
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int foo(int *a) {
  int x = 3;
  return __atomic_compare_exchange_n(a, &x, 0, 1, 2, 2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  mov    $0x3,%eax
  xor    %edx,%edx
  lock cmpxchg %edx,(%rdi)
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int foo(int *a) {
  int x = 3;
  return __atomic_compare_exchange_n(a, &x, 0, 1, 2, 2);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  mov    $0x3,%eax
  xor    %edx,%edx
  lock cmpxchg %edx,(%rdi)
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int foo(int *a) {
  int x = 3;
  return __atomic_compare_exchange_n(a, &x, 0, 1, 2, 2);
}
```