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

```asm
<serialWrite>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %sil,0x7(%rsp)
  lea    0x7(%rsp),%rsi
  call   <write@plt>
  mov    $0xa,%edi
  call   <waitms@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void serialWrite(int fd, unsigned char _byte) {
  unsigned char c[1];
  c[0] = _byte;
  write(fd, c, 1);
  waitms(10);
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<serialWrite>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %sil,0x7(%rsp)
  lea    0x7(%rsp),%rsi
  call   <write@plt>
  mov    $0xa,%edi
  call   <waitms@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void serialWrite(int fd, unsigned char _byte) {
  unsigned char c[1];
  c[0] = _byte;
  write(fd, c, 1);
  waitms(10);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<zeroStringArr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void zeroStringArr(char **stringArr, int numElements) {
  int i = 0;
  for (i; i < numElements; i++) {
    *(stringArr + i) = (char *)0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zeroStringArr>:
  endbr64
  test   %esi,%esi
  jle    L10
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
L2:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L2
L10:
  ret
```

assistant: ```c
void zeroStringArr(char **stringArr, int numElements) {
  int i = 0;
  for (i; i < numElements; i++) {
    *(stringArr + i) = (char *)0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<zeroStringArr>:
  endbr64
  test   %esi,%esi
  jle    L7
  movslq %esi,%rsi
  lea    0x0(,%rsi,8),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
void zeroStringArr(char **stringArr, int numElements) {
  int i = 0;
  for (i; i < numElements; i++) {
    *(stringArr + i) = (char *)0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<zeroStringArr>:
  endbr64
  test   %esi,%esi
  jle    L4
  movslq %esi,%rsi
  lea    0x0(,%rsi,8),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  ret
```

assistant: ```c
void zeroStringArr(char **stringArr, int numElements) {
  int i = 0;
  for (i; i < numElements; i++) {
    *(stringArr + i) = (char *)0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_s24le>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x10,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int32_t read_s24le(const char *buf) {
  const unsigned char *b = (const unsigned char *)buf;
  return b[0] | (b[1] << 8) | (((const signed char *)buf)[2] << 16);
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_s24le>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
int32_t read_s24le(const char *buf) {
  const unsigned char *b = (const unsigned char *)buf;
  return b[0] | (b[1] << 8) | (((const signed char *)buf)[2] << 16);
}
```
user: What is the c source code of the assembly code below?

```asm
<read_s24le>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
int32_t read_s24le(const char *buf) {
  const unsigned char *b = (const unsigned char *)buf;
  return b[0] | (b[1] << 8) | (((const signed char *)buf)[2] << 16);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_s24le>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
int32_t read_s24le(const char *buf) {
  const unsigned char *b = (const unsigned char *)buf;
  return b[0] | (b[1] << 8) | (((const signed char *)buf)[2] << 16);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<char_array_i>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L28:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L37
  mov    -0x4(%rbp),%eax
  jmp    L2
L37:
  addl   $0x1,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  cltq
  cmp    %rax,-0x20(%rbp)
  ja     L28
  mov    $0xffffffff,%eax
L2:
  leave
  ret
```

assistant: ```c
int char_array_i(const char *array[], size_t len, const char *s) {
  int i;
  for (i = 0; i < len; i++)
    if (!strcmp(s, array[i]))
      return i;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<char_array_i>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L23
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    $0x0,%ebx
L17:
  mov    %ebx,%r14d
  mov    (%r12,%rbx,8),%rsi
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L13
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L17
  mov    $0xffffffff,%r14d
L13:
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    $0xffffffff,%r14d
  jmp    L13
```

assistant: ```c
int char_array_i(const char *array[], size_t len, const char *s) {
  int i;
  for (i = 0; i < len; i++)
    if (!strcmp(s, array[i]))
      return i;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<char_array_i>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L26
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rdx,%r12
  xor    %ebx,%ebx
  jmp    L4
  xchg   %ax,%ax
L18:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L26
L4:
  mov    (%r14,%rbx,8),%rsi
  mov    %r12,%rdi
  mov    %ebx,%r13d
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L18
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  mov    $0xffffffff,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int char_array_i(const char *array[], size_t len, const char *s) {
  int i;
  for (i = 0; i < len; i++)
    if (!strcmp(s, array[i]))
      return i;
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<char_array_i>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L36
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rdx,%r12
  xor    %ebx,%ebx
  jmp    L4
  xchg   %ax,%ax
L1:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L36
L4:
  mov    (%r14,%rbx,8),%rsi
  mov    %r12,%rdi
  mov    %ebx,%r13d
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L1
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  mov    $0xffffffff,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int char_array_i(const char *array[], size_t len, const char *s) {
  int i;
  for (i = 0; i < len; i++)
    if (!strcmp(s, array[i]))
      return i;
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<CVDiagGetReturnFlagName>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x1e,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x7(%rbx),%edi
  cmp    $0x7,%edi
  ja     L15
  mov    %edi,%edi
  lea    D86(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D98(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
L86:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  lea    D50(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D96(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D74(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D18(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D75(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D40(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
  lea    D92(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L86
L15:
  lea    D90(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  jmp    L86
```

tool_calls: name: parse_data
arguments: label: D98
data_type: string
name: parse_data
arguments: label: D50
data_type: string
name: parse_data
arguments: label: D96
data_type: string
name: parse_data
arguments: label: D74
data_type: string
name: parse_data
arguments: label: D18
data_type: string
name: parse_data
arguments: label: D75
data_type: string
name: parse_data
arguments: label: D40
data_type: string
name: parse_data
arguments: label: D92
data_type: string
name: parse_data
arguments: label: D90
data_type: string

tool: D98: CVDIAG_SUCCESS

tool: D50: CVDIAG_MEM_NULL

tool: D96: CVDIAG_LMEM_NULL

tool: D74: CVDIAG_ILL_INPUT

tool: D18: CVDIAG_MEM_FAIL

tool: D75: CVDIAG_INV_FAIL

tool: D40: CVDIAG_RHSFUNC_UNRECVR

tool: D92: CVDIAG_RHSFUNC_RECVR

tool: D90: NONE

assistant: ```c
char *CVDiagGetReturnFlagName(int flag) {
  char *name;
  name = (char *)malloc(30 * sizeof(char));
  switch (flag) {
  case 0:
    sprintf(name, "CVDIAG_SUCCESS");
    break;
  case -1:
    sprintf(name, "CVDIAG_MEM_NULL");
    break;
  case -2:
    sprintf(name, "CVDIAG_LMEM_NULL");
    break;
  case -3:
    sprintf(name, "CVDIAG_ILL_INPUT");
    break;
  case -4:
    sprintf(name, "CVDIAG_MEM_FAIL");
    break;
  case -5:
    sprintf(name, "CVDIAG_INV_FAIL");
    break;
  case -6:
    sprintf(name, "CVDIAG_RHSFUNC_UNRECVR");
    break;
  case -7:
    sprintf(name, "CVDIAG_RHSFUNC_RECVR");
    break;
  default:
    sprintf(name, "NONE");
  }
  return (name);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x4,-0xc(%rbp)
  movl   $0x2,-0x8(%rbp)
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assume@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L51
L56:
  addl   $0x1,-0x14(%rbp)
  mov    $0x0,%eax
  call   <unknown2@plt>
  and    $0x1,%eax
  test   %eax,%eax
  jne    L22
  movl   $0x0,-0x18(%rbp)
  jmp    L39
L22:
  movl   $0x1,-0x18(%rbp)
L39:
  cmpl   $0x0,-0x18(%rbp)
  jne    L4
  mov    -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  jmp    L51
L4:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x10(%rbp)
L51:
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L56
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <static_assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c1 = 4;
  int c2 = 2;
  int n, v;
  int i, k, j;
  n = unknown1();
  assume(n > 0);
  k = 0;
  i = 0;
  while (i < n) {
    i++;
    if (unknown2() % 2 == 0)
      v = 0;
    else
      v = 1;
    if (v == 0)
      k += c1;
    else
      k += c2;
  }
  static_assert(k > n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  setg   %dil
  movzbl %dil,%edi
  call   <assume@plt>
  test   %r12d,%r12d
  jle    L27
  mov    $0x0,%ebx
  mov    $0x0,%ebp
L0:
  add    $0x1,%ebp
  mov    $0x0,%eax
  call   <unknown2@plt>
  mov    %eax,%edx
  lea    0x4(%rbx),%eax
  add    $0x2,%ebx
  test   $0x1,%dl
  cmove  %eax,%ebx
  cmp    %ebp,%r12d
  jne    L0
L22:
  cmp    %ebx,%r12d
  setl   %dil
  movzbl %dil,%edi
  call   <static_assert@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  mov    $0x0,%ebx
  jmp    L22
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c1 = 4;
  int c2 = 2;
  int n, v;
  int i, k, j;
  n = unknown1();
  assume(n > 0);
  k = 0;
  i = 0;
  while (i < n) {
    i++;
    if (unknown2() % 2 == 0)
      v = 0;
    else
      v = 1;
    if (v == 0)
      k += c1;
    else
      k += c2;
  }
  static_assert(k > n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <unknown1@plt>
  xor    %edi,%edi
  test   %eax,%eax
  mov    %eax,%r12d
  setg   %dil
  xor    %ebx,%ebx
  call   <assume@plt>
  test   %r12d,%r12d
  jle    L14
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <unknown2@plt>
  mov    %eax,%r8d
  lea    0x4(%rbx),%eax
  add    $0x2,%ebx
  and    $0x1,%r8d
  cmove  %eax,%ebx
  cmp    %ebp,%r12d
  jne    L13
L14:
  xor    %edi,%edi
  cmp    %ebx,%r12d
  setl   %dil
  call   <static_assert@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c1 = 4;
  int c2 = 2;
  int n, v;
  int i, k, j;
  n = unknown1();
  assume(n > 0);
  k = 0;
  i = 0;
  while (i < n) {
    i++;
    if (unknown2() % 2 == 0)
      v = 0;
    else
      v = 1;
    if (v == 0)
      k += c1;
    else
      k += c2;
  }
  static_assert(k > n);
  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
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <unknown1@plt>
  xor    %edi,%edi
  test   %eax,%eax
  mov    %eax,%r12d
  setg   %dil
  xor    %ebx,%ebx
  call   <assume@plt>
  test   %r12d,%r12d
  jle    L15
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L14:
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <unknown2@plt>
  mov    %eax,%r8d
  lea    0x4(%rbx),%eax
  add    $0x2,%ebx
  and    $0x1,%r8d
  cmove  %eax,%ebx
  cmp    %ebp,%r12d
  jne    L14
L15:
  xor    %edi,%edi
  cmp    %ebx,%r12d
  setl   %dil
  call   <static_assert@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c1 = 4;
  int c2 = 2;
  int n, v;
  int i, k, j;
  n = unknown1();
  assume(n > 0);
  k = 0;
  i = 0;
  while (i < n) {
    i++;
    if (unknown2() % 2 == 0)
      v = 0;
    else
      v = 1;
    if (v == 0)
      k += c1;
    else
      k += c2;
  }
  static_assert(k > n);
  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
<uc6_rdphalf_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <RDP@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <RDP_E@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type_guessed: string

tool: D19: uc6:rdphalf_1


assistant: ```c
void uc6_rdphalf_1() {
  RDP("uc6:rdphalf_1\n");
  RDP_E("uc6:rdphalf_1\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<uc6_rdphalf_1>:
  endbr64
  push   %rbx
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <RDP@plt>
  mov    %rbx,%rdi
  call   <RDP_E@plt>
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: string

tool: D7: uc6:rdphalf_1


assistant: ```c
void uc6_rdphalf_1() {
  RDP("uc6:rdphalf_1\n");
  RDP_E("uc6:rdphalf_1\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<uc6_rdphalf_1>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  mov    %rbp,%rdi
  call   <RDP@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <RDP_E@plt>
```

tool_calls: name: unpack_data
arguments: tag: D8
type_guessed: string

tool: D8: uc6:rdphalf_1


assistant: ```c
void uc6_rdphalf_1() {
  RDP("uc6:rdphalf_1\n");
  RDP_E("uc6:rdphalf_1\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<uc6_rdphalf_1>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  mov    %rbp,%rdi
  call   <RDP@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <RDP_E@plt>
```

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

tool: D5: uc6:rdphalf_1


assistant: ```c
void uc6_rdphalf_1() {
  RDP("uc6:rdphalf_1\n");
  RDP_E("uc6:rdphalf_1\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<give_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L11
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  neg    %eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  sub    -0x8(%rbp),%eax
  sub    $0x1,%eax
  jmp    L17
L11:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  neg    %eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  sub    -0x4(%rbp),%eax
  sub    $0x1,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int give_index(int i, int j, int n) {
  if (i > j)
    return (n * (j - 1) - j * (j - 1) / 2 + i - j - 1);
  else
    return (n * (i - 1) - i * (i - 1) / 2 + j - i - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<give_index>:
  endbr64
  mov    %esi,%r8d
  cmp    %esi,%edi
  jle    L6
  lea    -0x1(%rsi),%esi
  mov    %esi,%eax
  imul   %r8d,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  mov    %esi,%eax
  imul   %edx,%eax
  sub    %ecx,%eax
  add    %edi,%eax
  sub    %r8d,%eax
  sub    $0x1,%eax
  ret
L6:
  lea    -0x1(%rdi),%ecx
  mov    %ecx,%eax
  imul   %edi,%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  mov    %ecx,%eax
  imul   %edx,%eax
  sub    %esi,%eax
  add    %r8d,%eax
  sub    %edi,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int give_index(int i, int j, int n) {
  if (i > j)
    return (n * (j - 1) - j * (j - 1) / 2 + i - j - 1);
  else
    return (n * (i - 1) - i * (i - 1) / 2 + j - i - 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<give_index>:
  endbr64
  mov    %esi,%r8d
  cmp    %esi,%edi
  jle    L9
  lea    -0x1(%rsi),%esi
  mov    %esi,%eax
  imul   %r8d,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  mov    %esi,%eax
  imul   %edx,%eax
  sar    %ecx
  sub    %ecx,%eax
  add    %edi,%eax
  sub    %r8d,%eax
  sub    $0x1,%eax
  ret
  nopl   (%rax)
L9:
  lea    -0x1(%rdi),%ecx
  mov    %ecx,%eax
  imul   %edi,%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  mov    %ecx,%eax
  imul   %edx,%eax
  sar    %esi
  sub    %esi,%eax
  add    %r8d,%eax
  sub    %edi,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int give_index(int i, int j, int n) {
  if (i > j)
    return (n * (j - 1) - j * (j - 1) / 2 + i - j - 1);
  else
    return (n * (i - 1) - i * (i - 1) / 2 + j - i - 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<give_index>:
  endbr64
  mov    %esi,%r8d
  cmp    %esi,%edi
  jle    L5
  lea    -0x1(%rsi),%esi
  mov    %esi,%eax
  imul   %r8d,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  mov    %esi,%eax
  imul   %edx,%eax
  sar    %ecx
  sub    %ecx,%eax
  add    %edi,%eax
  sub    %r8d,%eax
  sub    $0x1,%eax
  ret
  nopl   (%rax)
L5:
  lea    -0x1(%rdi),%ecx
  mov    %ecx,%eax
  imul   %edi,%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  mov    %ecx,%eax
  imul   %edx,%eax
  sar    %esi
  sub    %esi,%eax
  add    %r8d,%eax
  sub    %edi,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int give_index(int i, int j, int n) {
  if (i > j)
    return (n * (j - 1) - j * (j - 1) / 2 + i - j - 1);
  else
    return (n * (i - 1) - i * (i - 1) / 2 + j - i - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L0
L9:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    -0x14(%rbp),%edx
  mov    %edx,(%rax)
  addq   $0x4,-0x10(%rbp)
  subq   $0x4,-0x8(%rbp)
L0:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L9
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse(int *data, int n) {
  int *lo = &data[0], *hi = &data[n];
  for (; lo < hi; ++lo, --hi) {
    int tmp = *lo;
    *lo = *hi;
    *hi = tmp;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rax
  cmp    %rax,%rdi
  jae    L1
L5:
  mov    (%rdi),%edx
  mov    (%rax),%ecx
  mov    %ecx,(%rdi)
  mov    %edx,(%rax)
  add    $0x4,%rdi
  sub    $0x4,%rax
  cmp    %rax,%rdi
  jb     L5
L1:
  ret
```

assistant: ```c
void reverse(int *data, int n) {
  int *lo = &data[0], *hi = &data[n];
  for (; lo < hi; ++lo, --hi) {
    int tmp = *lo;
    *lo = *hi;
    *hi = tmp;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rax
  cmp    %rax,%rdi
  jae    L11
L4:
  mov    (%rdi),%edx
  mov    (%rax),%ecx
  add    $0x4,%rdi
  sub    $0x4,%rax
  mov    %ecx,-0x4(%rdi)
  mov    %edx,0x4(%rax)
  cmp    %rax,%rdi
  jb     L4
L11:
  ret
```

assistant: ```c
void reverse(int *data, int n) {
  int *lo = &data[0], *hi = &data[n];
  for (; lo < hi; ++lo, --hi) {
    int tmp = *lo;
    *lo = *hi;
    *hi = tmp;
  }
}
```