ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nmea_get_item>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L5
L44:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  je     L42
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L44
L42:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L17
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L22
L17:
  mov    -0x8(%rbp),%rax
L22:
  pop    %rbp
  ret
```

assistant: ```c
const char *nmea_get_item(const char *nmea, char *item) {
  while (*nmea != ',' && *nmea != '*')
    *item++ = *nmea++;
  *item = '\0';
  if (*nmea == ',')
    return nmea + 1;
  return nmea;
}
```
user: Translate the below assembly code into C programming language:

```asm
<nmea_get_item>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  lea    -0x2a(%rdx),%ecx
  test   $0xfd,%cl
  je     L5
L17:
  add    $0x1,%rax
  add    $0x1,%rsi
  mov    %dl,-0x1(%rsi)
  movzbl (%rax),%edx
  lea    -0x2a(%rdx),%ecx
  test   $0xfd,%cl
  jne    L17
L5:
  movb   $0x0,(%rsi)
  cmpb   $0x2c,(%rax)
  sete   %dl
  movzbl %dl,%edx
  add    %rdx,%rax
  ret
```

assistant: ```c
const char *nmea_get_item(const char *nmea, char *item) {
  while (*nmea != ',' && *nmea != '*')
    *item++ = *nmea++;
  *item = '\0';
  if (*nmea == ',')
    return nmea + 1;
  return nmea;
}
```
user: Translate the below assembly code into C programming language:

```asm
<nmea_get_item>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  lea    -0x2a(%rdx),%ecx
  and    $0xfd,%ecx
  je     L17
  nopl   (%rax)
L5:
  add    $0x1,%rax
  mov    %dl,(%rsi)
  add    $0x1,%rsi
  movzbl (%rax),%edx
  lea    -0x2a(%rdx),%ecx
  and    $0xfd,%ecx
  jne    L5
L17:
  movb   $0x0,(%rsi)
  xor    %edx,%edx
  cmpb   $0x2c,(%rax)
  sete   %dl
  add    %rdx,%rax
  ret
```

assistant: ```c
const char *nmea_get_item(const char *nmea, char *item) {
  while (*nmea != ',' && *nmea != '*')
    *item++ = *nmea++;
  *item = '\0';
  if (*nmea == ',')
    return nmea + 1;
  return nmea;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<nmea_get_item>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  lea    -0x2a(%rdx),%ecx
  and    $0xfd,%ecx
  je     L15
  nopl   (%rax)
L10:
  add    $0x1,%rax
  mov    %dl,(%rsi)
  add    $0x1,%rsi
  movzbl (%rax),%edx
  lea    -0x2a(%rdx),%ecx
  and    $0xfd,%ecx
  jne    L10
L15:
  movb   $0x0,(%rsi)
  xor    %edx,%edx
  cmpb   $0x2c,(%rax)
  sete   %dl
  add    %rdx,%rax
  ret
```

assistant: ```c
const char *nmea_get_item(const char *nmea, char *item) {
  while (*nmea != ',' && *nmea != '*')
    *item++ = *nmea++;
  *item = '\0';
  if (*nmea == ',')
    return nmea + 1;
  return nmea;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<crc_update>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x18(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L28
L13:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  and    $0x8000,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  cmpl   $0x0,-0x4(%rbp)
  sete   %dl
  mov    -0x14(%rbp),%eax
  and    -0x8(%rbp),%eax
  test   %eax,%eax
  sete   %al
  xor    %edx,%eax
  test   %al,%al
  je     L28
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  xor    $0x8005,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
L28:
  shrl   -0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L13
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movzwl %ax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void crc_update(unsigned int data, unsigned int length, unsigned int *crc) {
  unsigned int masking, carry;
  masking = 1 << length;
  while ((masking >>= 1)) {
    carry = *crc & 0x8000;
    *crc <<= 1;
    if (!carry ^ !(data & masking))
      *crc ^= 0x8005;
  }
  *crc &= 0xffff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<crc_update>:
  endbr64
  mov    $0x1,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  shr    %ecx
  je     L14
  mov    (%rdx),%eax
L2:
  mov    %eax,%esi
  and    $0x8000,%esi
  add    %eax,%eax
  test   %esi,%esi
  sete   %r9b
  test   %ecx,%edi
  sete   %r8b
  mov    %eax,%esi
  xor    $0x8005,%esi
  cmp    %r8b,%r9b
  cmovne %esi,%eax
  shr    %ecx
  jne    L2
  mov    %eax,(%rdx)
L14:
  andl   $0xffff,(%rdx)
  ret
```

assistant: ```c
void crc_update(unsigned int data, unsigned int length, unsigned int *crc) {
  unsigned int masking, carry;
  masking = 1 << length;
  while ((masking >>= 1)) {
    carry = *crc & 0x8000;
    *crc <<= 1;
    if (!carry ^ !(data & masking))
      *crc ^= 0x8005;
  }
  *crc &= 0xffff;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<crc_update>:
  endbr64
  mov    %esi,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  mov    (%rdx),%eax
  shr    %ecx
  je     L5
  nopl   (%rax)
L2:
  mov    %eax,%esi
  add    %eax,%eax
  and    $0x8000,%esi
  test   %esi,%esi
  mov    %eax,%esi
  sete   %r9b
  test   %ecx,%edi
  sete   %r8b
  xor    $0x8005,%esi
  cmp    %r8b,%r9b
  cmovne %esi,%eax
  shr    %ecx
  jne    L2
L5:
  and    $0xffff,%eax
  mov    %eax,(%rdx)
  ret
```

assistant: ```c
void crc_update(unsigned int data, unsigned int length, unsigned int *crc) {
  unsigned int masking, carry;
  masking = 1 << length;
  while ((masking >>= 1)) {
    carry = *crc & 0x8000;
    *crc <<= 1;
    if (!carry ^ !(data & masking))
      *crc ^= 0x8005;
  }
  *crc &= 0xffff;
}
```
user: What is the c source code of the assembly code below:

```asm
<crc_update>:
  endbr64
  mov    %esi,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  mov    (%rdx),%eax
  shr    %ecx
  je     L6
  nopl   (%rax)
L12:
  mov    %eax,%esi
  add    %eax,%eax
  and    $0x8000,%esi
  test   %esi,%esi
  mov    %eax,%esi
  sete   %r9b
  test   %ecx,%edi
  sete   %r8b
  xor    $0x8005,%esi
  cmp    %r8b,%r9b
  cmovne %esi,%eax
  shr    %ecx
  jne    L12
L6:
  and    $0xffff,%eax
  mov    %eax,(%rdx)
  ret
```

assistant: ```c
void crc_update(unsigned int data, unsigned int length, unsigned int *crc) {
  unsigned int masking, carry;
  masking = 1 << length;
  while ((masking >>= 1)) {
    carry = *crc & 0x8000;
    *crc <<= 1;
    if (!carry ^ !(data & masking))
      *crc ^= 0x8005;
  }
  *crc &= 0xffff;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<MD5InitKey>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <MD5Init@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <MD5Update@plt>
  nop
  leave
  ret
```

assistant: ```c
void MD5InitKey(ctx, s) struct MD5Context *ctx;
const char *s;
{
  MD5Init(ctx);
  MD5Update(ctx, (unsigned char *)s, (unsigned)strlen(s));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MD5InitKey>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  call   <MD5Init@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <MD5Update@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void MD5InitKey(ctx, s) struct MD5Context *ctx;
const char *s;
{
  MD5Init(ctx);
  MD5Update(ctx, (unsigned char *)s, (unsigned)strlen(s));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MD5InitKey>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <MD5Init@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <MD5Update@plt>
```

assistant: ```c
void MD5InitKey(ctx, s) struct MD5Context *ctx;
const char *s;
{
  MD5Init(ctx);
  MD5Update(ctx, (unsigned char *)s, (unsigned)strlen(s));
}
```
user: Translate the below assembly code into C programming language:

```asm
<MD5InitKey>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <MD5Init@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <MD5Update@plt>
```

assistant: ```c
void MD5InitKey(ctx, s) struct MD5Context *ctx;
const char *s;
{
  MD5Init(ctx);
  MD5Update(ctx, (unsigned char *)s, (unsigned)strlen(s));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L18
L3:
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x28(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  sub    %rdx,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <ncr@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
L18:
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  jne    L3
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D26
type_infered: string

tool: D16: %lld

tool: D26: %lld


assistant: ```c
int main() {
  long long int n, r, t, i;
  scanf("%lld", &t);
  while (t != 0) {
    scanf("%lld", &n);
    scanf("%lld", &r);
    n = n - r;
    r--;
    long long int ans = ncr(n + r, n);
    printf("%lld\n", ans);
    t--;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  cmpq   $0x0,0x10(%rsp)
  je     L14
  mov    %rsp,%r13
  lea    D24(%rip),%rbx
  lea    0x8(%rsp),%r12
  lea    D29(%rip),%rbp
L27:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  mov    0x8(%rsp),%rax
  mov    %rdi,%rsi
  sub    %rax,%rsi
  mov    %rsi,(%rsp)
  sub    $0x1,%rax
  mov    %rax,0x8(%rsp)
  sub    $0x1,%rdi
  call   <ncr@plt>
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%rax
  sub    $0x1,%rax
  mov    %rax,0x10(%rsp)
  jne    L27
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D29
datatype: string

tool: D24: %lld

tool: D29: %lld


assistant: ```c
int main() {
  long long int n, r, t, i;
  scanf("%lld", &t);
  while (t != 0) {
    scanf("%lld", &n);
    scanf("%lld", &r);
    n = n - r;
    r--;
    long long int ans = ncr(n + r, n);
    printf("%lld\n", ans);
    t--;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x10(%rsp)
  je     L7
  mov    %rsp,%r13
  lea    0x8(%rsp),%r12
  lea    D6(%rip),%rbp
  nopl   0x0(%rax)
L13:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  mov    0x8(%rsp),%rax
  mov    %rdi,%rsi
  sub    $0x1,%rdi
  sub    %rax,%rsi
  sub    $0x1,%rax
  mov    %rsi,(%rsp)
  mov    %rax,0x8(%rsp)
  call   <ncr@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  subq   $0x1,0x10(%rsp)
  jne    L13
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D17
data_type: string
name: unpack_data
arguments: label: D6
data_type: string

tool: D17: %lld

tool: D6: %lld


assistant: ```c
int main() {
  long long int n, r, t, i;
  scanf("%lld", &t);
  while (t != 0) {
    scanf("%lld", &n);
    scanf("%lld", &r);
    n = n - r;
    r--;
    long long int ans = ncr(n + r, n);
    printf("%lld\n", ans);
    t--;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D19(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x10(%rsp)
  je     L16
  mov    %rsp,%r13
  lea    0x8(%rsp),%r12
  lea    D5(%rip),%rbp
  nopl   0x0(%rax)
L24:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  mov    0x8(%rsp),%rax
  mov    %rdi,%rsi
  sub    $0x1,%rdi
  sub    %rax,%rsi
  sub    $0x1,%rax
  mov    %rsi,(%rsp)
  mov    %rax,0x8(%rsp)
  call   <ncr@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  subq   $0x1,0x10(%rsp)
  jne    L24
L16:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D19
type_guessed: string
name: disassemble_data
arguments: label: D5
type_guessed: string

tool: D19: %lld

tool: D5: %lld


assistant: ```c
int main() {
  long long int n, r, t, i;
  scanf("%lld", &t);
  while (t != 0) {
    scanf("%lld", &n);
    scanf("%lld", &r);
    n = n - r;
    r--;
    long long int ans = ncr(n + r, n);
    printf("%lld\n", ans);
    t--;
  }
  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
<bst_contains>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L56
  mov    $0x0,%eax
  jmp    L42
L56:
  movl   $0x0,-0x4(%rbp)
  jmp    L63
L47:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L45
  mov    $0x1,%eax
  jmp    L42
L45:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L70
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L12
L70:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L12:
  mov    %eax,-0x4(%rbp)
L63:
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L47
  mov    $0x0,%eax
L42:
  pop    %rbp
  ret
```

assistant: ```c
int bst_contains(int item, int *items, int size, int *left, int *right) {
  if (size == 0) {
    return 0;
  }
  int c = 0;
  while (c != -1) {
    if (items[c] == item) {
      return 1;
    }
    c = (item < items[c]) ? left[c] : right[c];
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bst_contains>:
  endbr64
  mov    %edx,%eax
  test   %edx,%edx
  je     L39
  mov    $0x0,%eax
  jmp    L24
L49:
  mov    (%r8,%rax,4),%eax
L17:
  cmp    $0xffffffff,%eax
  je     L40
L24:
  cltq
  mov    (%rsi,%rax,4),%edx
  cmp    %edi,%edx
  je     L58
  jle    L49
  mov    (%rcx,%rax,4),%eax
  jmp    L17
L40:
  mov    $0x0,%eax
  ret
L58:
  mov    $0x1,%eax
L39:
  ret
```

assistant: ```c
int bst_contains(int item, int *items, int size, int *left, int *right) {
  if (size == 0) {
    return 0;
  }
  int c = 0;
  while (c != -1) {
    if (items[c] == item) {
      return 1;
    }
    c = (item < items[c]) ? left[c] : right[c];
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bst_contains>:
  endbr64
  xor    %eax,%eax
  test   %edx,%edx
  jne    L47
  jmp    L44
  nopl   0x0(%rax)
L58:
  movslq (%rcx,%rax,4),%rax
  cmp    $0xffffffff,%eax
  je     L53
L47:
  cmp    %edi,(%rsi,%rax,4)
  je     L9
  jg     L58
  movslq (%r8,%rax,4),%rax
  cmp    $0xffffffff,%eax
  jne    L47
L53:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L9:
  mov    $0x1,%eax
L44:
  ret
```

assistant: ```c
int bst_contains(int item, int *items, int size, int *left, int *right) {
  if (size == 0) {
    return 0;
  }
  int c = 0;
  while (c != -1) {
    if (items[c] == item) {
      return 1;
    }
    c = (item < items[c]) ? left[c] : right[c];
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bst_contains>:
  endbr64
  xor    %eax,%eax
  test   %edx,%edx
  je     L43
L36:
  cmp    %edi,(%rsi,%rax,4)
  je     L56
L55:
  jle    L2
  movslq (%rcx,%rax,4),%rax
  cmp    $0xffffffff,%eax
  jne    L36
L30:
  xor    %eax,%eax
L43:
  ret
  nopl   (%rax)
L2:
  movslq (%r8,%rax,4),%rax
  cmp    $0xffffffff,%eax
  je     L30
  cmp    %edi,(%rsi,%rax,4)
  jne    L55
L56:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int bst_contains(int item, int *items, int size, int *left, int *right) {
  if (size == 0) {
    return 0;
  }
  int c = 0;
  while (c != -1) {
    if (items[c] == item) {
      return 1;
    }
    c = (item < items[c]) ? left[c] : right[c];
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<imprimeModulo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L13
L10:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  mov    %rax,%rbx
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  add    %rbx,%rax
  mov    %rax,%rdi
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x2,-0x14(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L10
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
datatype: string

tool: D5: %f 

assistant: ```c
void imprimeModulo(float dadosImp[], int tamVet) {
  int posicao;
  for (posicao = 0; posicao < tamVet; posicao = posicao + 2) {
    printf("%f ", sqrt(pow(dadosImp[posicao], 2) + pow(dadosImp[posicao + 1], 2)));
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<imprimeModulo>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L9
  mov    %rdi,%rbp
  sub    $0x1,%esi
  shr    %esi
  mov    %esi,%esi
  lea    0x8(%rdi,%rsi,8),%r13
  lea    D7(%rip),%r12
L1:
  mov    $0x2,%edi
  movss  0x0(%rbp),%xmm0
  call   <pow@plt>
  mov    %rax,%rbx
  movss  0x4(%rbp),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x8,%rbp
  cmp    %r13,%rbp
  jne    L1
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type: string

tool: D7: %f 

assistant: ```c
void imprimeModulo(float dadosImp[], int tamVet) {
  int posicao;
  for (posicao = 0; posicao < tamVet; posicao = posicao + 2) {
    printf("%f ", sqrt(pow(dadosImp[posicao], 2) + pow(dadosImp[posicao + 1], 2)));
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<imprimeModulo>:
  endbr64
  test   %esi,%esi
  jle    L6
  push   %r13
  sub    $0x1,%esi
  push   %r12
  shr    %esi
  lea    D3(%rip),%r12
  push   %rbp
  lea    0x8(%rdi,%rsi,8),%r13
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L7:
  movss  0x0(%rbp),%xmm0
  mov    $0x2,%edi
  add    $0x8,%rbp
  call   <pow@plt>
  movss  -0x4(%rbp),%xmm0
  mov    $0x2,%edi
  mov    %rax,%rbx
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  cmp    %r13,%rbp
  jne    L7
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <putchar@plt>
  nopl   0x0(%rax)
L6:
  mov    $0xa,%edi
  jmp    1080 <putchar@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_infered: string

tool: D3: %f 

assistant: ```c
void imprimeModulo(float dadosImp[], int tamVet) {
  int posicao;
  for (posicao = 0; posicao < tamVet; posicao = posicao + 2) {
    printf("%f ", sqrt(pow(dadosImp[posicao], 2) + pow(dadosImp[posicao + 1], 2)));
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<imprimeModulo>:
  endbr64
  test   %esi,%esi
  jle    L16
  push   %r13
  sub    $0x1,%esi
  push   %r12
  shr    %esi
  lea    D8(%rip),%r12
  push   %rbp
  lea    0x8(%rdi,%rsi,8),%r13
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L10:
  movss  0x0(%rbp),%xmm0
  mov    $0x2,%edi
  add    $0x8,%rbp
  call   <pow@plt>
  movss  -0x4(%rbp),%xmm0
  mov    $0x2,%edi
  mov    %rax,%rbx
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  cmp    %r13,%rbp
  jne    L10
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <putchar@plt>
  nopl   0x0(%rax)
L16:
  mov    $0xa,%edi
  jmp    1080 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string

tool: D8: %f 

assistant: ```c
void imprimeModulo(float dadosImp[], int tamVet) {
  int posicao;
  for (posicao = 0; posicao < tamVet; posicao = posicao + 2) {
    printf("%f ", sqrt(pow(dadosImp[posicao], 2) + pow(dadosImp[posicao + 1], 2)));
  }
  printf("\n");
}
```