ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<longswap>:
  endbr64
  test   %rsi,%rsi
  je     L26
  movzbl (%rdi),%r8d
  movzbl 0x3(%rdi),%r9d
  xor    %eax,%eax
  movzbl 0x1(%rdi),%edx
  movzbl 0x2(%rdi),%ecx
  jmp    L6
  xchg   %ax,%ax
L20:
  mov    %edx,%r10d
  mov    %ecx,%edx
  mov    %r10d,%ecx
  mov    %r8d,%r10d
  mov    %r9d,%r8d
  mov    %r10d,%r9d
L6:
  add    $0x4,%rax
  cmp    %rax,%rsi
  ja     L20
  mov    %r9b,(%rdi)
  mov    %r8b,0x3(%rdi)
  mov    %cl,0x1(%rdi)
  mov    %dl,0x2(%rdi)
L26:
  ret
```

assistant: ```c
void longswap(void *_ptr, size_t len) {
  const union {
    long one;
    char little;
  } is_endian = {1};
  if (is_endian.little) {
    size_t i;
    unsigned char *p = _ptr, c;
    for (i = 0; i < len; i += 4) {
      c = p[0], p[0] = p[3], p[3] = c;
      c = p[1], p[1] = p[2], p[2] = c;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<longswap>:
  endbr64
  test   %rsi,%rsi
  je     L2
  movzbl (%rdi),%r8d
  movzbl 0x3(%rdi),%r9d
  xor    %eax,%eax
  movzbl 0x1(%rdi),%edx
  movzbl 0x2(%rdi),%ecx
  jmp    L17
  xchg   %ax,%ax
L1:
  mov    %edx,%r10d
  mov    %ecx,%edx
  mov    %r10d,%ecx
  mov    %r8d,%r10d
  mov    %r9d,%r8d
  mov    %r10d,%r9d
L17:
  add    $0x4,%rax
  cmp    %rax,%rsi
  ja     L1
  mov    %r9b,(%rdi)
  mov    %r8b,0x3(%rdi)
  mov    %cl,0x1(%rdi)
  mov    %dl,0x2(%rdi)
L2:
  ret
```

assistant: ```c
void longswap(void *_ptr, size_t len) {
  const union {
    long one;
    char little;
  } is_endian = {1};
  if (is_endian.little) {
    size_t i;
    unsigned char *p = _ptr, c;
    for (i = 0; i < len; i += 4) {
      c = p[0], p[0] = p[3], p[3] = c;
      c = p[1], p[1] = p[2], p[2] = c;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__atomic_load_c>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    $0x0,%eax
  call   <irq_disable@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <irq_restore@plt>
  nop
  leave
  ret
```

assistant: ```c
void __atomic_load_c(size_t size, const void *src, void *dest, int memorder) {
  (void)memorder;
  unsigned int mask = irq_disable();
  memcpy(dest, src, size);
  irq_restore(mask);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__atomic_load_c>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    $0x0,%eax
  call   <irq_disable@plt>
  mov    %eax,%ebx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %ebx,%edi
  call   <irq_restore@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void __atomic_load_c(size_t size, const void *src, void *dest, int memorder) {
  (void)memorder;
  unsigned int mask = irq_disable();
  memcpy(dest, src, size);
  irq_restore(mask);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__atomic_load_c>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <irq_disable@plt>
  mov    %r12,%rdi
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %eax,%ebp
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <irq_restore@plt>
```

assistant: ```c
void __atomic_load_c(size_t size, const void *src, void *dest, int memorder) {
  (void)memorder;
  unsigned int mask = irq_disable();
  memcpy(dest, src, size);
  irq_restore(mask);
}
```
user: Translate the below assembly code into C programming language:

```asm
<__atomic_load_c>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <irq_disable@plt>
  mov    %r12,%rdi
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %eax,%ebp
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <irq_restore@plt>
```

assistant: ```c
void __atomic_load_c(size_t size, const void *src, void *dest, int memorder) {
  (void)memorder;
  unsigned int mask = irq_disable();
  memcpy(dest, src, size);
  irq_restore(mask);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sleep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x4804a03c,%eax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movq   $0x4804a03c,-0x8(%rbp)
  nop
L4:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0xc(%rbp),%edx
  sub    %edx,%eax
  cmp    $0xffffff,%eax
  jbe    L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void sleep() {
  int start = (*(volatile unsigned int *)(0x4804A03C));
  volatile unsigned int *cur = (unsigned int *)0x4804A03C;
  while (*cur - start < 0x01000000)
    ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sleep>:
  endbr64
  mov    0x4804a03c,%edx
L0:
  mov    0x4804a03c,%eax
  sub    %edx,%eax
  cmp    $0xffffff,%eax
  jbe    L0
  ret
```

assistant: ```c
void sleep() {
  int start = (*(volatile unsigned int *)(0x4804A03C));
  volatile unsigned int *cur = (unsigned int *)0x4804A03C;
  while (*cur - start < 0x01000000)
    ;
}
```
user: What is the c source code of the assembly code below:

```asm
<sleep>:
  endbr64
  mov    0x4804a03c,%edx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    0x4804a03c,%eax
  sub    %edx,%eax
  cmp    $0xffffff,%eax
  jbe    L6
  ret
```

assistant: ```c
void sleep() {
  int start = (*(volatile unsigned int *)(0x4804A03C));
  volatile unsigned int *cur = (unsigned int *)0x4804A03C;
  while (*cur - start < 0x01000000)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sleep>:
  endbr64
  mov    0x4804a03c,%edx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    0x4804a03c,%eax
  sub    %edx,%eax
  cmp    $0xffffff,%eax
  jbe    L1
  ret
```

assistant: ```c
void sleep() {
  int start = (*(volatile unsigned int *)(0x4804A03C));
  volatile unsigned int *cur = (unsigned int *)0x4804A03C;
  while (*cur - start < 0x01000000)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<drandom_init_memo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x18(%rbp)
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  mov    D2(%rip),%rax
  movsd  (%rax),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <srand48@plt>
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand48@plt>
  nop
  leave
  ret
```

assistant: ```c
void drandom_init_memo() {
  int seed;
  clock_t start, end;
  double cpu_time_used;
  start = clock();
  seed = cpu_time_used;
  end = clock();
  cpu_time_used = ((double)(end - start)) / CLOCKS_PER_SEC;
  srand48(seed);
  srand48((unsigned int)time(NULL));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<drandom_init_memo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clock@plt>
  mov    $0x0,%eax
  call   <clock@plt>
  mov    $0x0,%edi
  call   <srand48@plt>
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand48@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void drandom_init_memo() {
  int seed;
  clock_t start, end;
  double cpu_time_used;
  start = clock();
  seed = cpu_time_used;
  end = clock();
  cpu_time_used = ((double)(end - start)) / CLOCKS_PER_SEC;
  srand48(seed);
  srand48((unsigned int)time(NULL));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<drandom_init_memo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clock@plt>
  xor    %eax,%eax
  call   <clock@plt>
  xor    %edi,%edi
  call   <srand48@plt>
  xor    %edi,%edi
  call   <time@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1090 <srand48@plt>
```

assistant: ```c
void drandom_init_memo() {
  int seed;
  clock_t start, end;
  double cpu_time_used;
  start = clock();
  seed = cpu_time_used;
  end = clock();
  cpu_time_used = ((double)(end - start)) / CLOCKS_PER_SEC;
  srand48(seed);
  srand48((unsigned int)time(NULL));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<drandom_init_memo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clock@plt>
  xor    %eax,%eax
  call   <clock@plt>
  xor    %edi,%edi
  call   <srand48@plt>
  xor    %edi,%edi
  call   <time@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1090 <srand48@plt>
```

assistant: ```c
void drandom_init_memo() {
  int seed;
  clock_t start, end;
  double cpu_time_used;
  start = clock();
  seed = cpu_time_used;
  end = clock();
  cpu_time_used = ((double)(end - start)) / CLOCKS_PER_SEC;
  srand48(seed);
  srand48((unsigned int)time(NULL));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atoft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  D17(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movsbl -0xd(%rbp),%eax
  sub    $0x48,%eax
  cmp    $0x2b,%eax
  seta   %dl
  test   %dl,%dl
  jne    L39
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  movabs $0x80000000800,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L6
  movabs $0x2000000020,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L27
  movabs $0x100000001,%rdx
  and    %rdx,%rax
  test   %rax,%rax
  setne  %al
  test   %al,%al
  je     L39
  movsd  -0x8(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
L27:
  movsd  -0x8(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movsd  -0x8(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  jmp    L31
L39:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
L31:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
datatype: f64
name: extract_symbolic_data
arguments: label: D14
datatype: byte[8]

tool: D17: 1

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0x4e,0x40]}

assistant: ```c
double atoft(char *s) {
  char last;
  int len;
  double suff = 1.0;
  len = strlen(s);
  last = s[len - 1];
  s[len - 1] = '\0';
  switch (last) {
  case 'h':
  case 'H':
    suff *= 60;
  case 'm':
  case 'M':
    suff *= 60;
  case 's':
  case 'S':
    suff *= atof(s);
    s[len - 1] = last;
    return suff;
  }
  s[len - 1] = last;
  return atof(s);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<atoft>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rbp
  movzbl 0x0(%rbp),%r12d
  movb   $0x0,0x0(%rbp)
  lea    -0x48(%r12),%ecx
  cmp    $0x2b,%cl
  ja     L50
  mov    $0x1,%eax
  shl    %cl,%rax
  movabs $0x2000000020,%rdx
  test   %rdx,%rax
  jne    L53
  movabs $0x80000000800,%rdx
  test   %rdx,%rax
  jne    L41
  movabs $0x100000001,%rdx
  movsd  D36(%rip),%xmm0
  test   %rdx,%rax
  jne    L42
L50:
  mov    %r12b,0x0(%rbp)
  mov    %rbx,%rdi
  call   <atof@plt>
  jmp    L29
L53:
  movsd  D18(%rip),%xmm0
L42:
  mulsd  D36(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
L1:
  mov    %rbx,%rdi
  call   <atof@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    %r12b,0x0(%rbp)
L29:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L41:
  movsd  D18(%rip),%xmm2
  movsd  %xmm2,0x8(%rsp)
  jmp    L1
```

tool_calls: name: retrieve_data
arguments: data_tag: D36
datatype: byte[8]
name: retrieve_data
arguments: data_tag: D18
datatype: f64

tool: {"D36": [0x00,0x00,0x00,0x00,0x00,0x00,0x4e,0x40]}

tool: D18: 1

assistant: ```c
double atoft(char *s) {
  char last;
  int len;
  double suff = 1.0;
  len = strlen(s);
  last = s[len - 1];
  s[len - 1] = '\0';
  switch (last) {
  case 'h':
  case 'H':
    suff *= 60;
  case 'm':
  case 'M':
    suff *= 60;
  case 's':
  case 'S':
    suff *= atof(s);
    s[len - 1] = last;
    return suff;
  }
  s[len - 1] = last;
  return atof(s);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atoft>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x10,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  movzbl (%rbx),%r12d
  movb   $0x0,(%rbx)
  lea    -0x48(%r12),%ecx
  cmp    $0x2b,%cl
  ja     L0
  movabs $0x100000001,%rdx
  mov    $0x1,%eax
  shl    %cl,%rax
  test   %rdx,%rax
  jne    L31
  movabs $0x80000000800,%rdx
  test   %rdx,%rax
  jne    L50
  movabs $0x2000000020,%rdx
  test   %rdx,%rax
  jne    L24
L0:
  mov    %r12b,(%rbx)
  add    $0x10,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <atof@plt>
  nopl   0x0(%rax)
L24:
  movsd  D1(%rip),%xmm2
  movsd  %xmm2,0x8(%rsp)
L45:
  mov    %rbp,%rdi
  call   <atof@plt>
  mov    %r12b,(%rbx)
  mulsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L50:
  movsd  D26(%rip),%xmm4
  movsd  %xmm4,0x8(%rsp)
  jmp    L45
L31:
  movsd  D20(%rip),%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L45
```

tool_calls: name: parse_data
arguments: label: D1
type_guessed: byte[8]
name: parse_data
arguments: label: D26
type_guessed: f64
name: parse_data
arguments: label: D20
type_guessed: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x4e,0x40]}

tool: D26: 1

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x20,0xac,0x40]}

assistant: ```c
double atoft(char *s) {
  char last;
  int len;
  double suff = 1.0;
  len = strlen(s);
  last = s[len - 1];
  s[len - 1] = '\0';
  switch (last) {
  case 'h':
  case 'H':
    suff *= 60;
  case 'm':
  case 'M':
    suff *= 60;
  case 's':
  case 'S':
    suff *= atof(s);
    s[len - 1] = last;
    return suff;
  }
  s[len - 1] = last;
  return atof(s);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<atoft>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x10,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  movzbl (%rbx),%r12d
  movb   $0x0,(%rbx)
  lea    -0x48(%r12),%ecx
  cmp    $0x2b,%cl
  ja     L37
  movabs $0x100000001,%rdx
  mov    $0x1,%eax
  shl    %cl,%rax
  test   %rdx,%rax
  jne    L10
  movabs $0x80000000800,%rdx
  test   %rdx,%rax
  jne    L48
  movabs $0x2000000020,%rdx
  test   %rdx,%rax
  jne    L12
L37:
  mov    %r12b,(%rbx)
  add    $0x10,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <atof@plt>
  nopl   0x0(%rax)
L12:
  movsd  D6(%rip),%xmm2
  movsd  %xmm2,0x8(%rsp)
L46:
  mov    %rbp,%rdi
  call   <atof@plt>
  mov    %r12b,(%rbx)
  mulsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L48:
  movsd  D10(%rip),%xmm4
  movsd  %xmm4,0x8(%rsp)
  jmp    L46
L10:
  movsd  D20(%rip),%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L46
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: byte[8]
name: unpack_data
arguments: data_tag: D10
data_type: f64
name: unpack_data
arguments: data_tag: D20
data_type: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x4e,0x40]}

tool: D10: 1

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x20,0xac,0x40]}

assistant: ```c
double atoft(char *s) {
  char last;
  int len;
  double suff = 1.0;
  len = strlen(s);
  last = s[len - 1];
  s[len - 1] = '\0';
  switch (last) {
  case 'h':
  case 'H':
    suff *= 60;
  case 'm':
  case 'M':
    suff *= 60;
  case 's':
  case 'S':
    suff *= atof(s);
    s[len - 1] = last;
    return suff;
  }
  s[len - 1] = last;
  return atof(s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<subt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int subt(int x, int y) {
  int sresult;
  sresult = (x - y);
  return sresult;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subt>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subt(int x, int y) {
  int sresult;
  sresult = (x - y);
  return sresult;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<subt>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subt(int x, int y) {
  int sresult;
  sresult = (x - y);
  return sresult;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<subt>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subt(int x, int y) {
  int sresult;
  sresult = (x - y);
  return sresult;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x3,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x9,0x18(%rsp)
  movl   $0xc,0x1c(%rsp)
  movl   $0xe,0x20(%rsp)
  movl   $0x11,0x24(%rsp)
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D32(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x3(%rax),%edx
  cmp    $0xe,%edx
  ja     L7
  mov    $0x2,%esi
  mov    $0x5,%ecx
  mov    $0x0,%edx
  movsd  D2(%rip),%xmm2
  movapd %xmm2,%xmm1
  jmp    L5
L7:
  lea    D59(%rip),%rdi
  call   <printf@plt>
L99:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L51:
  lea    0x1(%rsi),%edx
  lea    (%rdx,%rcx,1),%esi
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  mulsd  %xmm2,%xmm0
  cvttsd2si %xmm0,%esi
L17:
  cmp    %edx,%ecx
  jle    L108
L5:
  movslq %esi,%rdi
  mov    0x10(%rsp,%rdi,4),%edi
  cmp    %edi,%eax
  je     L49
  movslq %edx,%r8
  cmp    0x10(%rsp,%r8,4),%eax
  je     L49
  movslq %ecx,%r8
  cmp    0x10(%rsp,%r8,4),%eax
  je     L49
  cmp    %edi,%eax
  jge    L51
  lea    -0x1(%rsi),%ecx
  lea    (%rcx,%rdx,1),%esi
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  jmp    L17
L49:
  lea    D25(%rip),%rdi
  call   <printf@plt>
  jmp    L99
L108:
  lea    D59(%rip),%rdi
  call   <printf@plt>
  jmp    L99
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D36
datatype: string
name: parse_data
arguments: data_label: D32
datatype: string
name: parse_data
arguments: data_label: D2
datatype: byte[8]
name: parse_data
arguments: data_label: D59
datatype: string
name: parse_data
arguments: data_label: D25
datatype: string

tool: D36: Enter a number: 

tool: D32: %d

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D59: It is not found!


tool: D25: It is found!


assistant: ```c
int main() {
  int a[] = {3, 4, 9, 12, 14, 17}, i, n, f = 0, j, k;
  printf("Enter a number: ");
  scanf("%d", &n);
  i = 0;
  j = sizeof(a) / sizeof(int) - 1;
  k = (int)(i + j) / 2.0;
  if (n < a[i] || n > a[j]) {
    printf("It is not found!\n");
    return 0;
  }
  while (i < j) {
    if (a[k] == n || a[i] == n || a[j] == n) {
      f = 1;
      break;
    } else if (a[k] > n) {
      j = k - 1;
      k = (i + j) / 2.0;
    } else {
      i = k + 1;
      k = (i + j) / 2.0;
    }
  }
  if (f == 1)
    printf("It is found!\n");
  else
    printf("It is not found!\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x10(%rsp)
  movabs $0xc00000009,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x110000000e,%rax
  mov    %rax,0x20(%rsp)
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x3(%rax),%edx
  cmp    $0xe,%edx
  ja     L61
  movsd  D41(%rip),%xmm1
  mov    $0x2,%esi
  mov    $0x5,%ecx
  xor    %edx,%edx
  jmp    L42
  nopw   0x0(%rax,%rax,1)
L20:
  lea    -0x1(%rsi),%ecx
  pxor   %xmm0,%xmm0
  lea    (%rcx,%rdx,1),%esi
  cvtsi2sd %esi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  cmp    %ecx,%edx
  jge    L61
L42:
  movslq %esi,%rdi
  mov    0x10(%rsp,%rdi,4),%edi
  cmp    %edi,%eax
  je     L3
  movslq %edx,%r8
  cmp    0x10(%rsp,%r8,4),%eax
  je     L3
  movslq %ecx,%r8
  cmp    0x10(%rsp,%r8,4),%eax
  je     L3
  cmp    %edi,%eax
  jl     L20
  lea    0x1(%rsi),%edx
  pxor   %xmm0,%xmm0
  lea    (%rdx,%rcx,1),%esi
  cvtsi2sd %esi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  cmp    %ecx,%edx
  jl     L42
  nop
L61:
  lea    D16(%rip),%rdi
  call   <printf@plt>
L6:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L3:
  lea    D49(%rip),%rdi
  call   <printf@plt>
  jmp    L6
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
datatype: string
name: retrieve_data
arguments: label: D13
datatype: string
name: retrieve_data
arguments: label: D41
datatype: byte[8]
name: retrieve_data
arguments: label: D16
datatype: string
name: retrieve_data
arguments: label: D49
datatype: string

tool: D7: Enter a number: 

tool: D13: %d

tool: {"D41": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D16: It is not found!


tool: D49: It is found!


assistant: ```c
int main() {
  int a[] = {3, 4, 9, 12, 14, 17}, i, n, f = 0, j, k;
  printf("Enter a number: ");
  scanf("%d", &n);
  i = 0;
  j = sizeof(a) / sizeof(int) - 1;
  k = (int)(i + j) / 2.0;
  if (n < a[i] || n > a[j]) {
    printf("It is not found!\n");
    return 0;
  }
  while (i < j) {
    if (a[k] == n || a[i] == n || a[j] == n) {
      f = 1;
      break;
    } else if (a[k] > n) {
      j = k - 1;
      k = (i + j) / 2.0;
    } else {
      i = k + 1;
      k = (i + j) / 2.0;
    }
  }
  if (f == 1)
    printf("It is found!\n");
  else
    printf("It is not found!\n");
  return 0;
}
```