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

```asm
<summaryUsage>:
  endbr64
  cmpq   $0x0,(%rdi)
  je     L8
  lea    D31(%rip),%rdx
  lea    D6(%rip),%rsi
  mov    $0x3,%edi
  jmp    10c0 <writeError@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  lea    D25(%rip),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  mov    %r12,%rdi
  add    %eax,%ebp
  call   <strlen@plt>
  lea    0x1(%rbp,%rax,1),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %ebp,%edx
  xor    %esi,%esi
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  call   <memset@plt>
  mov    (%rbx),%rdi
  mov    %r14,%r8
  pop    %rbx
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %ebp,%esi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10a0 <snprintf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D31
type_guessed: string
name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string
name: disassemble_data
arguments: data_label: D13
type_guessed: string
name: disassemble_data
arguments: data_label: D25
type_guessed: string

tool: D31: vmauthd.mod

tool: D6: %s reports an error in summaryUsage() : ppszSummary must be NULL when called

tool: D1: 2.0

tool: D13: Brute force module for the VMware Authentication Daemon

tool: D25: %s : version %s

assistant: ```c
void summaryUsage(char **ppszSummary) {
  int iLength = 0;
  if (*ppszSummary == NULL) {
    iLength = strlen("Brute force module for the VMware Authentication Daemon") + strlen("2.0") + strlen("%s : version %s") + 1;
    *ppszSummary = (char *)malloc(iLength);
    memset(*ppszSummary, 0, iLength);
    snprintf(*ppszSummary, iLength, "%s : version %s", "Brute force module for the VMware Authentication Daemon", "2.0");
  } else {
    writeError(3, "%s reports an error in summaryUsage() : ppszSummary must be NULL when called", "vmauthd.mod");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L49
L12:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D3(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D13(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <erf@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L21
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L26
L21:
  addl   $0x1,-0x24(%rbp)
L49:
  cmpl   $0xf423f,-0x24(%rbp)
  jle    L12
L26:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L47
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L47:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
data_type: f64
name: parse_data
arguments: data_label: D13
data_type: f64
name: parse_data
arguments: data_label: D1
data_type: string

tool: D3: 1000

tool: D13: 500

tool: D1: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = erf(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<benchmark>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r12
  mov    $0xf4240,%ebx
L1:
  mov    $0x0,%eax
  call   <rand_double@plt>
  mulsd  D23(%rip),%xmm0
  subsd  D22(%rip),%xmm0
  call   <erf@plt>
  ucomisd %xmm0,%xmm0
  jp     L21
  sub    $0x1,%ebx
  jne    L1
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm2
  subsd  %xmm2,%xmm0
  movq   %xmm0,%rbx
  jmp    L8
L21:
  lea    D7(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
L8:
  movq   %rbx,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D23
type_guessed: f64
name: retrieve_data
arguments: data_label: D22
type_guessed: f64
name: retrieve_data
arguments: data_label: D7
type_guessed: string

tool: D23: 1000

tool: D22: 500

tool: D7: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = erf(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L2
  xchg   %ax,%ax
L15:
  sub    $0x1,%ebx
  je     L3
L2:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D23(%rip),%xmm0
  subsd  D16(%rip),%xmm0
  call   <erf@plt>
  ucomisd %xmm0,%xmm0
  jnp    L15
  lea    D20(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L3:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D23
datatype: f64
name: read_data
arguments: data_tag: D16
datatype: f64
name: read_data
arguments: data_tag: D20
datatype: string

tool: D23: 1000

tool: D16: 500

tool: D20: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = erf(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L7
  xchg   %ax,%ax
L25:
  sub    $0x1,%ebx
  je     L18
L7:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D3(%rip),%xmm0
  subsd  D27(%rip),%xmm0
  call   <erf@plt>
  ucomisd %xmm0,%xmm0
  jnp    L25
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L18:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type: f64
name: parse_data
arguments: data_tag: D27
type: f64
name: parse_data
arguments: data_tag: D17
type: string

tool: D3: 1000

tool: D27: 500

tool: D17: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (1000.0 * rand_double()) - 500.0;
    y = erf(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uint64t_strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x34(%rbp),%edx
  lea    -0x20(%rbp),%rax
  mov    %edx,%ecx
  lea    D3(%rip),%rdx
  mov    $0x15,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %eax,-0x28(%rbp)
  cmpl   $0x0,-0x28(%rbp)
  jns    L22
  mov    $0xffffffff,%eax
  jmp    L19
L22:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
datatype: string

tool: D3: %llu

assistant: ```c
int8_t uint64t_strlen(uint64_t i) {
  char buf[20 + 1];
  int retval = snprintf(buf, sizeof(buf), "%llu", i);
  if (retval < 0)
    return -1;
  int8_t len = strlen(buf);
  return len;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uint64t_strlen>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D9(%rip),%rdx
  mov    $0x15,%esi
  call   <snprintf@plt>
  test   %eax,%eax
  js     L19
  mov    %rsp,%rdi
  call   <strlen@plt>
L11:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L19:
  mov    $0xffffffff,%eax
  jmp    L11
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %llu

assistant: ```c
int8_t uint64t_strlen(uint64_t i) {
  char buf[20 + 1];
  int retval = snprintf(buf, sizeof(buf), "%llu", i);
  if (retval < 0)
    return -1;
  int8_t len = strlen(buf);
  return len;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<uint64t_strlen>:
  endbr64
  push   %rbp
  mov    %edi,%ecx
  lea    D9(%rip),%rdx
  mov    $0x15,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  test   %eax,%eax
  js     L7
  mov    %rbp,%rdi
  call   <strlen@plt>
L28:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x20,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L7:
  mov    $0xffffffff,%eax
  jmp    L28
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %llu

assistant: ```c
int8_t uint64t_strlen(uint64_t i) {
  char buf[20 + 1];
  int retval = snprintf(buf, sizeof(buf), "%llu", i);
  if (retval < 0)
    return -1;
  int8_t len = strlen(buf);
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uint64t_strlen>:
  endbr64
  push   %rbp
  mov    %edi,%ecx
  lea    D3(%rip),%rdx
  mov    $0x15,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  test   %eax,%eax
  js     L21
  mov    %rbp,%rdi
  call   <strlen@plt>
L29:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x20,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L21:
  mov    $0xffffffff,%eax
  jmp    L29
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_guessed: string

tool: D3: %llu

assistant: ```c
int8_t uint64t_strlen(uint64_t i) {
  char buf[20 + 1];
  int retval = snprintf(buf, sizeof(buf), "%llu", i);
  if (retval < 0)
    return -1;
  int8_t len = strlen(buf);
  return len;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<divide>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L8
  mov    $0x1,%r14d
  test   %esi,%esi
  jle    L8
L85:
  mov    $0x1,%r13d
  cmp    %esi,%edi
  je     L98
  mov    %edi,%r13d
  cmp    $0x1,%esi
  je     L98
  cmp    $0xffffffff,%esi
  je     L76
  mov    $0x0,%r13d
  cmp    $0x80000000,%esi
  je     L98
  cmp    $0x80000000,%edi
  je     L147
L153:
  mov    %edi,%r12d
  neg    %r12d
  cmovs  %edi,%r12d
  mov    %esi,%ebp
  neg    %ebp
  cmovs  %esi,%ebp
  cmp    %ebp,%r12d
  jge    L124
L198:
  mov    %r13d,%eax
  neg    %eax
  cmp    $0x1,%r14d
  cmovne %eax,%r13d
L98:
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  test   %esi,%edi
  js     L10
  mov    $0x7fffffff,%r13d
  mov    $0xffffffff,%r14d
  test   %esi,%esi
  je     L98
  jmp    L85
L10:
  mov    $0x1,%r14d
  jmp    L85
L76:
  neg    %r13d
  cmp    $0x80000000,%edi
  mov    $0x7fffffff,%eax
  cmove  %eax,%r13d
  jmp    L98
L147:
  mov    %esi,%edi
  neg    %edi
  cmp    $0x1,%r14d
  cmovne %esi,%edi
  add    $0x80000000,%edi
  mov    $0x1,%r13d
  jmp    L153
L43:
  test   %ecx,%ecx
  je     L198
  lea    -0x1(%rcx),%ebx
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  add    %eax,%r13d
  mov    %ebp,%eax
  mov    %ebx,%ecx
  shl    %cl,%eax
  sub    %eax,%r12d
  cmp    %r12d,%ebp
  jg     L198
L124:
  mov    $0x0,%ecx
L126:
  mov    %ebp,%eax
  shl    %cl,%eax
  cmp    %r12d,%eax
  jg     L43
  add    $0x1,%ecx
  cmp    $0x3ffffffe,%ebp
  jg     L43
  mov    %ebp,%eax
  shl    %cl,%eax
  cmp    $0x3ffffffe,%eax
  jle    L126
  jmp    L43
```

assistant: ```c
int divide(int A, int B) {
  int flags = ((A > 0 && B > 0) || (A < 0 && B < 0)) ? 1 : -1;
  int out = 0;
  if (B == 0)
    return ((int)(~0U >> 1));
  if (A == B)
    return 1;
  if (B == 1)
    return A;
  if (B == -1)
    return (A == (-((int)(~0U >> 1)) - 1)) ? ((int)(~0U >> 1)) : -A;
  if (B == (-((int)(~0U >> 1)) - 1))
    return 0;
  if (A == (-((int)(~0U >> 1)) - 1)) {
    A += (flags == 1) ? -B : B;
    out++;
  }
  A = abs(A);
  B = abs(B);
  int shift = 0;
  while (A >= B) {
    shift = 0;
    while (A >= B << shift) {
      shift++;
      if (B >= ((int)(~0U >> 1)) / 2 || B << shift >= ((int)(~0U >> 1)) / 2)
        break;
    }
    if (shift == 0)
      break;
    shift--;
    out += (int)pow(2, shift);
    A -= B << shift;
  }
  return (flags == 1) ? out : -out;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<divide>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L5
  mov    $0x1,%r12d
  test   %esi,%esi
  jle    L5
L83:
  mov    $0x1,%r14d
  cmp    %esi,%edi
  je     L8
  mov    %edi,%r14d
  cmp    $0x1,%esi
  je     L8
  cmp    $0xffffffff,%esi
  je     L109
  xor    %r14d,%r14d
  cmp    $0x80000000,%esi
  je     L8
  cmp    $0x80000000,%edi
  jne    L108
  mov    %esi,%edi
  mov    $0x1,%r14d
  neg    %edi
  cmp    $0x1,%r12d
  cmovne %esi,%edi
  add    $0x80000000,%edi
L108:
  mov    %edi,%r13d
  mov    %esi,%ebx
  neg    %r13d
  cmovs  %edi,%r13d
  neg    %ebx
  cmovs  %esi,%ebx
  cmp    %ebx,%r13d
  jl     L68
  nopl   0x0(%rax)
L0:
  xor    %esi,%esi
  jmp    L100
  nopl   0x0(%rax)
L61:
  lea    0x1(%rsi),%ecx
  cmp    $0x3ffffffe,%ebx
  jg     L168
  mov    %ebx,%eax
  shl    %cl,%eax
  cmp    $0x3ffffffe,%eax
  jg     L168
  mov    %ecx,%esi
L100:
  mov    %ebx,%ebp
  mov    %esi,%ecx
  shl    %cl,%ebp
  cmp    %r13d,%ebp
  jle    L61
  test   %esi,%esi
  je     L68
  sub    $0x1,%esi
  mov    %ebx,%ebp
  mov    %esi,%ecx
  shl    %cl,%ebp
L168:
  mov    $0x2,%edi
  sub    %ebp,%r13d
  call   <pow@plt>
  add    %eax,%r14d
  cmp    %r13d,%ebx
  jle    L0
L68:
  mov    %r14d,%eax
  neg    %eax
  cmp    $0x1,%r12d
  cmovne %eax,%r14d
L8:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  test   %esi,%edi
  jns    L37
  mov    $0x1,%r12d
  jmp    L83
L37:
  mov    $0x7fffffff,%r14d
  mov    $0xffffffff,%r12d
  test   %esi,%esi
  jne    L83
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L109:
  mov    %edi,%eax
  mov    $0x7fffffff,%r14d
  pop    %rbx
  pop    %rbp
  neg    %eax
  cmp    $0x80000000,%edi
  pop    %r12
  pop    %r13
  cmovne %eax,%r14d
  mov    %r14d,%eax
  pop    %r14
  ret
```

assistant: ```c
int divide(int A, int B) {
  int flags = ((A > 0 && B > 0) || (A < 0 && B < 0)) ? 1 : -1;
  int out = 0;
  if (B == 0)
    return ((int)(~0U >> 1));
  if (A == B)
    return 1;
  if (B == 1)
    return A;
  if (B == -1)
    return (A == (-((int)(~0U >> 1)) - 1)) ? ((int)(~0U >> 1)) : -A;
  if (B == (-((int)(~0U >> 1)) - 1))
    return 0;
  if (A == (-((int)(~0U >> 1)) - 1)) {
    A += (flags == 1) ? -B : B;
    out++;
  }
  A = abs(A);
  B = abs(B);
  int shift = 0;
  while (A >= B) {
    shift = 0;
    while (A >= B << shift) {
      shift++;
      if (B >= ((int)(~0U >> 1)) / 2 || B << shift >= ((int)(~0U >> 1)) / 2)
        break;
    }
    if (shift == 0)
      break;
    shift--;
    out += (int)pow(2, shift);
    A -= B << shift;
  }
  return (flags == 1) ? out : -out;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divide>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L161
  mov    $0x1,%r12d
  test   %esi,%esi
  jle    L161
L129:
  mov    $0x1,%r14d
  cmp    %esi,%edi
  je     L156
  mov    %edi,%r14d
  cmp    $0x1,%esi
  je     L156
  cmp    $0xffffffff,%esi
  je     L179
  xor    %r14d,%r14d
  cmp    $0x80000000,%esi
  je     L156
  cmp    $0x80000000,%edi
  jne    L9
  mov    %esi,%edi
  mov    $0x1,%r14d
  neg    %edi
  cmp    $0x1,%r12d
  cmovne %esi,%edi
  add    $0x80000000,%edi
L9:
  mov    %edi,%r13d
  mov    %esi,%ebx
  neg    %r13d
  cmovs  %edi,%r13d
  neg    %ebx
  cmovs  %esi,%ebx
  cmp    %ebx,%r13d
  jl     L108
  nopl   0x0(%rax)
L26:
  cmp    $0x3ffffffe,%ebx
  jg     L127
  xor    %esi,%esi
  jmp    L80
L164:
  lea    0x1(%rsi),%ecx
  mov    %ebx,%eax
  shl    %cl,%eax
  cmp    $0x3ffffffe,%eax
  jg     L53
  mov    %ecx,%esi
L80:
  mov    %ebx,%ebp
  mov    %esi,%ecx
  shl    %cl,%ebp
  cmp    %r13d,%ebp
  jle    L164
  test   %esi,%esi
  je     L108
  sub    $0x1,%esi
  mov    %ebx,%ebp
  mov    %esi,%ecx
  shl    %cl,%ebp
L53:
  mov    $0x2,%edi
  sub    %ebp,%r13d
  call   <pow@plt>
  add    %eax,%r14d
  cmp    %r13d,%ebx
  jle    L26
L108:
  mov    %r14d,%eax
  neg    %eax
  cmp    $0x1,%r12d
  cmovne %eax,%r14d
L156:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L161:
  test   %esi,%edi
  jns    L58
  mov    $0x1,%r12d
  jmp    L129
L58:
  mov    $0x7fffffff,%r14d
  mov    $0xffffffff,%r12d
  test   %esi,%esi
  jne    L129
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L127:
  cmp    %r13d,%ebx
  jg     L108
  mov    %ebx,%ebp
  xor    %esi,%esi
  jmp    L53
L179:
  mov    %edi,%eax
  mov    $0x7fffffff,%r14d
  pop    %rbx
  pop    %rbp
  neg    %eax
  cmp    $0x80000000,%edi
  pop    %r12
  pop    %r13
  cmovne %eax,%r14d
  mov    %r14d,%eax
  pop    %r14
  ret
```

assistant: ```c
int divide(int A, int B) {
  int flags = ((A > 0 && B > 0) || (A < 0 && B < 0)) ? 1 : -1;
  int out = 0;
  if (B == 0)
    return ((int)(~0U >> 1));
  if (A == B)
    return 1;
  if (B == 1)
    return A;
  if (B == -1)
    return (A == (-((int)(~0U >> 1)) - 1)) ? ((int)(~0U >> 1)) : -A;
  if (B == (-((int)(~0U >> 1)) - 1))
    return 0;
  if (A == (-((int)(~0U >> 1)) - 1)) {
    A += (flags == 1) ? -B : B;
    out++;
  }
  A = abs(A);
  B = abs(B);
  int shift = 0;
  while (A >= B) {
    shift = 0;
    while (A >= B << shift) {
      shift++;
      if (B >= ((int)(~0U >> 1)) / 2 || B << shift >= ((int)(~0U >> 1)) / 2)
        break;
    }
    if (shift == 0)
      break;
    shift--;
    out += (int)pow(2, shift);
    A -= B << shift;
  }
  return (flags == 1) ? out : -out;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ibetad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  %xmm2,-0x28(%rbp)
  movsd  -0x28(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <betapdf@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double ibetad(double x, double a, double b) {
  double oup;
  oup = betapdf(x, a, b);
  return oup;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ibetad>:
  endbr64
  sub    $0x8,%rsp
  call   <betapdf@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
double ibetad(double x, double a, double b) {
  double oup;
  oup = betapdf(x, a, b);
  return oup;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ibetad>:
  endbr64
  jmp    1050 <betapdf@plt>
```

assistant: ```c
double ibetad(double x, double a, double b) {
  double oup;
  oup = betapdf(x, a, b);
  return oup;
}
```
user: What is the c source code of the assembly code below:

```asm
<ibetad>:
  endbr64
  jmp    1050 <betapdf@plt>
```

assistant: ```c
double ibetad(double x, double a, double b) {
  double oup;
  oup = betapdf(x, a, b);
  return oup;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L1
  mov    $0x0,%eax
  call   <abort@plt>
L1:
  nop
  leave
  ret
```

assistant: ```c
void test6(const char *ptr) {
  if (strcmp(ptr, ptr) != 0)
    abort();
}
```
user: What is the c source code of the assembly code below?

```asm
<test6>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L4
L13:
  add    $0x8,%rsp
  ret
L4:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L13
```

assistant: ```c
void test6(const char *ptr) {
  if (strcmp(ptr, ptr) != 0)
    abort();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test6>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L9
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <abort@plt>
```

assistant: ```c
void test6(const char *ptr) {
  if (strcmp(ptr, ptr) != 0)
    abort();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test6>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L4
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <abort@plt>
```

assistant: ```c
void test6(const char *ptr) {
  if (strcmp(ptr, ptr) != 0)
    abort();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<exit1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x7b,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void exit1() {
  int returnVal = 123;
  exit(returnVal);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<exit1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7b,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void exit1() {
  int returnVal = 123;
  exit(returnVal);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<exit1>:
  endbr64
  mov    $0x7b,%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
void exit1() {
  int returnVal = 123;
  exit(returnVal);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<exit1>:
  endbr64
  mov    $0x7b,%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
void exit1() {
  int returnVal = 123;
  exit(returnVal);
}
```