ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x40c(%rbp)
  mov    -0x40c(%rbp),%eax
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x408(%rbp)
  movl   $0x0,-0x410(%rbp)
  jmp    L19
L25:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x404(%rbp)
  mov    -0x404(%rbp),%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x404(%rbp)
  mov    -0x410(%rbp),%eax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x404(%rbp),%edx
  lea    -0x200(%rbp),%rax
  lea    D54(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x400(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x400(%rbp),%rax
  lea    D59(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  lea    -0x400(%rbp),%rdx
  lea    -0x200(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L29
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x200(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L29:
  addl   $0x1,-0x410(%rbp)
L19:
  cmpl   $0x31,-0x410(%rbp)
  jle    L25
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D26
data_type: string
name: unpack_data
arguments: tag: D54
data_type: string
name: unpack_data
arguments: tag: D49
data_type: string
name: unpack_data
arguments: tag: D59
data_type: string
name: unpack_data
arguments: tag: D58
data_type: string
name: unpack_data
arguments: tag: D20
data_type: string
name: unpack_data
arguments: tag: D35
data_type: string

tool: D26: %d days without an incident.


tool: D54: %d

tool: D49:  %10s

tool: D59: 


tool: D58: Well that didn't take long.

tool: D20: You should have used %s.


tool: D35: How very unpredictable. Level Cleared

assistant: ```c
int main() {
  int i;
  time_t var0 = time(NULL);
  srand(var0);
  int var1 = 0;
  char buf0[500];
  char buf1[500];
  for (i = 0; i < 50; i++) {
    int var2 = rand() % 100;
    printf("%d days without an incident.\n", i);
    sprintf(buf1, "%d", var2);
    scanf(" %10s", buf0);
    strtok(buf0, "\n");
    if (strcmp(buf1, buf0) != 0) {
      printf("Well that didn't take long.\n");
      printf("You should have used %s.\n", buf1);
      exit(0);
    }
  }
  printf("How very unpredictable. Level Cleared\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%ebx
  lea    D29(%rip),%r15
  lea    D41(%rip),%r14
  lea    D3(%rip),%r13
  jmp    L15
L43:
  lea    D60(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L48:
  add    $0x1,%ebx
  cmp    $0x32,%ebx
  je     L10
L15:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ebp
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movslq %ebp,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x64,%eax,%eax
  mov    %ebp,%edx
  sub    %eax,%edx
  lea    0x200(%rsp),%r12
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L48
  jmp    L43
L10:
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x408,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D29
type: string
name: extract_symbolic_data
arguments: label: D41
type: string
name: extract_symbolic_data
arguments: label: D3
type: string
name: extract_symbolic_data
arguments: label: D60
type: string
name: extract_symbolic_data
arguments: label: D33
type: string
name: extract_symbolic_data
arguments: label: D44
type: string
name: extract_symbolic_data
arguments: label: D18
type: string

tool: D29: %d days without an incident.


tool: D41: %d

tool: D3:  %10s

tool: D60: Well that didn't take long.

tool: D33: You should have used %s.


tool: D44: 


tool: D18: How very unpredictable. Level Cleared

assistant: ```c
int main() {
  int i;
  time_t var0 = time(NULL);
  srand(var0);
  int var1 = 0;
  char buf0[500];
  char buf1[500];
  for (i = 0; i < 50; i++) {
    int var2 = rand() % 100;
    printf("%d days without an incident.\n", i);
    sprintf(buf1, "%d", var2);
    scanf(" %10s", buf0);
    strtok(buf0, "\n");
    if (strcmp(buf1, buf0) != 0) {
      printf("Well that didn't take long.\n");
      printf("You should have used %s.\n", buf1);
      exit(0);
    }
  }
  printf("How very unpredictable. Level Cleared\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D61(%rip),%r15
  push   %r14
  lea    D19(%rip),%r14
  push   %r13
  lea    D37(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x210(%rsp),%r12
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  jmp    L39
  nopl   0x0(%rax)
L14:
  add    $0x1,%ebx
  cmp    $0x32,%ebx
  je     L37
L39:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    %eax,0xc(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0xc(%rsp),%rax
  mov    %r14,%rsi
  mov    %r12,%rdi
  mov    %rax,%rdx
  imul   $0x51eb851f,%rax,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sar    $0x25,%rax
  sub    %ecx,%eax
  imul   $0x64,%eax,%eax
  sub    %eax,%edx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L14
  lea    D23(%rip),%rdi
  add    $0x1,%ebx
  call   <puts@plt>
  mov    %r12,%rsi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  cmp    $0x32,%ebx
  jne    L39
L37:
  lea    D49(%rip),%rdi
  call   <puts@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x418,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D61
type_infered: string
name: parse_data
arguments: label: D19
type_infered: string
name: parse_data
arguments: label: D37
type_infered: string
name: parse_data
arguments: label: D3
type_infered: string
name: parse_data
arguments: label: D23
type_infered: string
name: parse_data
arguments: label: D13
type_infered: string
name: parse_data
arguments: label: D49
type_infered: string

tool: D61: %d days without an incident.


tool: D19: %d

tool: D37:  %10s

tool: D3: 


tool: D23: Well that didn't take long.

tool: D13: You should have used %s.


tool: D49: How very unpredictable. Level Cleared

assistant: ```c
int main() {
  int i;
  time_t var0 = time(NULL);
  srand(var0);
  int var1 = 0;
  char buf0[500];
  char buf1[500];
  for (i = 0; i < 50; i++) {
    int var2 = rand() % 100;
    printf("%d days without an incident.\n", i);
    sprintf(buf1, "%d", var2);
    scanf(" %10s", buf0);
    strtok(buf0, "\n");
    if (strcmp(buf1, buf0) != 0) {
      printf("Well that didn't take long.\n");
      printf("You should have used %s.\n", buf1);
      exit(0);
    }
  }
  printf("How very unpredictable. Level Cleared\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D38(%rip),%r15
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x210(%rsp),%r12
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  jmp    L6
  nopl   0x0(%rax)
L9:
  add    $0x1,%ebx
  cmp    $0x32,%ebx
  je     L32
L6:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    %eax,0xc(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0xc(%rsp),%rax
  mov    %r14,%rsi
  mov    %r12,%rdi
  mov    %rax,%rdx
  imul   $0x51eb851f,%rax,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sar    $0x25,%rax
  sub    %ecx,%eax
  imul   $0x64,%eax,%eax
  sub    %eax,%edx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D51(%rip),%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L9
  lea    D10(%rip),%rdi
  add    $0x1,%ebx
  call   <puts@plt>
  mov    %r12,%rsi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  cmp    $0x32,%ebx
  jne    L6
L32:
  lea    D67(%rip),%rdi
  call   <puts@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x418,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D38
type_guessed: string
name: unpack_data
arguments: tag: D2
type_guessed: string
name: unpack_data
arguments: tag: D24
type_guessed: string
name: unpack_data
arguments: tag: D51
type_guessed: string
name: unpack_data
arguments: tag: D10
type_guessed: string
name: unpack_data
arguments: tag: D7
type_guessed: string
name: unpack_data
arguments: tag: D67
type_guessed: string

tool: D38: %d days without an incident.


tool: D2: %d

tool: D24:  %10s

tool: D51: 


tool: D10: Well that didn't take long.

tool: D7: You should have used %s.


tool: D67: How very unpredictable. Level Cleared

assistant: ```c
int main() {
  int i;
  time_t var0 = time(NULL);
  srand(var0);
  int var1 = 0;
  char buf0[500];
  char buf1[500];
  for (i = 0; i < 50; i++) {
    int var2 = rand() % 100;
    printf("%d days without an incident.\n", i);
    sprintf(buf1, "%d", var2);
    scanf(" %10s", buf0);
    strtok(buf0, "\n");
    if (strcmp(buf1, buf0) != 0) {
      printf("Well that didn't take long.\n");
      printf("You should have used %s.\n", buf1);
      exit(0);
    }
  }
  printf("How very unpredictable. Level Cleared\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<sqrtf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movss  %xmm0,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  ucomiss -0x24(%rbp),%xmm0
  jp     L21
  pxor   %xmm0,%xmm0
  ucomiss -0x24(%rbp),%xmm0
  jne    L21
  mov    -0x24(%rbp),%eax
  jmp    L40
L21:
  movss  D32(%rip),%xmm0
  ucomiss -0x24(%rbp),%xmm0
  jp     L30
  movss  D32(%rip),%xmm0
  ucomiss -0x24(%rbp),%xmm0
  jne    L30
  mov    D32(%rip),%eax
  jmp    L40
L30:
  pxor   %xmm0,%xmm0
  comiss -0x24(%rbp),%xmm0
  jbe    L33
  mov    D6(%rip),%eax
  jmp    L40
L33:
  lea    -0x14(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rdi
  movd   %eax,%xmm0
  call   <frexpf@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0xc(%rbp),%xmm1
  movsd  D39(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  D9(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  -0xc(%rbp),%xmm0
  divss  -0x10(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    $0xfffffffe,%edi
  movd   %eax,%xmm0
  call   <ldexpf@plt>
  movss  -0xc(%rbp),%xmm1
  divss  -0x10(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L28
  pxor   %xmm1,%xmm1
  cvtss2sd -0x10(%rbp),%xmm1
  movsd  D53(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
L28:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%edi
  movd   %eax,%xmm0
  call   <ldexpf@plt>
  movd   %xmm0,%eax
L40:
  movd   %eax,%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D32
data_type: f32
name: unpack_data
arguments: data_label: D6
data_type: f32
name: unpack_data
arguments: data_label: D39
data_type: f64
name: unpack_data
arguments: data_label: D9
data_type: f64
name: unpack_data
arguments: data_label: D53
data_type: f64

tool: D32: 1

tool: D6: 0

tool: D39: 0.59016

tool: D9: 0.41731

tool: D53: 0.7071067812

assistant: ```c
float sqrtf(const float x) {
  float f, y;
  int n;
  if (x == 0.0)
    return x;
  else if (x == 1.0)
    return 1.0;
  else if (x < 0.0) {
    return 0.0;
  }
  f = frexpf(x, &n);
  y = 0.41731 + 0.59016 * f;
  y += f / y;
  y = ldexpf(y, -2) + f / y;
  if (n & 1) {
    y *= 0.7071067812;
    ++n;
  }
  return ldexpf(y, n / 2);
}
```
user: What is the c source code of the assembly code below?

```asm
<sqrtf>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  ucomiss D35(%rip),%xmm0
  jp     L34
  je     L9
L34:
  ucomiss D28(%rip),%xmm0
  jp     L30
  je     L9
L30:
  pxor   %xmm1,%xmm1
  comiss %xmm0,%xmm1
  ja     L3
  lea    0x14(%rsp),%rdi
  call   <frexpf@plt>
  movaps %xmm0,%xmm2
  cvtss2sd %xmm0,%xmm0
  mulsd  D26(%rip),%xmm0
  addsd  D40(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm2,0x8(%rsp)
  movaps %xmm2,%xmm1
  divss  %xmm0,%xmm1
  movaps %xmm1,%xmm3
  addss  %xmm0,%xmm3
  mov    $0xfffffffe,%edi
  movss  %xmm3,0xc(%rsp)
  movaps %xmm3,%xmm0
  call   <ldexpf@plt>
  movss  0x8(%rsp),%xmm1
  divss  0xc(%rsp),%xmm1
  addss  %xmm1,%xmm0
  mov    0x14(%rsp),%eax
  test   $0x1,%al
  je     L44
  cvtss2sd %xmm0,%xmm0
  mulsd  D12(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  add    $0x1,%eax
  mov    %eax,0x14(%rsp)
L44:
  mov    0x14(%rsp),%eax
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  call   <ldexpf@plt>
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x28,%rsp
  ret
L3:
  pxor   %xmm0,%xmm0
  jmp    L9
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D35
type: f32
name: retrieve_data
arguments: tag: D28
type: f32
name: retrieve_data
arguments: tag: D26
type: f64
name: retrieve_data
arguments: tag: D40
type: f64
name: retrieve_data
arguments: tag: D12
type: f64

tool: D35: 0

tool: D28: 1

tool: D26: 0.59016

tool: D40: 0.41731

tool: D12: 0.7071067812

assistant: ```c
float sqrtf(const float x) {
  float f, y;
  int n;
  if (x == 0.0)
    return x;
  else if (x == 1.0)
    return 1.0;
  else if (x < 0.0) {
    return 0.0;
  }
  f = frexpf(x, &n);
  y = 0.41731 + 0.59016 * f;
  y += f / y;
  y = ldexpf(y, -2) + f / y;
  if (n & 1) {
    y *= 0.7071067812;
    ++n;
  }
  return ldexpf(y, n / 2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sqrtf>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm1,%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  ucomiss %xmm1,%xmm0
  jp     L43
  je     L19
L43:
  ucomiss D34(%rip),%xmm0
  jp     L80
  je     L19
L80:
  comiss %xmm0,%xmm1
  ja     L70
  lea    0x14(%rsp),%rdi
  call   <frexpf@plt>
  pxor   %xmm1,%xmm1
  mov    $0xfffffffe,%edi
  cvtss2sd %xmm0,%xmm1
  mulsd  D22(%rip),%xmm1
  movss  %xmm0,0xc(%rsp)
  addsd  D38(%rip),%xmm1
  cvtsd2ss %xmm1,%xmm1
  divss  %xmm1,%xmm0
  addss  %xmm0,%xmm1
  movaps %xmm1,%xmm0
  movss  %xmm1,0x8(%rsp)
  call   <ldexpf@plt>
  movss  0x8(%rsp),%xmm1
  movss  0xc(%rsp),%xmm2
  mov    0x14(%rsp),%eax
  divss  %xmm1,%xmm2
  addss  %xmm2,%xmm0
  test   $0x1,%al
  jne    L58
L56:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L79
  mov    %eax,%edi
  add    $0x28,%rsp
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  jmp    1090 <ldexpf@plt>
  nopl   0x0(%rax)
L70:
  movaps %xmm1,%xmm0
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  add    $0x28,%rsp
  ret
L58:
  cvtss2sd %xmm0,%xmm0
  mulsd  D19(%rip),%xmm0
  add    $0x1,%eax
  cvtsd2ss %xmm0,%xmm0
  jmp    L56
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D34
type_guessed: f32
name: unpack_data
arguments: tag: D22
type_guessed: f64
name: unpack_data
arguments: tag: D38
type_guessed: f64
name: unpack_data
arguments: tag: D19
type_guessed: f64

tool: D34: 1

tool: D22: 0.59016

tool: D38: 0.41731

tool: D19: 0.7071067812

assistant: ```c
float sqrtf(const float x) {
  float f, y;
  int n;
  if (x == 0.0)
    return x;
  else if (x == 1.0)
    return 1.0;
  else if (x < 0.0) {
    return 0.0;
  }
  f = frexpf(x, &n);
  y = 0.41731 + 0.59016 * f;
  y += f / y;
  y = ldexpf(y, -2) + f / y;
  if (n & 1) {
    y *= 0.7071067812;
    ++n;
  }
  return ldexpf(y, n / 2);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sqrtf>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm1,%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  ucomiss %xmm1,%xmm0
  jp     L2
  je     L25
L2:
  ucomiss D10(%rip),%xmm0
  jp     L73
  je     L25
L73:
  comiss %xmm0,%xmm1
  ja     L7
  lea    0x14(%rsp),%rdi
  call   <frexpf@plt>
  pxor   %xmm1,%xmm1
  mov    $0xfffffffe,%edi
  cvtss2sd %xmm0,%xmm1
  mulsd  D39(%rip),%xmm1
  movss  %xmm0,0xc(%rsp)
  addsd  D33(%rip),%xmm1
  cvtsd2ss %xmm1,%xmm1
  divss  %xmm1,%xmm0
  addss  %xmm0,%xmm1
  movaps %xmm1,%xmm0
  movss  %xmm1,0x8(%rsp)
  call   <ldexpf@plt>
  movss  0x8(%rsp),%xmm1
  movss  0xc(%rsp),%xmm2
  mov    0x14(%rsp),%eax
  divss  %xmm1,%xmm2
  addss  %xmm2,%xmm0
  test   $0x1,%al
  jne    L56
L69:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  mov    %eax,%edi
  add    $0x28,%rsp
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  jmp    1090 <ldexpf@plt>
  nopl   0x0(%rax)
L7:
  movaps %xmm1,%xmm0
L25:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x28,%rsp
  ret
L56:
  cvtss2sd %xmm0,%xmm0
  mulsd  D2(%rip),%xmm0
  add    $0x1,%eax
  cvtsd2ss %xmm0,%xmm0
  jmp    L69
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
type_infered: f32
name: retrieve_data
arguments: data_tag: D39
type_infered: f64
name: retrieve_data
arguments: data_tag: D33
type_infered: f64
name: retrieve_data
arguments: data_tag: D2
type_infered: f64

tool: D10: 1

tool: D39: 0.59016

tool: D33: 0.41731

tool: D2: 0.7071067812

assistant: ```c
float sqrtf(const float x) {
  float f, y;
  int n;
  if (x == 0.0)
    return x;
  else if (x == 1.0)
    return 1.0;
  else if (x < 0.0) {
    return 0.0;
  }
  f = frexpf(x, &n);
  y = 0.41731 + 0.59016 * f;
  y += f / y;
  y = ldexpf(y, -2) + f / y;
  if (n & 1) {
    y *= 0.7071067812;
    ++n;
  }
  return ldexpf(y, n / 2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<find_character>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <string_length@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L2
L12:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L35
  mov    -0x8(%rbp),%eax
  jmp    L4
L35:
  addl   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L12
  mov    $0xffffffff,%eax
L4:
  leave
  ret
```

assistant: ```c
int find_character(char buffer[], char ch) {
  int length = string_length(buffer), count;
  for (count = 0; count < length; count++) {
    if (buffer[count] == ch)
      return count;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<find_character>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <string_length@plt>
  test   %eax,%eax
  jle    L35
  mov    %eax,%eax
  mov    $0x0,%edx
L10:
  mov    %edx,%ecx
  cmp    %bpl,(%rbx,%rdx,1)
  je     L14
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L10
  mov    $0xffffffff,%ecx
L14:
  mov    %ecx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L35:
  mov    $0xffffffff,%ecx
  jmp    L14
```

assistant: ```c
int find_character(char buffer[], char ch) {
  int length = string_length(buffer), count;
  for (count = 0; count < length; count++) {
    if (buffer[count] == ch)
      return count;
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<find_character>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <string_length@plt>
  test   %eax,%eax
  jle    L9
  cltq
  xor    %edx,%edx
  jmp    L29
  xchg   %ax,%ax
L38:
  add    $0x1,%rdx
  cmp    %rax,%rdx
  je     L9
L29:
  mov    %edx,%r8d
  cmp    %bpl,(%rbx,%rdx,1)
  jne    L38
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  add    $0x8,%rsp
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int find_character(char buffer[], char ch) {
  int length = string_length(buffer), count;
  for (count = 0; count < length; count++) {
    if (buffer[count] == ch)
      return count;
  }
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<find_character>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <string_length@plt>
  test   %eax,%eax
  jle    L9
  cltq
  xor    %edx,%edx
  jmp    L21
  xchg   %ax,%ax
L3:
  add    $0x1,%rdx
  cmp    %rax,%rdx
  je     L9
L21:
  mov    %edx,%r8d
  cmp    %bpl,(%rbx,%rdx,1)
  jne    L3
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  add    $0x8,%rsp
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int find_character(char buffer[], char ch) {
  int length = string_length(buffer), count;
  for (count = 0; count < length; count++) {
    if (buffer[count] == ch)
      return count;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_func_01>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *test_func_01(char *p, int *x) {
  int r = 0;
  r += *p++;
  r += *p++;
  r += *p++;
  *x = r;
  return p;
}
```