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
<main>:
  endbr64
  push   %rbx
  mov    $0x5a,%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D66(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x32,%esi
  lea    D88(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D1(%rip),%rbx
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    $0x1b,%edi
  call   <f@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  movsd  D22(%rip),%xmm1
  movsd  D58(%rip),%xmm0
  call   <g@plt>
  cvtss2sd %xmm0,%xmm0
  lea    D83(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    $0x24,%esi
  movsd  D25(%rip),%xmm0
  mov    $0x4,%edi
  call   <h@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D8
type_infered: string
name: read_data
arguments: label: D66
type_infered: string
name: read_data
arguments: label: D17
type_infered: string
name: read_data
arguments: label: D88
type_infered: string
name: read_data
arguments: label: D1
type_infered: string
name: read_data
arguments: label: D22
type_infered: f64
name: read_data
arguments: label: D58
type_infered: f64
name: read_data
arguments: label: D83
type_infered: string
name: read_data
arguments: label: D25
type_infered: f64

tool: D8: (a + b) * c / d = %d


tool: D66: ((a + b) * c) / d = %d


tool: D17: (a + b) * (c / d) = %d


tool: D88: a + (b * c) / d = %d


tool: D1: 


tool: D22: 6.6

tool: D58: 1.9

tool: D83: %f


tool: D25: 2.5

assistant: ```c
int main(int argc, char **argv) {
  int a = 20;
  int b = 10;
  int c = 15;
  int d = 5;
  int e;
  e = (a + b) * c / d;
  printf("(a + b) * c / d = %d\n", e);
  e = ((a + b) * c) / d;
  printf("((a + b) * c) / d = %d\n", e);
  e = (a + b) * (c / d);
  printf("(a + b) * (c / d) = %d\n", e);
  e = a + (b * c) / d;
  printf("a + (b * c) / d = %d\n", e);
  printf("\n\n");
  f(27);
  printf("\n\n");
  float x = g(1.9, 6.6);
  printf("%f\n", x);
  printf("\n\n");
  char *y = h(4, 2.5, '$');
  printf("%s\n", y);
  free(y);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x5a,%esi
  lea    D87(%rip),%rdi
  xor    %eax,%eax
  lea    D15(%rip),%rbp
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D73(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x32,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1b,%edi
  call   <f@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  movsd  D29(%rip),%xmm1
  movsd  D76(%rip),%xmm0
  call   <g@plt>
  lea    D41(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x24,%esi
  mov    $0x4,%edi
  movsd  D52(%rip),%xmm0
  call   <h@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D87
type_infered: string
name: retrieve_data
arguments: label: D15
type_infered: string
name: retrieve_data
arguments: label: D73
type_infered: string
name: retrieve_data
arguments: label: D58
type_infered: string
name: retrieve_data
arguments: label: D9
type_infered: string
name: retrieve_data
arguments: label: D29
type_infered: f64
name: retrieve_data
arguments: label: D76
type_infered: f64
name: retrieve_data
arguments: label: D41
type_infered: string
name: retrieve_data
arguments: label: D52
type_infered: f64

tool: D87: (a + b) * c / d = %d


tool: D15: 


tool: D73: ((a + b) * c) / d = %d


tool: D58: (a + b) * (c / d) = %d


tool: D9: a + (b * c) / d = %d


tool: D29: 6.6

tool: D76: 1.9

tool: D41: %f


tool: D52: 2.5

assistant: ```c
int main(int argc, char **argv) {
  int a = 20;
  int b = 10;
  int c = 15;
  int d = 5;
  int e;
  e = (a + b) * c / d;
  printf("(a + b) * c / d = %d\n", e);
  e = ((a + b) * c) / d;
  printf("((a + b) * c) / d = %d\n", e);
  e = (a + b) * (c / d);
  printf("(a + b) * (c / d) = %d\n", e);
  e = a + (b * c) / d;
  printf("a + (b * c) / d = %d\n", e);
  printf("\n\n");
  f(27);
  printf("\n\n");
  float x = g(1.9, 6.6);
  printf("%f\n", x);
  printf("\n\n");
  char *y = h(4, 2.5, '$');
  printf("%s\n", y);
  free(y);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x5a,%esi
  lea    D80(%rip),%rdi
  xor    %eax,%eax
  lea    D24(%rip),%rbp
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5a,%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x32,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1b,%edi
  call   <f@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  movsd  D48(%rip),%xmm1
  movsd  D40(%rip),%xmm0
  call   <g@plt>
  lea    D55(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x24,%esi
  mov    $0x4,%edi
  movsd  D71(%rip),%xmm0
  call   <h@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D80
datatype: string
name: unpack_data
arguments: data_label: D24
datatype: string
name: unpack_data
arguments: data_label: D77
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D15
datatype: string
name: unpack_data
arguments: data_label: D48
datatype: f64
name: unpack_data
arguments: data_label: D40
datatype: f64
name: unpack_data
arguments: data_label: D55
datatype: string
name: unpack_data
arguments: data_label: D71
datatype: f64

tool: D80: (a + b) * c / d = %d


tool: D24: 


tool: D77: ((a + b) * c) / d = %d


tool: D32: (a + b) * (c / d) = %d


tool: D15: a + (b * c) / d = %d


tool: D48: 6.6

tool: D40: 1.9

tool: D55: %f


tool: D71: 2.5

assistant: ```c
int main(int argc, char **argv) {
  int a = 20;
  int b = 10;
  int c = 15;
  int d = 5;
  int e;
  e = (a + b) * c / d;
  printf("(a + b) * c / d = %d\n", e);
  e = ((a + b) * c) / d;
  printf("((a + b) * c) / d = %d\n", e);
  e = (a + b) * (c / d);
  printf("(a + b) * (c / d) = %d\n", e);
  e = a + (b * c) / d;
  printf("a + (b * c) / d = %d\n", e);
  printf("\n\n");
  f(27);
  printf("\n\n");
  float x = g(1.9, 6.6);
  printf("%f\n", x);
  printf("\n\n");
  char *y = h(4, 2.5, '$');
  printf("%s\n", y);
  free(y);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<perror_msg_and_die>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <verror_msg@plt>
  mov    $0x0,%eax
  call   <die@plt>
  nop
  leave
  ret
```

assistant: ```c
void perror_msg_and_die(const char *fmt, ...) {
  va_list p;
  va_start(p, fmt);
  verror_msg(errno, fmt, p);
  die();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<perror_msg_and_die>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <verror_msg@plt>
  mov    $0x0,%eax
  call   <die@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void perror_msg_and_die(const char *fmt, ...) {
  va_list p;
  va_start(p, fmt);
  verror_msg(errno, fmt, p);
  die();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<perror_msg_and_die>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D2(%rip),%rax
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <verror_msg@plt>
  xor    %eax,%eax
  call   <die@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void perror_msg_and_die(const char *fmt, ...) {
  va_list p;
  va_start(p, fmt);
  verror_msg(errno, fmt, p);
  die();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<perror_msg_and_die>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D9(%rip),%rax
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <verror_msg@plt>
  xor    %eax,%eax
  call   <die@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void perror_msg_and_die(const char *fmt, ...) {
  va_list p;
  va_start(p, fmt);
  verror_msg(errno, fmt, p);
  die();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x8c(%rbp)
L27:
  cmpl   $0x0,-0x8c(%rbp)
  jg     L24
  addl   $0x1,-0x8c(%rbp)
  jmp    L27
L24:
  nop
  movl   $0x0,-0x8c(%rbp)
L53:
  cmpl   $0x0,-0x8c(%rbp)
  jg     L97
  addl   $0x1,-0x8c(%rbp)
  jmp    L53
L97:
  nop
  movl   $0x0,-0x8c(%rbp)
L123:
  addl   $0x1,-0x8c(%rbp)
  cmpl   $0x3,-0x8c(%rbp)
  jle    L123
  jmp    L214
L224:
  addl   $0x1,-0x8c(%rbp)
L214:
  cmpl   $0x7,-0x8c(%rbp)
  jle    L224
  movsd  D26(%rip),%xmm0
  movsd  %xmm0,-0x88(%rbp)
  jmp    L102
L84:
  movsd  -0x88(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x8c(%rbp)
  mov    -0x8c(%rbp),%eax
  cltq
  movl   $0x1,-0x80(%rbp,%rax,4)
  movsd  -0x88(%rbp),%xmm1
  movsd  D8(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x88(%rbp)
L102:
  movsd  D2(%rip),%xmm0
  comisd -0x88(%rbp),%xmm0
  ja     L84
  movl   $0x1,-0x8c(%rbp)
  jmp    L99
L92:
  cmpl   $0x10,-0x8c(%rbp)
  je     L82
  addl   $0x1,-0x8c(%rbp)
L99:
  cmpl   $0x1d,-0x8c(%rbp)
  jle    L92
  jmp    L236
L82:
  nop
L236:
  movl   $0x1,-0x8c(%rbp)
  jmp    L85
L172:
  addl   $0x1,-0x8c(%rbp)
L85:
  cmpl   $0x9,-0x8c(%rbp)
  jle    L172
  movl   $0x1,-0x8c(%rbp)
  jmp    L0
L79:
  mov    -0x8c(%rbp),%eax
  cltq
  mov    -0x8c(%rbp),%edx
  mov    %edx,-0x80(%rbp,%rax,4)
  addl   $0x2,-0x8c(%rbp)
L0:
  cmpl   $0x9,-0x8c(%rbp)
  jle    L79
  movl   $0x1,-0x7c(%rbp)
  movl   $0x1,-0x8c(%rbp)
  jmp    L55
L23:
  addl   $0x1,-0x8c(%rbp)
L55:
  cmpl   $0x9,-0x8c(%rbp)
  jle    L23
  movl   $0x1,-0x8c(%rbp)
  jmp    L5
L42:
  addl   $0x1,-0x8c(%rbp)
L5:
  cmpl   $0x9,-0x8c(%rbp)
  jle    L42
  movl   $0x1,-0x8c(%rbp)
  jmp    L51
L37:
  addl   $0x1,-0x8c(%rbp)
L51:
  cmpl   $0x9,-0x8c(%rbp)
  jle    L37
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L175
  call   <__stack_chk_fail@plt>
L175:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D26
type: byte[8]
name: retrieve_data
arguments: label: D8
type: byte[8]
name: retrieve_data
arguments: label: D2
type: byte[8]

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x18,0x40]}

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x3e,0x40]}

assistant: ```c
int test1() {
  int i, j, k, b[10];
  int a[30];
  double d;
  float r;
  i = 0;
#pragma acc loop
  while (1) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  for (;;) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  do {
    i = i + 1;
  } while (i < 4);
#pragma acc loop
  while (i < 8) {
    i = i + 1;
  }
#pragma acc loop
  for (d = 1; d < 30; d += 6) {
    i = d;
    a[i] = 1;
  }
#pragma acc loop
  for (i = 1; i < 30; i++)
    if (i == 16)
      break;
#pragma acc loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc loop
  for (i = 1; i < 10; i += 2) {
    a[i] = i;
  }
#pragma acc loop
  a[1] = 1;
  for (i = 1; i < 10; i++)
    ;
#pragma acc parallel loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc kernels loop
  for (i = 1; i < 10; i++) {
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test1>:
  endbr64
  movsd  D20(%rip),%xmm0
  movsd  D12(%rip),%xmm2
  movsd  D9(%rip),%xmm1
L29:
  addsd  %xmm2,%xmm0
  comisd %xmm0,%xmm1
  ja     L29
  mov    $0x1,%eax
L7:
  add    $0x1,%eax
  cmp    $0x10,%eax
  je     L26
  cmp    $0x1d,%eax
  jle    L7
L26:
  mov    $0x0,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D20
type_guessed: byte[8]
name: parse_data
arguments: data_tag: D12
type_guessed: byte[8]
name: parse_data
arguments: data_tag: D9
type_guessed: byte[8]

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0x18,0x40]}

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x3e,0x40]}

assistant: ```c
int test1() {
  int i, j, k, b[10];
  int a[30];
  double d;
  float r;
  i = 0;
#pragma acc loop
  while (1) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  for (;;) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  do {
    i = i + 1;
  } while (i < 4);
#pragma acc loop
  while (i < 8) {
    i = i + 1;
  }
#pragma acc loop
  for (d = 1; d < 30; d += 6) {
    i = d;
    a[i] = 1;
  }
#pragma acc loop
  for (i = 1; i < 30; i++)
    if (i == 16)
      break;
#pragma acc loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc loop
  for (i = 1; i < 10; i += 2) {
    a[i] = i;
  }
#pragma acc loop
  a[1] = 1;
  for (i = 1; i < 10; i++)
    ;
#pragma acc parallel loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc kernels loop
  for (i = 1; i < 10; i++) {
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test1>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int test1() {
  int i, j, k, b[10];
  int a[30];
  double d;
  float r;
  i = 0;
#pragma acc loop
  while (1) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  for (;;) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  do {
    i = i + 1;
  } while (i < 4);
#pragma acc loop
  while (i < 8) {
    i = i + 1;
  }
#pragma acc loop
  for (d = 1; d < 30; d += 6) {
    i = d;
    a[i] = 1;
  }
#pragma acc loop
  for (i = 1; i < 30; i++)
    if (i == 16)
      break;
#pragma acc loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc loop
  for (i = 1; i < 10; i += 2) {
    a[i] = i;
  }
#pragma acc loop
  a[1] = 1;
  for (i = 1; i < 10; i++)
    ;
#pragma acc parallel loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc kernels loop
  for (i = 1; i < 10; i++) {
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test1>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int test1() {
  int i, j, k, b[10];
  int a[30];
  double d;
  float r;
  i = 0;
#pragma acc loop
  while (1) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  for (;;) {
    if (i > 0)
      break;
    i = i + 1;
  }
  i = 0;
#pragma acc loop
  do {
    i = i + 1;
  } while (i < 4);
#pragma acc loop
  while (i < 8) {
    i = i + 1;
  }
#pragma acc loop
  for (d = 1; d < 30; d += 6) {
    i = d;
    a[i] = 1;
  }
#pragma acc loop
  for (i = 1; i < 30; i++)
    if (i == 16)
      break;
#pragma acc loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc loop
  for (i = 1; i < 10; i += 2) {
    a[i] = i;
  }
#pragma acc loop
  a[1] = 1;
  for (i = 1; i < 10; i++)
    ;
#pragma acc parallel loop
  for (i = 1; i < 10; i++) {
  }
#pragma acc kernels loop
  for (i = 1; i < 10; i++) {
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<satsb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xffffff80,-0x4(%rbp)
  jge    L23
  mov    $0xffffff80,%eax
  jmp    L27
L23:
  cmpl   $0x7f,-0x4(%rbp)
  jle    L15
  mov    $0x7f,%eax
  jmp    L27
L15:
  mov    -0x4(%rbp),%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int satsb(int x) {
  if (x < -128)
    return -128;
  else if (x > 127)
    return 127;
  else
    return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<satsb>:
  endbr64
  mov    $0x7f,%eax
  cmp    %eax,%edi
  cmovle %edi,%eax
  mov    $0xffffff80,%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
```

assistant: ```c
int satsb(int x) {
  if (x < -128)
    return -128;
  else if (x > 127)
    return 127;
  else
    return x;
}
```
user: What is the c source code of the assembly code below?

```asm
<satsb>:
  endbr64
  mov    $0x7f,%eax
  mov    $0xffffff80,%edx
  cmp    %eax,%edi
  cmovle %edi,%eax
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
```

assistant: ```c
int satsb(int x) {
  if (x < -128)
    return -128;
  else if (x > 127)
    return 127;
  else
    return x;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<satsb>:
  endbr64
  mov    $0x7f,%eax
  mov    $0xffffff80,%edx
  cmp    %eax,%edi
  cmovle %edi,%eax
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
```

assistant: ```c
int satsb(int x) {
  if (x < -128)
    return -128;
  else if (x > 127)
    return 127;
  else
    return x;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<roundl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  fldt   0x10(%rbp)
  fstpt  -0x10(%rbp)
  mov    -0xc(%rbp),%eax
  sar    $0x14,%eax
  and    $0x7ff,%eax
  sub    $0x3ff,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x34,-0x14(%rbp)
  jle    L78
  fldt   0x10(%rbp)
  jmp    L20
L78:
  cmpl   $0xffffffff,-0x14(%rbp)
  jge    L33
  mov    -0x10(%rbp),%rax
  movabs $0x8000000000000000,%rdx
  and    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  fldt   -0x10(%rbp)
  jmp    L20
L33:
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L81
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jns    L52
  fld1
  fchs
  jmp    L20
L52:
  fld1
  jmp    L20
L81:
  mov    -0x10(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  movabs $0x8000000000000,%rdx
  mov    %eax,%ecx
  sar    %cl,%rdx
  mov    %rdx,%rax
  and    %rsi,%rax
  test   %rax,%rax
  je     L51
  mov    -0x10(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  movabs $0x10000000000000,%rdx
  mov    %eax,%ecx
  sar    %cl,%rdx
  mov    %rdx,%rax
  add    %rsi,%rax
  mov    %rax,-0x10(%rbp)
L51:
  mov    -0x10(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  movabs $0xfffffffffffff,%rdx
  mov    %eax,%ecx
  sar    %cl,%rdx
  mov    %rdx,%rax
  not    %rax
  and    %rsi,%rax
  mov    %rax,-0x10(%rbp)
  fldt   -0x10(%rbp)
L20:
  pop    %rbp
  ret
```

assistant: ```c
long double roundl(long double x) {
  union {
    long double f;
    long i;
    struct {
      int i0, i1;
    };
  } u = {x};
  int exp = ((u.i1 >> 20) & 0x7FF) - 1023;
  if (exp >= 53)
    return x;
  if (exp < -1)
    return u.i &= 0x8000000000000000, u.f;
  if (exp == -1)
    return u.i1 & 0x80000000 ? -1.0l : 1.0l;
  if (u.i & (0x0008000000000000 >> exp))
    u.i += 0x0010000000000000 >> exp;
  return u.i &= ~(0x000FFFFFFFFFFFFF >> exp), u.f;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<roundl>:
  endbr64
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%edx
  mov    %rax,-0x18(%rsp)
  mov    %edx,-0x10(%rsp)
  mov    %rax,%rcx
  mov    %rax,%rsi
  sar    $0x20,%rsi
  sar    $0x34,%rcx
  and    $0x7ff,%ecx
  sub    $0x3ff,%ecx
  mov    %rax,-0x28(%rsp)
  mov    %edx,-0x20(%rsp)
  fldt   -0x28(%rsp)
  cmp    $0x34,%ecx
  jg     L1
  fstp   %st(0)
  cmp    $0xffffffff,%ecx
  jl     L40
  je     L46
  movabs $0x8000000000000,%rdx
  sar    %cl,%rdx
  test   %rax,%rdx
  je     L9
  movabs $0x10000000000000,%rdx
  sar    %cl,%rdx
  add    %rax,%rdx
  mov    %rdx,-0x18(%rsp)
L9:
  movabs $0xfff0000000000000,%rax
  sar    %cl,%rax
  and    %rax,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
L40:
  movabs $0x8000000000000000,%rdx
  and    %rdx,%rax
  mov    %rax,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
L46:
  test   %esi,%esi
  js     L43
  fld1
  ret
L43:
  fld1
  fchs
L1:
  ret
```

assistant: ```c
long double roundl(long double x) {
  union {
    long double f;
    long i;
    struct {
      int i0, i1;
    };
  } u = {x};
  int exp = ((u.i1 >> 20) & 0x7FF) - 1023;
  if (exp >= 53)
    return x;
  if (exp < -1)
    return u.i &= 0x8000000000000000, u.f;
  if (exp == -1)
    return u.i1 & 0x80000000 ? -1.0l : 1.0l;
  if (u.i & (0x0008000000000000 >> exp))
    u.i += 0x0010000000000000 >> exp;
  return u.i &= ~(0x000FFFFFFFFFFFFF >> exp), u.f;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<roundl>:
  endbr64
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%edx
  mov    %rax,%rcx
  mov    %rax,%rdi
  mov    %rax,-0x28(%rsp)
  mov    %rax,%rsi
  sar    $0x34,%rcx
  mov    %edx,-0x20(%rsp)
  sar    $0x20,%rdi
  mov    %rax,-0x18(%rsp)
  fldt   -0x28(%rsp)
  and    $0x7ff,%ecx
  sub    $0x3ff,%ecx
  mov    %edx,-0x10(%rsp)
  cmp    $0x34,%ecx
  jg     L31
  fstp   %st(0)
  cmp    $0xffffffff,%ecx
  jl     L40
  je     L32
  movabs $0x8000000000000,%rax
  sar    %cl,%rax
  test   %rsi,%rax
  je     L44
  movabs $0x10000000000000,%rax
  sar    %cl,%rax
  add    %rax,%rsi
L44:
  movabs $0xfff0000000000000,%rax
  sar    %cl,%rax
  and    %rsi,%rax
  mov    %rax,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  fchs
L31:
  ret
  nopl   0x0(%rax,%rax,1)
L40:
  movabs $0x8000000000000000,%rax
  and    %rax,%rsi
  mov    %rsi,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
  nop
L32:
  fld1
  test   %edi,%edi
  js     L43
  ret
```

assistant: ```c
long double roundl(long double x) {
  union {
    long double f;
    long i;
    struct {
      int i0, i1;
    };
  } u = {x};
  int exp = ((u.i1 >> 20) & 0x7FF) - 1023;
  if (exp >= 53)
    return x;
  if (exp < -1)
    return u.i &= 0x8000000000000000, u.f;
  if (exp == -1)
    return u.i1 & 0x80000000 ? -1.0l : 1.0l;
  if (u.i & (0x0008000000000000 >> exp))
    u.i += 0x0010000000000000 >> exp;
  return u.i &= ~(0x000FFFFFFFFFFFFF >> exp), u.f;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<roundl>:
  endbr64
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%edx
  mov    %rax,%rcx
  mov    %rax,%rdi
  mov    %rax,-0x28(%rsp)
  mov    %rax,%rsi
  sar    $0x34,%rcx
  mov    %edx,-0x20(%rsp)
  sar    $0x20,%rdi
  mov    %rax,-0x18(%rsp)
  fldt   -0x28(%rsp)
  and    $0x7ff,%ecx
  sub    $0x3ff,%ecx
  mov    %edx,-0x10(%rsp)
  cmp    $0x34,%ecx
  jg     L30
  fstp   %st(0)
  cmp    $0xffffffff,%ecx
  jl     L12
  je     L8
  movabs $0x8000000000000,%rax
  sar    %cl,%rax
  test   %rsi,%rax
  je     L32
  movabs $0x10000000000000,%rax
  sar    %cl,%rax
  add    %rax,%rsi
L32:
  movabs $0xfff0000000000000,%rax
  sar    %cl,%rax
  and    %rsi,%rax
  mov    %rax,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  fchs
L30:
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  movabs $0x8000000000000000,%rax
  and    %rax,%rsi
  mov    %rsi,-0x18(%rsp)
  fldt   -0x18(%rsp)
  ret
  nop
L8:
  fld1
  test   %edi,%edi
  js     L38
  ret
```

assistant: ```c
long double roundl(long double x) {
  union {
    long double f;
    long i;
    struct {
      int i0, i1;
    };
  } u = {x};
  int exp = ((u.i1 >> 20) & 0x7FF) - 1023;
  if (exp >= 53)
    return x;
  if (exp < -1)
    return u.i &= 0x8000000000000000, u.f;
  if (exp == -1)
    return u.i1 & 0x80000000 ? -1.0l : 1.0l;
  if (u.i & (0x0008000000000000 >> exp))
    u.i += 0x0010000000000000 >> exp;
  return u.i &= ~(0x000FFFFFFFFFFFFF >> exp), u.f;
}
```