ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  pxor   %xmm1,%xmm1
  push   %r12
  lea    D47(%rip),%r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x38(%rsp),%r13
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rdi
  xor    %eax,%eax
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm1
  addss  -0x4(%rbx),%xmm1
  cmp    %r13,%rbx
  jne    L18
  lea    D29(%rip),%rdi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  divss  D66(%rip),%xmm1
  cvtss2sd %xmm1,%xmm0
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  movss  0x38(%rsp),%xmm0
  xor    %eax,%eax
  movss  0xc(%rsp),%xmm1
  lea    D41(%rip),%rdi
  comiss %xmm1,%xmm0
  pxor   %xmm0,%xmm0
  seta   %al
  cvtsi2ss %eax,%xmm0
  mov    $0x1,%eax
  mulss  D66(%rip),%xmm0
  divss  D7(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D47
type: string
name: read_data
arguments: label: D2
type: string
name: read_data
arguments: label: D29
type: string
name: read_data
arguments: label: D66
type: byte[4]
name: read_data
arguments: label: D41
type: string
name: read_data
arguments: label: D7
type: byte[4]

tool: D47: Digite o salario:

tool: D2: %f

tool: D29: %.2f


tool: {"D66": [0x00,0x00,0x20,0x41]}

tool: D41: %.2f%% recebem acima da media

tool: {"D7": [0x00,0x00,0xc8,0x42]}

assistant: ```c
int main() {
  float sal[10], cont = 0, media = 0, porcentagem = 0;
  int i, porcent = 0;
  for (i = 0; i <= 9; i++) {
    printf("Digite o salario:");
    scanf("%f", &sal[i]);
    cont = cont + sal[i];
  }
  media = cont / 10;
  printf("%.2f\n", media);
  if (sal[i] > media) {
    porcent++;
  }
  porcentagem = ((float)porcent * 10) / 100;
  printf("%.2f%% recebem acima da media", porcentagem);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  pxor   %xmm1,%xmm1
  push   %r12
  lea    D49(%rip),%r12
  push   %rbp
  lea    D35(%rip),%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x38(%rsp),%r13
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %r12,%rdi
  xor    %eax,%eax
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm1
  addss  -0x4(%rbx),%xmm1
  cmp    %r13,%rbx
  jne    L7
  lea    D55(%rip),%rdi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  divss  D39(%rip),%xmm1
  cvtss2sd %xmm1,%xmm0
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  movss  0x38(%rsp),%xmm0
  xor    %eax,%eax
  movss  0xc(%rsp),%xmm1
  lea    D24(%rip),%rdi
  comiss %xmm1,%xmm0
  pxor   %xmm0,%xmm0
  seta   %al
  cvtsi2ss %eax,%xmm0
  mov    $0x1,%eax
  mulss  D39(%rip),%xmm0
  divss  D23(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D49
type: string
name: retrieve_data
arguments: tag: D35
type: string
name: retrieve_data
arguments: tag: D55
type: string
name: retrieve_data
arguments: tag: D39
type: byte[4]
name: retrieve_data
arguments: tag: D24
type: string
name: retrieve_data
arguments: tag: D23
type: byte[4]

tool: D49: Digite o salario:

tool: D35: %f

tool: D55: %.2f


tool: {"D39": [0x00,0x00,0x20,0x41]}

tool: D24: %.2f%% recebem acima da media

tool: {"D23": [0x00,0x00,0xc8,0x42]}

assistant: ```c
int main() {
  float sal[10], cont = 0, media = 0, porcentagem = 0;
  int i, porcent = 0;
  for (i = 0; i <= 9; i++) {
    printf("Digite o salario:");
    scanf("%f", &sal[i]);
    cont = cont + sal[i];
  }
  media = cont / 10;
  printf("%.2f\n", media);
  if (sal[i] > media) {
    porcent++;
  }
  porcentagem = ((float)porcent * 10) / 100;
  printf("%.2f%% recebem acima da media", porcentagem);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  'a' && 'b';
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  'a' && 'b';
  return 0;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int main(void) {
  'a' && 'b';
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

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

assistant: ```c
int main(void) {
  'a' && 'b';
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<process>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L17
L4:
  addq   $0x1,-0x18(%rbp)
L17:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L4
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
  jmp    L3
L11:
  addq   $0x1,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L11
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <output@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int process(char *line) {
  int first;
  int second;
  int end;
  first = 0;
  second = 0;
  end = 0;
  first = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  second = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  end = atoi(line);
  output(first, second, end);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<process>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <atoi@plt>
  mov    %eax,%r12d
  cmpb   $0x20,(%rbx)
  je     L12
L31:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  jne    L31
L12:
  lea    0x1(%rbx),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  mov    %eax,%r13d
  cmpb   $0x20,0x1(%rbx)
  je     L1
L23:
  add    $0x1,%rbp
  cmpb   $0x20,0x0(%rbp)
  jne    L23
L1:
  lea    0x1(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <output@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int process(char *line) {
  int first;
  int second;
  int end;
  first = 0;
  second = 0;
  end = 0;
  first = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  second = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  end = atoi(line);
  output(first, second, end);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<process>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <atoi@plt>
  cmpb   $0x20,(%rbx)
  mov    %eax,%r12d
  je     L33
  xchg   %ax,%ax
L10:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  jne    L10
L33:
  lea    0x1(%rbx),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  cmpb   $0x20,0x1(%rbx)
  mov    %eax,%r13d
  je     L4
  xchg   %ax,%ax
L7:
  add    $0x1,%rbp
  cmpb   $0x20,0x0(%rbp)
  jne    L7
L4:
  lea    0x1(%rbp),%rdi
  call   <atoi@plt>
  mov    %r13d,%esi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <output@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int process(char *line) {
  int first;
  int second;
  int end;
  first = 0;
  second = 0;
  end = 0;
  first = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  second = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  end = atoi(line);
  output(first, second, end);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<process>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <atoi@plt>
  cmpb   $0x20,(%rbx)
  mov    %eax,%r12d
  je     L25
  xchg   %ax,%ax
L12:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  jne    L12
L25:
  lea    0x1(%rbx),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  cmpb   $0x20,0x1(%rbx)
  mov    %eax,%r13d
  je     L2
  xchg   %ax,%ax
L6:
  add    $0x1,%rbp
  cmpb   $0x20,0x0(%rbp)
  jne    L6
L2:
  lea    0x1(%rbp),%rdi
  call   <atoi@plt>
  mov    %r13d,%esi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <output@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int process(char *line) {
  int first;
  int second;
  int end;
  first = 0;
  second = 0;
  end = 0;
  first = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  second = atoi(line);
  while (*line != ' ')
    line++;
  line++;
  end = atoi(line);
  output(first, second, end);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isnota_num>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x30,-0x4(%rbp)
  je     L57
  cmpb   $0x31,-0x4(%rbp)
  je     L57
  cmpb   $0x32,-0x4(%rbp)
  je     L57
  cmpb   $0x33,-0x4(%rbp)
  je     L57
  cmpb   $0x34,-0x4(%rbp)
  je     L57
  cmpb   $0x35,-0x4(%rbp)
  je     L57
  cmpb   $0x36,-0x4(%rbp)
  je     L57
  cmpb   $0x37,-0x4(%rbp)
  je     L57
  cmpb   $0x38,-0x4(%rbp)
  je     L57
  cmpb   $0x39,-0x4(%rbp)
  je     L57
  mov    $0x1,%eax
  jmp    L62
L57:
  mov    $0x0,%eax
L62:
  pop    %rbp
  ret
```

assistant: ```c
int isnota_num(char d) {
  if (d != '0' && d != '1' && d != '2' && d != '3' && d != '4' && d != '5' && d != '6' && d != '7' && d != '8' && d != '9') {
    return 1;
  } else
    return 0;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<isnota_num>:
  endbr64
  sub    $0x30,%edi
  cmp    $0x9,%dil
  seta   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int isnota_num(char d) {
  if (d != '0' && d != '1' && d != '2' && d != '3' && d != '4' && d != '5' && d != '6' && d != '7' && d != '8' && d != '9') {
    return 1;
  } else
    return 0;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<isnota_num>:
  endbr64
  sub    $0x30,%edi
  xor    %eax,%eax
  cmp    $0x9,%dil
  seta   %al
  ret
```

assistant: ```c
int isnota_num(char d) {
  if (d != '0' && d != '1' && d != '2' && d != '3' && d != '4' && d != '5' && d != '6' && d != '7' && d != '8' && d != '9') {
    return 1;
  } else
    return 0;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isnota_num>:
  endbr64
  sub    $0x30,%edi
  xor    %eax,%eax
  cmp    $0x9,%dil
  seta   %al
  ret
```

assistant: ```c
int isnota_num(char d) {
  if (d != '0' && d != '1' && d != '2' && d != '3' && d != '4' && d != '5' && d != '6' && d != '7' && d != '8' && d != '9') {
    return 1;
  } else
    return 0;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<func3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  D4(%rip),%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  movss  D3(%rip),%xmm1
  comiss %xmm1,%xmm0
  jbe    L2
  movl   $0x2,-0x8(%rbp)
  jmp    L16
L2:
  movl   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type: f32
name: disassemble_data
arguments: tag: D3
type: byte[4]

tool: D4: 0.3

tool: {"D3": [0x00,0x00,0x80,0x3f]}

assistant: ```c
int func3() {
  int bar;
  float func1 = 0.3;
  if (func1 > 1)
    bar = 2;
  else
    bar = 1;
  return bar;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<func3>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int func3() {
  int bar;
  float func1 = 0.3;
  if (func1 > 1)
    bar = 2;
  else
    bar = 1;
  return bar;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<func3>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int func3() {
  int bar;
  float func1 = 0.3;
  if (func1 > 1)
    bar = 2;
  else
    bar = 1;
  return bar;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<func3>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int func3() {
  int bar;
  float func1 = 0.3;
  if (func1 > 1)
    bar = 2;
  else
    bar = 1;
  return bar;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <foo@plt>
  cmp    $0x1a,%eax
  je     L9
  mov    $0x0,%eax
  call   <abort@plt>
L9:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (foo() != 26)
    abort();
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <foo@plt>
  cmp    $0x1a,%eax
  jne    L19
L17:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L19:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L17
```

assistant: ```c
int main() {
  if (foo() != 26)
    abort();
  exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  cmp    $0x1a,%eax
  je     L2
  xor    %eax,%eax
  call   <abort@plt>
L2:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  if (foo() != 26)
    abort();
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  cmp    $0x1a,%eax
  je     L5
  xor    %eax,%eax
  call   <abort@plt>
L5:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  if (foo() != 26)
    abort();
  exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<NvMemWaitReady>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x14,-0xc(%rbp)
  jmp    L34
L58:
  mov    $0x1,%edi
  call   <Delay@plt>
L34:
  subl   $0x1,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L2
  lea    -0xd(%rbp),%rax
  mov    $0x0,%r9d
  mov    $0x1,%r8d
  mov    %rax,%rcx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x57,%edi
  call   <TwMasterTransact@plt>
  cmp    $0xffffffff,%eax
  je     L58
L2:
  cmpl   $0x0,-0xc(%rbp)
  je     L36
  mov    $0x0,%eax
  jmp    L48
L36:
  mov    $0xffffffff,%eax
L48:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
int NvMemWaitReady(void) {
  unsigned char poll;
  int cnt = 20;
  while (--cnt && TwMasterTransact(0x57, 0, 0, &poll, 1, 0) == -1) {
    Delay(1);
  }
  return cnt ? 0 : -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<NvMemWaitReady>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x13,%ebx
  lea    0x7(%rsp),%rbp
L22:
  mov    $0x0,%r9d
  mov    $0x1,%r8d
  mov    %rbp,%rcx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x57,%edi
  call   <TwMasterTransact@plt>
  cmp    $0xffffffff,%eax
  jne    L24
  mov    $0x1,%edi
  call   <Delay@plt>
  sub    $0x1,%ebx
  jne    L22
L24:
  cmp    $0x1,%ebx
  sbb    %eax,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int NvMemWaitReady(void) {
  unsigned char poll;
  int cnt = 20;
  while (--cnt && TwMasterTransact(0x57, 0, 0, &poll, 1, 0) == -1) {
    Delay(1);
  }
  return cnt ? 0 : -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<NvMemWaitReady>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x13,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbp
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L16:
  mov    $0x1,%edi
  call   <Delay@plt>
  sub    $0x1,%ebx
  je     L4
L19:
  xor    %r9d,%r9d
  mov    $0x1,%r8d
  mov    %rbp,%rcx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x57,%edi
  call   <TwMasterTransact@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L16
  xor    %r12d,%r12d
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int NvMemWaitReady(void) {
  unsigned char poll;
  int cnt = 20;
  while (--cnt && TwMasterTransact(0x57, 0, 0, &poll, 1, 0) == -1) {
    Delay(1);
  }
  return cnt ? 0 : -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<NvMemWaitReady>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x13,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbp
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L28:
  mov    $0x1,%edi
  call   <Delay@plt>
  sub    $0x1,%ebx
  je     L11
L32:
  xor    %r9d,%r9d
  mov    $0x1,%r8d
  mov    %rbp,%rcx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x57,%edi
  call   <TwMasterTransact@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L28
  xor    %r12d,%r12d
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int NvMemWaitReady(void) {
  unsigned char poll;
  int cnt = 20;
  while (--cnt && TwMasterTransact(0x57, 0, 0, &poll, 1, 0) == -1) {
    Delay(1);
  }
  return cnt ? 0 : -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cms_cb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L12
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x83,-0x14(%rbp)
  je     L125
  jmp    L54
L12:
  mov    $0x1,%eax
  jmp    L152
L54:
  cmpl   $0x83,-0x14(%rbp)
  jg     L47
  cmpl   $0x82,-0x14(%rbp)
  je     L13
  cmpl   $0x82,-0x14(%rbp)
  jg     L47
  cmpl   $0x80,-0x14(%rbp)
  je     L78
  cmpl   $0x81,-0x14(%rbp)
  je     L125
  jmp    L47
L78:
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <CMS_stream@plt>
  test   %eax,%eax
  jg     L13
  mov    $0x0,%eax
  jmp    L152
L13:
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <CMS_dataInit@plt>
  mov    -0x10(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L49
  mov    $0x0,%eax
  jmp    L152
L125:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <CMS_dataFinal@plt>
  test   %eax,%eax
  jg     L137
  mov    $0x0,%eax
  jmp    L152
L49:
  nop
  jmp    L47
L137:
  nop
L47:
  mov    $0x1,%eax
L152:
  leave
  ret
```

assistant: ```c
int cms_cb(int operation, ASN1_VALUE **pval, const ASN1_ITEM *it, void *exarg) {
  ASN1_STREAM_ARG *sarg = exarg;
  CMS_ContentInfo *cms = NULL;
  if (pval)
    cms = (CMS_ContentInfo *)*pval;
  else
    return 1;
  switch (operation) {
  case ASN1_OP_STREAM_PRE:
    if (CMS_stream(&sarg->boundary, cms) <= 0)
      return 0;
  case ASN1_OP_DETACHED_PRE:
    sarg->ndef_bio = CMS_dataInit(cms, sarg->out);
    if (!sarg->ndef_bio)
      return 0;
    break;
  case ASN1_OP_STREAM_POST:
  case ASN1_OP_DETACHED_POST:
    if (CMS_dataFinal(cms, sarg->ndef_bio) <= 0)
      return 0;
    break;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cms_cb>:
  endbr64
  test   %rsi,%rsi
  je     L28
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rcx,%rbx
  mov    (%rsi),%rbp
  cmp    $0x82,%edi
  je     L68
  jg     L8
  cmp    $0x80,%edi
  je     L61
  cmp    $0x81,%edi
  jne    L7
L37:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <CMS_dataFinal@plt>
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  jmp    L31
L7:
  mov    $0x1,%eax
  jmp    L31
L8:
  mov    $0x1,%eax
  cmp    $0x83,%edi
  je     L37
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L61:
  lea    0x8(%rcx),%rdi
  mov    %rbp,%rsi
  call   <CMS_stream@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  test   %edx,%edx
  jle    L31
L68:
  mov    0x4(%rbx),%esi
  mov    %rbp,%rdi
  call   <CMS_dataInit@plt>
  mov    %eax,(%rbx)
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  jmp    L31
L28:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int cms_cb(int operation, ASN1_VALUE **pval, const ASN1_ITEM *it, void *exarg) {
  ASN1_STREAM_ARG *sarg = exarg;
  CMS_ContentInfo *cms = NULL;
  if (pval)
    cms = (CMS_ContentInfo *)*pval;
  else
    return 1;
  switch (operation) {
  case ASN1_OP_STREAM_PRE:
    if (CMS_stream(&sarg->boundary, cms) <= 0)
      return 0;
  case ASN1_OP_DETACHED_PRE:
    sarg->ndef_bio = CMS_dataInit(cms, sarg->out);
    if (!sarg->ndef_bio)
      return 0;
    break;
  case ASN1_OP_STREAM_POST:
  case ASN1_OP_DETACHED_POST:
    if (CMS_dataFinal(cms, sarg->ndef_bio) <= 0)
      return 0;
    break;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cms_cb>:
  endbr64
  test   %rsi,%rsi
  je     L16
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rbp
  cmp    $0x82,%edi
  je     L27
  jg     L66
  cmp    $0x80,%edi
  je     L59
  cmp    $0x81,%edi
  jne    L23
L35:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <CMS_dataFinal@plt>
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
L67:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L66:
  cmp    $0x83,%edi
  je     L35
L23:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L59:
  lea    0x8(%rcx),%rdi
  mov    %rbp,%rsi
  call   <CMS_stream@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  test   %r8d,%r8d
  jle    L67
L27:
  mov    0x4(%rbx),%esi
  mov    %rbp,%rdi
  call   <CMS_dataInit@plt>
  test   %eax,%eax
  mov    %eax,(%rbx)
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int cms_cb(int operation, ASN1_VALUE **pval, const ASN1_ITEM *it, void *exarg) {
  ASN1_STREAM_ARG *sarg = exarg;
  CMS_ContentInfo *cms = NULL;
  if (pval)
    cms = (CMS_ContentInfo *)*pval;
  else
    return 1;
  switch (operation) {
  case ASN1_OP_STREAM_PRE:
    if (CMS_stream(&sarg->boundary, cms) <= 0)
      return 0;
  case ASN1_OP_DETACHED_PRE:
    sarg->ndef_bio = CMS_dataInit(cms, sarg->out);
    if (!sarg->ndef_bio)
      return 0;
    break;
  case ASN1_OP_STREAM_POST:
  case ASN1_OP_DETACHED_POST:
    if (CMS_dataFinal(cms, sarg->ndef_bio) <= 0)
      return 0;
    break;
  }
  return 1;
}
```