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

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  lea    0x38(%rsp),%rdx
L14:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L14
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L23
  movabs $0x6666666666666667,%rdi
L54:
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  addl   $0x1,0x10(%rsp,%rcx,4)
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L54
  movq   $0x0,0x8(%rsp)
L23:
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D41(%rip),%rbp
  jmp    L19
L18:
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  je     L56
L19:
  mov    %ebx,%esi
  cmpl   $0x1,0x10(%rsp,%rbx,4)
  jle    L18
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L18
L56:
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D33
datatype: string
name: extract_symbolic_data
arguments: data_label: D39
datatype: string
name: extract_symbolic_data
arguments: data_label: D25
datatype: string
name: extract_symbolic_data
arguments: data_label: D41
datatype: string
name: extract_symbolic_data
arguments: data_label: D15
datatype: string

tool: D33: 
Vavedi chislo:

tool: D39: %lli

tool: D25: 
Tsifrite, koito se povtariat sa: 

tool: D41: %d, 

tool: D15: .

assistant: ```c
void main() {
  int i, a[10];
  long long int n;
  for (i = 0; i < 10; i++)
    a[i] = 0;
  printf("\nVavedi chislo:");
  scanf("%lli", &n);
  while (n != 0) {
    a[n % 10]++;
    n /= 10;
  }
  printf("\nTsifrite, koito se povtariat sa: ");
  for (i = 0; i < 10; i++)
    if (a[i] > 1)
      printf("%d, ", i);
  printf("\b\b.");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  lea    D19(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L8
  movabs $0x6666666666666667,%rdi
  nopl   0x0(%rax)
L42:
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rcx
  addl   $0x1,0x10(%rsp,%rcx,4)
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L42
  movq   $0x0,0x8(%rsp)
L8:
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    D18(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L35:
  cmpl   $0x1,0x0(%rbp,%rbx,4)
  mov    %ebx,%esi
  jle    L31
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L31:
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  jne    L35
  xor    %eax,%eax
  lea    D43(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D19
type_guessed: string
name: retrieve_data
arguments: label: D29
type_guessed: string
name: retrieve_data
arguments: label: D28
type_guessed: string
name: retrieve_data
arguments: label: D18
type_guessed: string
name: retrieve_data
arguments: label: D43
type_guessed: string

tool: D19: 
Vavedi chislo:

tool: D29: %lli

tool: D28: 
Tsifrite, koito se povtariat sa: 

tool: D18: %d, 

tool: D43: .

assistant: ```c
void main() {
  int i, a[10];
  long long int n;
  for (i = 0; i < 10; i++)
    a[i] = 0;
  printf("\nVavedi chislo:");
  scanf("%lli", &n);
  while (n != 0) {
    a[n % 10]++;
    n /= 10;
  }
  printf("\nTsifrite, koito se povtariat sa: ");
  for (i = 0; i < 10; i++)
    if (a[i] > 1)
      printf("%d, ", i);
  printf("\b\b.");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  pxor   %xmm0,%xmm0
  lea    D138(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L67
  movabs $0x6666666666666667,%rdi
L113:
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rcx
  addl   $0x1,0x10(%rsp,%rcx,4)
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L113
  movq   $0x0,0x8(%rsp)
L67:
  xor    %eax,%eax
  lea    D139(%rip),%rdi
  call   <printf@plt>
  cmpl   $0x1,0x10(%rsp)
  jle    L75
  xor    %esi,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L75:
  cmpl   $0x1,0x14(%rsp)
  jle    L57
  mov    $0x1,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L57:
  cmpl   $0x1,0x18(%rsp)
  jle    L69
  mov    $0x2,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L69:
  cmpl   $0x1,0x1c(%rsp)
  jle    L103
  mov    $0x3,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L103:
  cmpl   $0x1,0x20(%rsp)
  jle    L5
  mov    $0x4,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L5:
  cmpl   $0x1,0x24(%rsp)
  jle    L16
  mov    $0x5,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L16:
  cmpl   $0x1,0x28(%rsp)
  jle    L123
  mov    $0x6,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L123:
  cmpl   $0x1,0x2c(%rsp)
  jle    L65
  mov    $0x7,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L65:
  cmpl   $0x1,0x30(%rsp)
  jle    L90
  mov    $0x8,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L90:
  cmpl   $0x1,0x34(%rsp)
  jle    L76
  mov    $0x9,%esi
  lea    D104(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L76:
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L102
  add    $0x48,%rsp
  ret
L102:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D138
type_infered: string
name: read_data
arguments: tag: D20
type_infered: string
name: read_data
arguments: tag: D139
type_infered: string
name: read_data
arguments: tag: D104
type_infered: string
name: read_data
arguments: tag: D1
type_infered: string

tool: D138: 
Vavedi chislo:

tool: D20: %lli

tool: D139: 
Tsifrite, koito se povtariat sa: 

tool: D104: %d, 

tool: D1: .

assistant: ```c
void main() {
  int i, a[10];
  long long int n;
  for (i = 0; i < 10; i++)
    a[i] = 0;
  printf("\nVavedi chislo:");
  scanf("%lli", &n);
  while (n != 0) {
    a[n % 10]++;
    n /= 10;
  }
  printf("\nTsifrite, koito se povtariat sa: ");
  for (i = 0; i < 10; i++)
    if (a[i] > 1)
      printf("%d, ", i);
  printf("\b\b.");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strLower>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L27
L45:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L14
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L14
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  lea    0x20(%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L27
L14:
  mov    -0x18(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L27:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L45
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strLower(const char *in, char *out) {
  char *scan = out;
  while (*in != '\0')
    if ((*in >= 'A') && (*in <= 'Z'))
      *scan++ = *in++ + 'a' - 'A';
    else
      *scan++ = *in++;
  *scan = '\0';
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strLower>:
  endbr64
  mov    %rsi,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L47
  add    $0x1,%rdi
  lea    0x1(%rsi),%rcx
  jmp    L5
L33:
  mov    %rdi,%r8
  mov    %rcx,%r9
L43:
  mov    %dl,-0x1(%rcx)
  movzbl (%r8),%edx
  add    $0x1,%rdi
  add    $0x1,%rcx
  test   %dl,%dl
  je     L0
L5:
  lea    -0x41(%rdx),%r8d
  cmp    $0x19,%r8b
  ja     L33
  mov    %rdi,%r8
  mov    %rcx,%r9
  add    $0x20,%edx
  jmp    L43
L47:
  mov    %rsi,%r9
L0:
  movb   $0x0,(%r9)
  ret
```

assistant: ```c
char *strLower(const char *in, char *out) {
  char *scan = out;
  while (*in != '\0')
    if ((*in >= 'A') && (*in <= 'Z'))
      *scan++ = *in++ + 'a' - 'A';
    else
      *scan++ = *in++;
  *scan = '\0';
  return out;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strLower>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rsi,%rax
  mov    %rsi,%rcx
  test   %dl,%dl
  je     L19
  nopl   0x0(%rax)
L7:
  lea    -0x41(%rdx),%r9d
  add    $0x1,%rdi
  lea    0x20(%rdx),%r8d
  add    $0x1,%rcx
  cmp    $0x1a,%r9b
  cmovb  %r8d,%edx
  mov    %dl,-0x1(%rcx)
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L7
L19:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *strLower(const char *in, char *out) {
  char *scan = out;
  while (*in != '\0')
    if ((*in >= 'A') && (*in <= 'Z'))
      *scan++ = *in++ + 'a' - 'A';
    else
      *scan++ = *in++;
  *scan = '\0';
  return out;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strLower>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rsi,%rax
  mov    %rsi,%rcx
  test   %dl,%dl
  je     L33
  nopl   0x0(%rax)
L21:
  lea    -0x41(%rdx),%r8d
  add    $0x1,%rdi
  add    $0x1,%rcx
  cmp    $0x19,%r8b
  ja     L22
  add    $0x20,%edx
  mov    %dl,-0x1(%rcx)
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L21
L33:
  movb   $0x0,(%rcx)
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %dl,-0x1(%rcx)
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L21
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *strLower(const char *in, char *out) {
  char *scan = out;
  while (*in != '\0')
    if ((*in >= 'A') && (*in <= 'Z'))
      *scan++ = *in++ + 'a' - 'A';
    else
      *scan++ = *in++;
  *scan = '\0';
  return out;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<trim_seq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L10
L4:
  addq   $0x1,-0x18(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L55
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L4
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L4
L55:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L47
L65:
  subl   $0x1,-0x4(%rbp)
L47:
  cmpl   $0x0,-0x4(%rbp)
  js     L34
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L65
L34:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *trim_seq(char string[]) {
  int i;
  while ((*string) != '\0' && (isspace(*string) || isdigit(*string))) {
    string++;
  }
  i = strlen(string) - 1;
  while (i >= 0 && isspace(string[i])) {
    i--;
  }
  string[i + 1] = '\0';
  return (string);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<trim_seq>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  jne    L27
  jmp    L54
L35:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L54
L27:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L35
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L35
L54:
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,%ebp
  js     L12
  movslq %eax,%r12
  add    %rbx,%r12
L20:
  movsbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L12
  sub    $0x1,%ebp
  sub    $0x1,%r12
  cmp    $0xffffffff,%ebp
  jne    L20
L12:
  movslq %ebp,%rbp
  movb   $0x0,0x1(%rbx,%rbp,1)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *trim_seq(char string[]) {
  int i;
  while ((*string) != '\0' && (isspace(*string) || isdigit(*string))) {
    string++;
  }
  i = strlen(string) - 1;
  while (i >= 0 && isspace(string[i])) {
    i--;
  }
  string[i + 1] = '\0';
  return (string);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<trim_seq>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L76
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L92:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L38
L76:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L92
  movsbl (%r12),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L92
L38:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  sub    $0x1,%ebp
  js     L46
  movslq %ebp,%rbx
  add    %r12,%rbx
  jmp    L58
  nop
L72:
  sub    $0x1,%ebp
  sub    $0x1,%rbx
  cmp    $0xffffffff,%ebp
  je     L16
L58:
  movsbl (%rbx),%edi
  mov    %rbx,%r13
  call   <isspace@plt>
  test   %rax,%rax
  jne    L72
  add    $0x1,%ebp
  movslq %ebp,%rbp
  lea    (%r12,%rbp,1),%r13
L16:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  cltq
  lea    (%r12,%rax,1),%r13
  jmp    L16
```

assistant: ```c
char *trim_seq(char string[]) {
  int i;
  while ((*string) != '\0' && (isspace(*string) || isdigit(*string))) {
    string++;
  }
  i = strlen(string) - 1;
  while (i >= 0 && isspace(string[i])) {
    i--;
  }
  string[i + 1] = '\0';
  return (string);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<trim_seq>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L6
  jmp    L96
  nopl   0x0(%rax,%rax,1)
L20:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L96
L6:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
  movsbl (%r12),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L20
L96:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  sub    $0x1,%ebp
  js     L92
  movslq %ebp,%rbx
  add    %r12,%rbx
  jmp    L98
  nop
L49:
  sub    $0x1,%ebp
  sub    $0x1,%rbx
  cmp    $0xffffffff,%ebp
  je     L35
L98:
  movsbl (%rbx),%edi
  mov    %rbx,%r13
  call   <isspace@plt>
  test   %rax,%rax
  jne    L49
  add    $0x1,%ebp
  movslq %ebp,%rbp
  lea    (%r12,%rbp,1),%r13
L35:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L92:
  cltq
  lea    (%r12,%rax,1),%r13
  jmp    L35
```

assistant: ```c
char *trim_seq(char string[]) {
  int i;
  while ((*string) != '\0' && (isspace(*string) || isdigit(*string))) {
    string++;
  }
  i = strlen(string) - 1;
  while (i >= 0 && isspace(string[i])) {
    i--;
  }
  string[i + 1] = '\0';
  return (string);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<endofgame>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L13
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L13
  mov    $0x3,%eax
  jmp    L56
L13:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L25
  mov    $0x2,%eax
  jmp    L56
L25:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L51
  mov    $0x1,%eax
  jmp    L56
L51:
  mov    $0x0,%eax
L56:
  leave
  ret
```

assistant: ```c
int endofgame(struct queue *player1, struct queue *player2) {
  if (emptyQ(player1) && emptyQ(player2))
    return 3;
  if (emptyQ(player1))
    return 2;
  if (emptyQ(player2))
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<endofgame>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L11
  mov    %rbp,%rdi
  call   <emptyQ@plt>
  mov    %rax,%rdx
  mov    $0x3,%eax
  test   %rdx,%rdx
  jne    L0
L11:
  mov    %rbx,%rdi
  call   <emptyQ@plt>
  mov    %rax,%rdx
  mov    $0x2,%eax
  test   %rdx,%rdx
  jne    L0
  mov    %rbp,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int endofgame(struct queue *player1, struct queue *player2) {
  if (emptyQ(player1) && emptyQ(player2))
    return 3;
  if (emptyQ(player1))
    return 2;
  if (emptyQ(player2))
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<endofgame>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L17
  mov    %r12,%rdi
  call   <emptyQ@plt>
  mov    %rax,%r8
  mov    $0x3,%eax
  test   %r8,%r8
  jne    L8
L17:
  mov    %rbp,%rdi
  call   <emptyQ@plt>
  mov    %rax,%r8
  mov    $0x2,%eax
  test   %r8,%r8
  jne    L8
  mov    %r12,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L8:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int endofgame(struct queue *player1, struct queue *player2) {
  if (emptyQ(player1) && emptyQ(player2))
    return 3;
  if (emptyQ(player1))
    return 2;
  if (emptyQ(player2))
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<endofgame>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <emptyQ@plt>
  test   %rax,%rax
  je     L17
  mov    %r12,%rdi
  call   <emptyQ@plt>
  mov    %rax,%r8
  mov    $0x3,%eax
  test   %r8,%r8
  jne    L10
L17:
  mov    %rbp,%rdi
  call   <emptyQ@plt>
  mov    %rax,%r8
  mov    $0x2,%eax
  test   %r8,%r8
  jne    L10
  mov    %r12,%rdi
  call   <emptyQ@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L10:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int endofgame(struct queue *player1, struct queue *player2) {
  if (emptyQ(player1) && emptyQ(player2))
    return 3;
  if (emptyQ(player1))
    return 2;
  if (emptyQ(player2))
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <get_callback@plt>
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x5,%edi
  call   *%rax
  leave
  ret
```

assistant: ```c
int main() {
  int (*f)(int) = get_callback();
  return f(5);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_callback@plt>
  cltq
  mov    $0x5,%edi
  call   *%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int (*f)(int) = get_callback();
  return f(5);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_callback@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  cltq
  jmp    *%rax
```

assistant: ```c
int main() {
  int (*f)(int) = get_callback();
  return f(5);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_callback@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  cltq
  jmp    *%rax
```

assistant: ```c
int main() {
  int (*f)(int) = get_callback();
  return f(5);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<who_mode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  je     L16
  mov    -0x14(%rbp),%eax
  or     %eax,-0x4(%rbp)
L16:
  mov    -0x18(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L10
  mov    -0x14(%rbp),%eax
  shl    $0x3,%eax
  or     %eax,-0x4(%rbp)
L10:
  mov    -0x18(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L7
  mov    -0x14(%rbp),%eax
  shl    $0x6,%eax
  or     %eax,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int who_mode(int mode, int who) {
  int ret = 0;
  if (who & (1 << 2))
    ret |= mode;
  if (who & (1 << 1))
    ret |= mode << 3;
  if (who & (1 << 0))
    ret |= mode << 6;
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<who_mode>:
  endbr64
  mov    %esi,%eax
  and    $0x4,%eax
  cmovne %edi,%eax
  lea    0x0(,%rdi,8),%edx
  or     %eax,%edx
  test   $0x2,%sil
  cmovne %edx,%eax
  shl    $0x6,%edi
  or     %eax,%edi
  test   $0x1,%sil
  cmovne %edi,%eax
  ret
```

assistant: ```c
int who_mode(int mode, int who) {
  int ret = 0;
  if (who & (1 << 2))
    ret |= mode;
  if (who & (1 << 1))
    ret |= mode << 3;
  if (who & (1 << 0))
    ret |= mode << 6;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<who_mode>:
  endbr64
  mov    %esi,%eax
  lea    0x0(,%rdi,8),%edx
  and    $0x4,%eax
  cmovne %edi,%eax
  or     %eax,%edx
  test   $0x2,%sil
  cmovne %edx,%eax
  shl    $0x6,%edi
  or     %eax,%edi
  and    $0x1,%esi
  cmovne %edi,%eax
  ret
```

assistant: ```c
int who_mode(int mode, int who) {
  int ret = 0;
  if (who & (1 << 2))
    ret |= mode;
  if (who & (1 << 1))
    ret |= mode << 3;
  if (who & (1 << 0))
    ret |= mode << 6;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<who_mode>:
  endbr64
  mov    %esi,%eax
  lea    0x0(,%rdi,8),%edx
  and    $0x4,%eax
  cmovne %edi,%eax
  or     %eax,%edx
  test   $0x2,%sil
  cmovne %edx,%eax
  shl    $0x6,%edi
  or     %eax,%edi
  and    $0x1,%esi
  cmovne %edi,%eax
  ret
```

assistant: ```c
int who_mode(int mode, int who) {
  int ret = 0;
  if (who & (1 << 2))
    ret |= mode;
  if (who & (1 << 1))
    ret |= mode << 3;
  if (who & (1 << 0))
    ret |= mode << 6;
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mallocEscapeFoo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0xc,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <myfoo@plt>
  nop
  leave
  ret
```

assistant: ```c
void mallocEscapeFoo() {
  int *p = malloc(12);
  myfoo(p);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<mallocEscapeFoo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <myfoo@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void mallocEscapeFoo() {
  int *p = malloc(12);
  myfoo(p);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mallocEscapeFoo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <myfoo@plt>
```

assistant: ```c
void mallocEscapeFoo() {
  int *p = malloc(12);
  myfoo(p);
  return;
}
```