ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<strerr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x14,%eax
  cmp    $0x18,%eax
  ja     L106
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D119(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D119(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D75(%rip),%rax
  jmp    L105
  lea    D72(%rip),%rax
  jmp    L105
  lea    D139(%rip),%rax
  jmp    L105
  lea    D159(%rip),%rax
  jmp    L105
  lea    D80(%rip),%rax
  jmp    L105
  lea    D67(%rip),%rax
  jmp    L105
  lea    D150(%rip),%rax
  jmp    L105
  lea    D45(%rip),%rax
  jmp    L105
  lea    D110(%rip),%rax
  jmp    L105
  lea    D133(%rip),%rax
  jmp    L105
  lea    D141(%rip),%rax
  jmp    L105
  lea    D64(%rip),%rax
  jmp    L105
  lea    D53(%rip),%rax
  jmp    L105
L106:
  lea    D82(%rip),%rax
L105:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D75
type: string
name: disassemble_data
arguments: tag: D72
type: string
name: disassemble_data
arguments: tag: D139
type: string
name: disassemble_data
arguments: tag: D159
type: string
name: disassemble_data
arguments: tag: D80
type: string
name: disassemble_data
arguments: tag: D67
type: string
name: disassemble_data
arguments: tag: D150
type: string
name: disassemble_data
arguments: tag: D45
type: string
name: disassemble_data
arguments: tag: D110
type: string
name: disassemble_data
arguments: tag: D133
type: string
name: disassemble_data
arguments: tag: D141
type: string
name: disassemble_data
arguments: tag: D64
type: string
name: disassemble_data
arguments: tag: D53
type: string
name: disassemble_data
arguments: tag: D82
type: string

tool: D75: Out of memory

tool: D72: No such file/directory

tool: D139: Permission denied

tool: D159: File exists

tool: D80: Not a directory

tool: D67: Is a directory

tool: D150: File name too long

tool: D45: Invalid path/device name

tool: D110: No such device

tool: D133: Device already registered

tool: D141: I/O error

tool: D64: Not a block device

tool: D53: Device busy

tool: D82: Unknown error

assistant: ```c
const char *strerr(int error_code) {
  switch (error_code) {
  case 20:
    return "Out of memory";
  case 30:
    return "No such file/directory";
  case 31:
    return "Permission denied";
  case 32:
    return "File exists";
  case 33:
    return "Not a directory";
  case 34:
    return "Is a directory";
  case 35:
    return "File name too long";
  case 36:
    return "Invalid path/device name";
  case 40:
    return "No such device";
  case 41:
    return "Device already registered";
  case 42:
    return "I/O error";
  case 43:
    return "Not a block device";
  case 44:
    return "Device busy";
  default:
    return "Unknown error";
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<strerr>:
  endbr64
  sub    $0x14,%edi
  cmp    $0x18,%edi
  ja     L1
  mov    %edi,%edi
  lea    D37(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D113(%rip),%rax
  ret
  lea    D143(%rip),%rax
  ret
  lea    D9(%rip),%rax
  ret
  lea    D4(%rip),%rax
  ret
  lea    D2(%rip),%rax
  ret
  lea    D65(%rip),%rax
  ret
  lea    D146(%rip),%rax
  ret
  lea    D23(%rip),%rax
  ret
  lea    D72(%rip),%rax
  ret
  lea    D64(%rip),%rax
  ret
  lea    D21(%rip),%rax
  ret
  lea    D45(%rip),%rax
  ret
L1:
  lea    D38(%rip),%rax
  ret
  lea    D50(%rip),%rax
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D113
type_infered: string
name: retrieve_data
arguments: data_label: D143
type_infered: string
name: retrieve_data
arguments: data_label: D9
type_infered: string
name: retrieve_data
arguments: data_label: D4
type_infered: string
name: retrieve_data
arguments: data_label: D2
type_infered: string
name: retrieve_data
arguments: data_label: D65
type_infered: string
name: retrieve_data
arguments: data_label: D146
type_infered: string
name: retrieve_data
arguments: data_label: D23
type_infered: string
name: retrieve_data
arguments: data_label: D72
type_infered: string
name: retrieve_data
arguments: data_label: D64
type_infered: string
name: retrieve_data
arguments: data_label: D21
type_infered: string
name: retrieve_data
arguments: data_label: D45
type_infered: string
name: retrieve_data
arguments: data_label: D38
type_infered: string
name: retrieve_data
arguments: data_label: D50
type_infered: string

tool: D113: No such file/directory

tool: D143: Permission denied

tool: D9: File exists

tool: D4: Not a directory

tool: D2: Is a directory

tool: D65: File name too long

tool: D146: Invalid path/device name

tool: D23: No such device

tool: D72: Device already registered

tool: D64: I/O error

tool: D21: Not a block device

tool: D45: Device busy

tool: D38: Unknown error

tool: D50: Out of memory

assistant: ```c
const char *strerr(int error_code) {
  switch (error_code) {
  case 20:
    return "Out of memory";
  case 30:
    return "No such file/directory";
  case 31:
    return "Permission denied";
  case 32:
    return "File exists";
  case 33:
    return "Not a directory";
  case 34:
    return "Is a directory";
  case 35:
    return "File name too long";
  case 36:
    return "Invalid path/device name";
  case 40:
    return "No such device";
  case 41:
    return "Device already registered";
  case 42:
    return "I/O error";
  case 43:
    return "Not a block device";
  case 44:
    return "Device busy";
  default:
    return "Unknown error";
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<LZ4_compress>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x1000a,-0x14(%rbp)
  jg     L11
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <LZ4_compress64k_stack@plt>
  jmp    L12
L11:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <LZ4_compress_stack@plt>
L12:
  leave
  ret
```

assistant: ```c
int LZ4_compress(const char *source, char *dest, int inputSize) {
  if (inputSize < (int)((1 << 16) + ((8 + 4) - 1)))
    return LZ4_compress64k_stack(source, dest, inputSize);
  return LZ4_compress_stack(source, dest, inputSize);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<LZ4_compress>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1000a,%edx
  jle    L16
  call   <LZ4_compress_stack@plt>
L10:
  add    $0x8,%rsp
  ret
L16:
  call   <LZ4_compress64k_stack@plt>
  jmp    L10
```

assistant: ```c
int LZ4_compress(const char *source, char *dest, int inputSize) {
  if (inputSize < (int)((1 << 16) + ((8 + 4) - 1)))
    return LZ4_compress64k_stack(source, dest, inputSize);
  return LZ4_compress_stack(source, dest, inputSize);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<LZ4_compress>:
  endbr64
  cmp    $0x1000a,%edx
  jle    L2
  jmp    1070 <LZ4_compress_stack@plt>
  nopl   0x0(%rax)
L2:
  jmp    1060 <LZ4_compress64k_stack@plt>
```

assistant: ```c
int LZ4_compress(const char *source, char *dest, int inputSize) {
  if (inputSize < (int)((1 << 16) + ((8 + 4) - 1)))
    return LZ4_compress64k_stack(source, dest, inputSize);
  return LZ4_compress_stack(source, dest, inputSize);
}
```
user: What is the c source code of the assembly code below?

```asm
<LZ4_compress>:
  endbr64
  cmp    $0x1000a,%edx
  jle    L8
  jmp    1070 <LZ4_compress_stack@plt>
  nopl   0x0(%rax)
L8:
  jmp    1060 <LZ4_compress64k_stack@plt>
```

assistant: ```c
int LZ4_compress(const char *source, char *dest, int inputSize) {
  if (inputSize < (int)((1 << 16) + ((8 + 4) - 1)))
    return LZ4_compress64k_stack(source, dest, inputSize);
  return LZ4_compress_stack(source, dest, inputSize);
}
```
user: Translate the below assembly code into C programming language:

```asm
<maxArea>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0xffffffff,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L77
L91:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  cmovg  %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jg     L19
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x18(%rbp)
  jmp    L38
L98:
  addl   $0x1,-0x18(%rbp)
L38:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L22
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jge    L98
  jmp    L22
L19:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x14(%rbp)
  jmp    L69
L3:
  subl   $0x1,-0x14(%rbp)
L69:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L22
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jge    L3
L22:
  addl   $0x1,-0x10(%rbp)
L77:
  mov    -0x10(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L91
  mov    -0x1c(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int maxArea(int a[], int n) {
  int max = -1;
  int left = 0, right = n - 1;
  for (int i = 1; i < n; ++i) {
    int area = (a[left] <= a[right] ? a[left] : a[right]) * (right - left);
    max = max < area ? area : max;
    if (a[left] <= a[right]) {
      int l = a[left];
      left++;
      while (left < right && a[left] <= l) {
        left++;
      }
    } else {
      int r = a[right];
      right--;
      while (left < right && a[right] <= r) {
        right--;
      }
    }
  }
  return max;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<maxArea>:
  endbr64
  cmp    $0x1,%esi
  jle    L100
  push   %rbx
  lea    -0x1(%rsi),%edx
  mov    $0x1,%r11d
  mov    $0x0,%ecx
  mov    $0xffffffff,%r10d
  jmp    L36
L7:
  sub    $0x1,%edx
  cmp    %ecx,%edx
  jle    L64
  movslq %edx,%rax
L95:
  mov    %eax,%edx
  cmp    (%rdi,%rax,4),%r8d
  jl     L64
  lea    -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jl     L95
L64:
  add    $0x1,%r11d
  cmp    %r11d,%esi
  je     L50
L36:
  movslq %ecx,%rax
  mov    (%rdi,%rax,4),%r9d
  movslq %edx,%rax
  mov    (%rdi,%rax,4),%r8d
  cmp    %r8d,%r9d
  mov    %r8d,%eax
  cmovle %r9d,%eax
  mov    %edx,%ebx
  sub    %ecx,%ebx
  imul   %ebx,%eax
  cmp    %eax,%r10d
  cmovl  %eax,%r10d
  cmp    %r8d,%r9d
  jg     L7
  add    $0x1,%ecx
  cmp    %edx,%ecx
  jge    L64
  movslq %ecx,%rax
L94:
  mov    %eax,%ecx
  cmp    (%rdi,%rax,4),%r9d
  jl     L64
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L94
  jmp    L64
L100:
  mov    $0xffffffff,%r10d
  mov    %r10d,%eax
  ret
L50:
  mov    %r10d,%eax
  pop    %rbx
  ret
```

assistant: ```c
int maxArea(int a[], int n) {
  int max = -1;
  int left = 0, right = n - 1;
  for (int i = 1; i < n; ++i) {
    int area = (a[left] <= a[right] ? a[left] : a[right]) * (right - left);
    max = max < area ? area : max;
    if (a[left] <= a[right]) {
      int l = a[left];
      left++;
      while (left < right && a[left] <= l) {
        left++;
      }
    } else {
      int r = a[right];
      right--;
      while (left < right && a[right] <= r) {
        right--;
      }
    }
  }
  return max;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<maxArea>:
  endbr64
  cmp    $0x1,%esi
  jle    L94
  push   %rbx
  lea    -0x1(%rsi),%edx
  mov    $0x1,%r11d
  xor    %ecx,%ecx
  mov    $0xffffffff,%r10d
  nop
L11:
  movslq %ecx,%rax
  mov    %edx,%ebx
  mov    (%rdi,%rax,4),%r9d
  movslq %edx,%rax
  mov    (%rdi,%rax,4),%r8d
  cmp    %r8d,%r9d
  mov    %r8d,%eax
  cmovle %r9d,%eax
  sub    %ecx,%ebx
  imul   %ebx,%eax
  cmp    %eax,%r10d
  cmovl  %eax,%r10d
  cmp    %r8d,%r9d
  jg     L122
  add    $0x1,%ecx
  cmp    %edx,%ecx
  jge    L65
  movslq %ecx,%rax
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L78:
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %eax,%edx
  jle    L65
L28:
  mov    %eax,%ecx
  cmp    (%rdi,%rax,4),%r9d
  jge    L78
L65:
  add    $0x1,%r11d
  cmp    %r11d,%esi
  jne    L11
L43:
  mov    %r10d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L122:
  sub    $0x1,%edx
  cmp    %ecx,%edx
  jle    L65
  movslq %edx,%rax
  jmp    L0
  nopl   0x0(%rax)
L73:
  lea    -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jge    L65
L0:
  mov    %eax,%edx
  cmp    (%rdi,%rax,4),%r8d
  jge    L73
  add    $0x1,%r11d
  cmp    %r11d,%esi
  jne    L11
  jmp    L43
L94:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int maxArea(int a[], int n) {
  int max = -1;
  int left = 0, right = n - 1;
  for (int i = 1; i < n; ++i) {
    int area = (a[left] <= a[right] ? a[left] : a[right]) * (right - left);
    max = max < area ? area : max;
    if (a[left] <= a[right]) {
      int l = a[left];
      left++;
      while (left < right && a[left] <= l) {
        left++;
      }
    } else {
      int r = a[right];
      right--;
      while (left < right && a[right] <= r) {
        right--;
      }
    }
  }
  return max;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<maxArea>:
  endbr64
  cmp    $0x1,%esi
  jle    L120
  push   %rbx
  lea    -0x1(%rsi),%edx
  mov    $0x1,%r11d
  xor    %ecx,%ecx
  mov    $0xffffffff,%r10d
  nop
L115:
  movslq %ecx,%rax
  mov    %edx,%ebx
  mov    (%rdi,%rax,4),%r9d
  movslq %edx,%rax
  mov    (%rdi,%rax,4),%r8d
  cmp    %r8d,%r9d
  mov    %r8d,%eax
  cmovle %r9d,%eax
  sub    %ecx,%ebx
  imul   %ebx,%eax
  cmp    %eax,%r10d
  cmovl  %eax,%r10d
  cmp    %r8d,%r9d
  jg     L53
  add    $0x1,%ecx
  cmp    %edx,%ecx
  jge    L11
  movslq %ecx,%rax
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L93:
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %eax,%edx
  jle    L11
L76:
  mov    %eax,%ecx
  cmp    (%rdi,%rax,4),%r9d
  jge    L93
L11:
  add    $0x1,%r11d
  cmp    %r11d,%esi
  jne    L115
L90:
  mov    %r10d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L53:
  sub    $0x1,%edx
  cmp    %ecx,%edx
  jle    L11
  movslq %edx,%rax
  jmp    L99
  nopl   0x0(%rax)
L57:
  lea    -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    %eax,%ecx
  jge    L11
L99:
  mov    %eax,%edx
  cmp    (%rdi,%rax,4),%r8d
  jge    L57
  add    $0x1,%r11d
  cmp    %r11d,%esi
  jne    L115
  jmp    L90
L120:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int maxArea(int a[], int n) {
  int max = -1;
  int left = 0, right = n - 1;
  for (int i = 1; i < n; ++i) {
    int area = (a[left] <= a[right] ? a[left] : a[right]) * (right - left);
    max = max < area ? area : max;
    if (a[left] <= a[right]) {
      int l = a[left];
      left++;
      while (left < right && a[left] <= l) {
        left++;
      }
    } else {
      int r = a[right];
      right--;
      while (left < right && a[right] <= r) {
        right--;
      }
    }
  }
  return max;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x400,-0x8(%rbp)
  movl   $0x54713,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  add    $0xe03,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3a23c,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  add    $0x2bcc,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D11
type: string

tool: D11: %d


assistant: ```c
int main() {
  int a = 1024;
  int b = 345875;
  printf("%d\n", a + 3587);
  printf("%d\n", 234587 + 3553);
  printf("%d\n", 11212 + b);
  printf("%d\n", 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
  push   %rbx
  mov    $0x1203,%esi
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3a23c,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x572df,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x54b13,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_guessed: string

tool: D6: %d


assistant: ```c
int main() {
  int a = 1024;
  int b = 345875;
  printf("%d\n", a + 3587);
  printf("%d\n", 234587 + 3553);
  printf("%d\n", 11212 + b);
  printf("%d\n", a + b);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  mov    $0x1203,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3a23c,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x572df,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x54b13,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_guessed: string

tool: D8: %d


assistant: ```c
int main() {
  int a = 1024;
  int b = 345875;
  printf("%d\n", a + 3587);
  printf("%d\n", 234587 + 3553);
  printf("%d\n", 11212 + b);
  printf("%d\n", a + b);
  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
  lea    D1(%rip),%rbp
  mov    $0x1203,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3a23c,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x572df,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x54b13,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D1
type_guessed: string

tool: D1: %d


assistant: ```c
int main() {
  int a = 1024;
  int b = 345875;
  printf("%d\n", a + 3587);
  printf("%d\n", 234587 + 3553);
  printf("%d\n", 11212 + b);
  printf("%d\n", a + b);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<exist_in_row>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %edx,-0x20(%rbp)
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L31:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L23
  mov    $0x1,%eax
  jmp    L34
L23:
  cmpl   $0x7,-0x4(%rbp)
  jle    L31
  mov    $0x0,%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
int exist_in_row(char **tab, char n, int i) {
  int j;
  j = 0;
  while (j < 8) {
    if (tab[i][j] == n)
      return (1);
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<exist_in_row>:
  endbr64
  movslq %edx,%rdx
  mov    (%rdi,%rdx,8),%rax
  movzbl (%rax),%eax
L5:
  cmp    %sil,%al
  jne    L5
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exist_in_row(char **tab, char n, int i) {
  int j;
  j = 0;
  while (j < 8) {
    if (tab[i][j] == n)
      return (1);
  }
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<exist_in_row>:
  endbr64
  movslq %edx,%rdx
  mov    (%rdi,%rdx,8),%rax
  movzbl (%rax),%eax
  xchg   %ax,%ax
L5:
  cmp    %sil,%al
  jne    L5
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exist_in_row(char **tab, char n, int i) {
  int j;
  j = 0;
  while (j < 8) {
    if (tab[i][j] == n)
      return (1);
  }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<exist_in_row>:
  endbr64
  movslq %edx,%rdx
  mov    (%rdi,%rdx,8),%rax
  cmp    %sil,(%rax)
  je     L19
L0:
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L19:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exist_in_row(char **tab, char n, int i) {
  int j;
  j = 0;
  while (j < 8) {
    if (tab[i][j] == n)
      return (1);
  }
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getRand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    D8(%rip),%eax
  test   %eax,%eax
  jne    L6
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x1,D8(%rip)
L6:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x18(%rbp),%edx
  sub    -0x14(%rbp),%edx
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getRand(int min, int max) {
  static int Init = 0;
  int rc;
  if (Init == 0) {
    srand(time(NULL));
    Init = 1;
  }
  rc = (rand() % (max - min + 1) + min);
  return (rc);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getRand>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  cmpl   $0x0,D7(%rip)
  je     L8
L19:
  mov    $0x0,%eax
  call   <rand@plt>
  sub    %ebp,%ebx
  lea    0x1(%rbx),%esi
  cltd
  idiv   %esi
  lea    (%rdx,%rbp,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x1,D7(%rip)
  jmp    L19
```

assistant: ```c
int getRand(int min, int max) {
  static int Init = 0;
  int rc;
  if (Init == 0) {
    srand(time(NULL));
    Init = 1;
  }
  rc = (rand() % (max - min + 1) + min);
  return (rc);
}
```
user: What is the c source code of the assembly code below?

```asm
<getRand>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  mov    D14(%rip),%eax
  test   %eax,%eax
  je     L10
L11:
  xor    %eax,%eax
  sub    %ebp,%ebx
  call   <rand@plt>
  lea    0x1(%rbx),%esi
  add    $0x8,%rsp
  cltd
  pop    %rbx
  idiv   %esi
  lea    (%rdx,%rbp,1),%eax
  pop    %rbp
  ret
  nopl   0x0(%rax)
L10:
  xor    %edi,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x1,D14(%rip)
  jmp    L11
```

assistant: ```c
int getRand(int min, int max) {
  static int Init = 0;
  int rc;
  if (Init == 0) {
    srand(time(NULL));
    Init = 1;
  }
  rc = (rand() % (max - min + 1) + min);
  return (rc);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getRand>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  mov    D12(%rip),%eax
  test   %eax,%eax
  je     L1
L9:
  xor    %eax,%eax
  sub    %ebp,%ebx
  call   <rand@plt>
  lea    0x1(%rbx),%esi
  add    $0x8,%rsp
  cltd
  pop    %rbx
  idiv   %esi
  lea    (%rdx,%rbp,1),%eax
  pop    %rbp
  ret
  nopl   0x0(%rax)
L1:
  xor    %edi,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x1,D12(%rip)
  jmp    L9
```

assistant: ```c
int getRand(int min, int max) {
  static int Init = 0;
  int rc;
  if (Init == 0) {
    srand(time(NULL));
    Init = 1;
  }
  rc = (rand() % (max - min + 1) + min);
  return (rc);
}
```