ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ziplistNext>:
  endbr64
  cmpb   $0xff,(%rsi)
  je     L19
  push   %rbx
  mov    %rsi,%rdi
  mov    %rsi,%rbx
  call   <zipRawEntryLength@plt>
  cltq
  add    %rbx,%rax
  cmpb   $0xff,(%rax)
  je     L2
  pop    %rbx
  ret
  nopl   0x0(%rax)
L2:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L19:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned char *ziplistNext(unsigned char *zl, unsigned char *p) {
  ((void)zl);
  if (p[0] == 255) {
    return NULL;
  } else {
    p = p + zipRawEntryLength(p);
    return (p[0] == 255) ? NULL : p;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ziplistNext>:
  endbr64
  cmpb   $0xff,(%rsi)
  je     L3
  push   %rbx
  mov    %rsi,%rdi
  mov    %rsi,%rbx
  call   <zipRawEntryLength@plt>
  cltq
  add    %rbx,%rax
  cmpb   $0xff,(%rax)
  je     L13
  pop    %rbx
  ret
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned char *ziplistNext(unsigned char *zl, unsigned char *p) {
  ((void)zl);
  if (p[0] == 255) {
    return NULL;
  } else {
    p = p + zipRawEntryLength(p);
    return (p[0] == 255) ? NULL : p;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<checkTipoTrama>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0x3,%eax
  cmp    $0x3,%eax
  je     L17
  cmp    $0x3,%eax
  jg     L8
  cmp    $0x2,%eax
  je     L75
  cmp    $0x2,%eax
  jg     L8
  test   %eax,%eax
  je     L71
  cmp    $0x1,%eax
  je     L63
  jmp    L8
L71:
  mov    $0x49,%eax
  jmp    L8
L63:
  mov    $0x49,%eax
  jmp    L8
L75:
  mov    $0x53,%eax
  jmp    L8
L17:
  mov    $0x55,%eax
  jmp    L8
L8:
  pop    %rbp
  ret
```

assistant: ```c
char checkTipoTrama(const unsigned char *tr) {
  switch (tr[16] & 3) {
  case 0:
    return 'I';
    break;
  case 1:
    return 'I';
    break;
  case 2:
    return 'S';
    break;
  case 3:
    return 'U';
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<checkTipoTrama>:
  endbr64
  movzbl 0x10(%rdi),%eax
  and    $0x3,%eax
  mov    $0x53,%edx
  cmp    $0x2,%al
  je     L11
  mov    $0x55,%edx
  ja     L11
  mov    $0x49,%edx
L11:
  mov    %edx,%eax
  ret
```

assistant: ```c
char checkTipoTrama(const unsigned char *tr) {
  switch (tr[16] & 3) {
  case 0:
    return 'I';
    break;
  case 1:
    return 'I';
    break;
  case 2:
    return 'S';
    break;
  case 3:
    return 'U';
    break;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<checkTipoTrama>:
  endbr64
  movzbl 0x10(%rdi),%edx
  mov    $0x53,%eax
  and    $0x3,%edx
  cmp    $0x2,%dl
  je     L9
  cmp    $0x3,%dl
  mov    $0x49,%eax
  mov    $0x55,%edx
  cmove  %edx,%eax
L9:
  ret
```

assistant: ```c
char checkTipoTrama(const unsigned char *tr) {
  switch (tr[16] & 3) {
  case 0:
    return 'I';
    break;
  case 1:
    return 'I';
    break;
  case 2:
    return 'S';
    break;
  case 3:
    return 'U';
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<checkTipoTrama>:
  endbr64
  movzbl 0x10(%rdi),%edx
  mov    $0x53,%eax
  and    $0x3,%edx
  cmp    $0x2,%dl
  je     L6
  cmp    $0x3,%dl
  mov    $0x49,%eax
  mov    $0x55,%edx
  cmove  %edx,%eax
L6:
  ret
```

assistant: ```c
char checkTipoTrama(const unsigned char *tr) {
  switch (tr[16] & 3) {
  case 0:
    return 'I';
    break;
  case 1:
    return 'I';
    break;
  case 2:
    return 'S';
    break;
  case 3:
    return 'U';
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L19
  mov    $0x0,%eax
  call   <printUsage@plt>
  mov    $0xffffffff,%eax
  jmp    L16
L19:
  mov    -0x10(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <printLink@plt>
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printUsage();
    return -1;
  } else {
    printLink(argv[1], argv[2]);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L18
  mov    0x10(%rsi),%rax
  mov    0x8(%rsi),%rdi
  mov    %rax,%rsi
  call   <printLink@plt>
  mov    $0x0,%eax
L16:
  add    $0x8,%rsp
  ret
L18:
  mov    $0x0,%eax
  call   <printUsage@plt>
  mov    $0xffffffff,%eax
  jmp    L16
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printUsage();
    return -1;
  } else {
    printLink(argv[1], argv[2]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L11
  mov    0x10(%rsi),%r8
  mov    0x8(%rsi),%rdi
  mov    %r8,%rsi
  call   <printLink@plt>
  xor    %eax,%eax
L6:
  add    $0x8,%rsp
  ret
L11:
  xor    %eax,%eax
  call   <printUsage@plt>
  or     $0xffffffff,%eax
  jmp    L6
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printUsage();
    return -1;
  } else {
    printLink(argv[1], argv[2]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jne    L6
  mov    0x10(%rsi),%r8
  mov    0x8(%rsi),%rdi
  mov    %r8,%rsi
  call   <printLink@plt>
  xor    %eax,%eax
L8:
  add    $0x8,%rsp
  ret
L6:
  xor    %eax,%eax
  call   <printUsage@plt>
  or     $0xffffffff,%eax
  jmp    L8
```

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printUsage();
    return -1;
  } else {
    printLink(argv[1], argv[2]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<setenv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getenv@plt>
  test   %eax,%eax
  je     L16
  cmpl   $0x0,-0x34(%rbp)
  je     L21
L16:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    $0x2,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <putenv@plt>
  jmp    L3
L21:
  mov    $0x0,%eax
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type_infered: string

tool: D9: %s=%s

assistant: ```c
int setenv(const char *name, const char *value, int overwrite) {
  char *n;
  if (!getenv(name) || overwrite) {
    n = (char *)malloc(strlen(name) + strlen(value) + 2);
    sprintf(n, "%s=%s", name, value);
    return putenv(n);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<setenv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %edx,%ebp
  call   <getenv@plt>
  test   %ebp,%ebp
  jne    L1
  mov    $0x0,%edx
  test   %eax,%eax
  je     L1
L21:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x2(%rbp,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12,%rcx
  mov    %rbx,%rdx
  lea    D0(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <putenv@plt>
  mov    %eax,%edx
  jmp    L21
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string

tool: D0: %s=%s

assistant: ```c
int setenv(const char *name, const char *value, int overwrite) {
  char *n;
  if (!getenv(name) || overwrite) {
    n = (char *)malloc(strlen(name) + strlen(value) + 2);
    sprintf(n, "%s=%s", name, value);
    return putenv(n);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<setenv>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <getenv@plt>
  test   %eax,%eax
  je     L11
  test   %ebx,%ebx
  jne    L11
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  lea    0x2(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <putenv@plt>
```

tool_calls: name: read_data
arguments: label: D4
data_type: string

tool: D4: %s=%s

assistant: ```c
int setenv(const char *name, const char *value, int overwrite) {
  char *n;
  if (!getenv(name) || overwrite) {
    n = (char *)malloc(strlen(name) + strlen(value) + 2);
    sprintf(n, "%s=%s", name, value);
    return putenv(n);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<setenv>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <getenv@plt>
  test   %eax,%eax
  je     L3
  test   %ebx,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  lea    0x2(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    D3(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <putenv@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string

tool: D3: %s=%s

assistant: ```c
int setenv(const char *name, const char *value, int overwrite) {
  char *n;
  if (!getenv(name) || overwrite) {
    n = (char *)malloc(strlen(name) + strlen(value) + 2);
    sprintf(n, "%s=%s", name, value);
    return putenv(n);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<parity>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  xor    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x8,%rax
  xor    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x4,%rax
  xor    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x2,%rax
  xor    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    %rax
  xor    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  and    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long parity(unsigned long in) {
  in ^= in >> 16;
  in ^= in >> 8;
  in ^= in >> 4;
  in ^= in >> 2;
  in ^= in >> 1;
  return in & 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parity>:
  endbr64
  mov    %rdi,%rax
  shr    $0x10,%rax
  xor    %rdi,%rax
  mov    %rax,%rdx
  shr    $0x8,%rdx
  xor    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x4,%rax
  xor    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x2,%rdx
  xor    %rax,%rdx
  mov    %rdx,%rax
  shr    %rax
  xor    %rdx,%rax
  and    $0x1,%eax
  ret
```

assistant: ```c
unsigned long parity(unsigned long in) {
  in ^= in >> 16;
  in ^= in >> 8;
  in ^= in >> 4;
  in ^= in >> 2;
  in ^= in >> 1;
  return in & 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<parity>:
  endbr64
  mov    %rdi,%rax
  shr    $0x10,%rax
  xor    %rdi,%rax
  mov    %rax,%rdx
  shr    $0x8,%rdx
  xor    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x4,%rax
  xor    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x2,%rdx
  xor    %rax,%rdx
  mov    %rdx,%rax
  shr    %rax
  xor    %rdx,%rax
  and    $0x1,%eax
  ret
```

assistant: ```c
unsigned long parity(unsigned long in) {
  in ^= in >> 16;
  in ^= in >> 8;
  in ^= in >> 4;
  in ^= in >> 2;
  in ^= in >> 1;
  return in & 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
L6:
  movl   $0x0,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type: string
name: extract_symbolic_data
arguments: label: D16
type: string
name: extract_symbolic_data
arguments: label: D27
type: string

tool: D12: %d

tool: D16: clear

tool: D27: %d


assistant: ```c
int main(void) {
  int dec = 1;
  while (1) {
    int in = 0;
    scanf("%d", &in);
    system("clear");
    int bin = (1 << in);
    printf("%d\n", bin);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  lea    D4(%rip),%r12
  lea    D12(%rip),%rbp
  lea    D26(%rip),%rbx
L7:
  movl   $0x0,0x4(%rsp)
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x4(%rsp),%ecx
  mov    $0x1,%esi
  shl    %cl,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L7
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string
name: read_data
arguments: data_tag: D12
data_type: string
name: read_data
arguments: data_tag: D26
data_type: string

tool: D4: %d

tool: D12: clear

tool: D26: %d


assistant: ```c
int main(void) {
  int dec = 1;
  while (1) {
    int in = 0;
    scanf("%d", &in);
    system("clear");
    int bin = (1 << in);
    printf("%d\n", bin);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  lea    D22(%rip),%r12
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r14
  nop
L3:
  mov    %r14,%rsi
  mov    %r13,%rdi
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <system@plt>
  mov    0x4(%rsp),%ecx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  shl    %cl,%esi
  call   <printf@plt>
  jmp    L3
```

tool_calls: name: unpack_data
arguments: data_label: D0
data_type: string
name: unpack_data
arguments: data_label: D22
data_type: string
name: unpack_data
arguments: data_label: D11
data_type: string

tool: D0: %d

tool: D22: clear

tool: D11: %d


assistant: ```c
int main(void) {
  int dec = 1;
  while (1) {
    int in = 0;
    scanf("%d", &in);
    system("clear");
    int bin = (1 << in);
    printf("%d\n", bin);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  lea    D26(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r14
  nop
L8:
  mov    %r14,%rsi
  mov    %r13,%rdi
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <system@plt>
  mov    0x4(%rsp),%ecx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  shl    %cl,%esi
  call   <printf@plt>
  jmp    L8
```

tool_calls: name: parse_data
arguments: data_label: D13
type_guessed: string
name: parse_data
arguments: data_label: D9
type_guessed: string
name: parse_data
arguments: data_label: D26
type_guessed: string

tool: D13: %d

tool: D9: clear

tool: D26: %d


assistant: ```c
int main(void) {
  int dec = 1;
  while (1) {
    int in = 0;
    scanf("%d", &in);
    system("clear");
    int bin = (1 << in);
    printf("%d\n", bin);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<checkSimChars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L11
L35:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L16
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  jmp    L8
L16:
  addq   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
char checkSimChars(char *str1, char *str2) {
  for (; *str1; str1++)
    if (strchr(str2, *str1))
      return *str1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<checkSimChars>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L17
L10:
  movsbl %al,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L26
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L10
  jmp    L17
L26:
  movzbl (%rbx),%eax
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char checkSimChars(char *str1, char *str2) {
  for (; *str1; str1++)
    if (strchr(str2, *str1))
      return *str1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<checkSimChars>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L3
  mov    %rdi,%rbx
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L26:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  je     L3
L33:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L26
  movzbl (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char checkSimChars(char *str1, char *str2) {
  for (; *str1; str1++)
    if (strchr(str2, *str1))
      return *str1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<checkSimChars>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L6
  mov    %rdi,%rbx
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L17:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  je     L6
L27:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L17
  movzbl (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L6:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char checkSimChars(char *str1, char *str2) {
  for (; *str1; str1++)
    if (strchr(str2, *str1))
      return *str1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<malloc_recursive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0xa,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L10
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <malloc_recursive>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L12
L10:
  nop
L12:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string

tool: D4: addr=%p, size=%d


assistant: ```c
void malloc_recursive(int i) {
  size_t size = i * 1024 * sizeof(char);
  void *ptr;
  if (i == 0)
    return;
  ptr = malloc(size);
  printf("addr=%p, size=%d\n", ptr, size);
  malloc_recursive(i - 1);
  free(ptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<malloc_recursive>:
  endbr64
  test   %edi,%edi
  jne    L9
  ret
L9:
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %edi,%r12d
  shl    $0xa,%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12,%rdx
  mov    %rax,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    -0x1(%rbx),%edi
  call   <malloc_recursive>
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
data_type: string

tool: D8: addr=%p, size=%d


assistant: ```c
void malloc_recursive(int i) {
  size_t size = i * 1024 * sizeof(char);
  void *ptr;
  if (i == 0)
    return;
  ptr = malloc(size);
  printf("addr=%p, size=%d\n", ptr, size);
  malloc_recursive(i - 1);
  free(ptr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<malloc_recursive>:
  endbr64
  test   %edi,%edi
  jne    L7
  ret
  nopl   0x0(%rax)
L7:
  push   %r15
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  shl    $0xa,%r12d
  push   %rbx
  movslq %r12d,%r13
  mov    %edi,%ebx
  mov    %r13,%rdi
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r14,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <printf@plt>
  cmp    $0x1,%ebx
  je     L46
  lea    -0x400(%r12),%r15d
  movslq %r15d,%r15
  mov    %r15,%rdi
  call   <malloc@plt>
  mov    %r15,%rdx
  mov    %r14,%rdi
  mov    %rax,%rsi
  mov    %rax,%r13
  call   <printf@plt>
  cmp    $0x2,%ebx
  jne    L22
L45:
  mov    %r13,%rdi
  call   <free@plt>
L46:
  add    $0x18,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <free@plt>
  cs nopw 0x0(%rax,%rax,1)
L22:
  lea    -0x800(%r12),%edx
  movslq %edx,%rdx
  mov    %rdx,%rdi
  mov    %rdx,0x8(%rsp)
  call   <malloc@plt>
  mov    0x8(%rsp),%rdx
  mov    %r14,%rdi
  mov    %rax,%rsi
  mov    %rax,%r15
  call   <printf@plt>
  cmp    $0x3,%ebx
  je     L30
  sub    $0xc00,%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %r12,%rdx
  mov    %r14,%rdi
  mov    %rax,%rsi
  mov    %rax,0x8(%rsp)
  call   <printf@plt>
  lea    -0x4(%rbx),%edi
  call   <malloc_recursive>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
L30:
  mov    %r15,%rdi
  call   <free@plt>
  jmp    L45
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string

tool: D1: addr=%p, size=%d


assistant: ```c
void malloc_recursive(int i) {
  size_t size = i * 1024 * sizeof(char);
  void *ptr;
  if (i == 0)
    return;
  ptr = malloc(size);
  printf("addr=%p, size=%d\n", ptr, size);
  malloc_recursive(i - 1);
  free(ptr);
}
```