ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<int_pow>:
  endbr64
  test   %esi,%esi
  jle    L15
  add    $0x1,%esi
  mov    $0x1,%eax
  mov    $0x1,%r8d
  cs nopw 0x0(%rax,%rax,1)
L10:
  add    $0x1,%eax
  imul   %edi,%r8d
  cmp    %esi,%eax
  jne    L10
  mov    %r8d,%eax
  ret
  nop
L15:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int int_pow(int x, int power) {
  int y = 1;
  int i = 1;
  for (i = 1; i <= power; i++) {
    y *= x;
  }
  return y;
}
```
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
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x74736574,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xd(%rbp),%rax
  mov    $0x74,%esi
  mov    %rax,%rdi
  call   <squeeze@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
type_infered: string

tool: D8: %s


assistant: ```c
main() {
  char s[] = "test";
  squeeze(s, 't');
  printf("%s\n", s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x74736574,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  lea    0x3(%rsp),%rbx
  mov    $0x74,%esi
  mov    %rbx,%rdi
  call   <squeeze@plt>
  mov    %rbx,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D8
type_infered: string

tool: D8: %s


assistant: ```c
main() {
  char s[] = "test";
  squeeze(s, 't');
  printf("%s\n", s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x74,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  movl   $0x74736574,0x3(%rsp)
  mov    %rbp,%rdi
  movb   $0x0,0x7(%rsp)
  call   <squeeze@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type: string

tool: D2: %s


assistant: ```c
main() {
  char s[] = "test";
  squeeze(s, 't');
  printf("%s\n", s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x74,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  movl   $0x74736574,0x3(%rsp)
  mov    %rbp,%rdi
  movb   $0x0,0x7(%rsp)
  call   <squeeze@plt>
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type: string

tool: D4: %s


assistant: ```c
main() {
  char s[] = "test";
  squeeze(s, 't');
  printf("%s\n", s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
datatype: string

tool: D1: hi

assistant: ```c
int main(void) {
  printf("hi");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
type: string

tool: D2: hi

assistant: ```c
int main(void) {
  printf("hi");
  return 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
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
datatype: string

tool: D6: hi

assistant: ```c
int main(void) {
  printf("hi");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
data_type: string

tool: D4: hi

assistant: ```c
int main(void) {
  printf("hi");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L14
  mov    $0x0,%eax
  jmp    L12
L14:
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L12:
  leave
  ret
```

assistant: ```c
char *strndup(const char *s, size_t n) {
  char *r;
  if (s == 0)
    return 0;
  r = (char *)malloc(n + 1);
  strncpy(r, s, n);
  r[n] = 0;
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  lea    0x1(%rsi),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *s, size_t n) {
  char *r;
  if (s == 0)
    return 0;
  r = (char *)malloc(n + 1);
  strncpy(r, s, n);
  r[n] = 0;
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  push   %rbx
  mov    %rsi,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *s, size_t n) {
  char *r;
  if (s == 0)
    return 0;
  r = (char *)malloc(n + 1);
  strncpy(r, s, n);
  r[n] = 0;
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  push   %rbx
  mov    %rsi,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *s, size_t n) {
  char *r;
  if (s == 0)
    return 0;
  r = (char *)malloc(n + 1);
  strncpy(r, s, n);
  r[n] = 0;
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bfdtx_cal_new_txinterval>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %al,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x30(%rbp)
  je     L52
  cmpq   $0x0,-0x38(%rbp)
  je     L52
  lea    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BFD_TIMEVAL2LONGUS@plt>
  cmpb   $0x1,-0x24(%rbp)
  jne    L50
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1c,%edx
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L8
L50:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,-0x14(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  add    $0x4b,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  imul   %rdx,%rax
  shr    $0x2,%rax
  movabs $0x28f5c28f5c28f5c3,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x2,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BFD_LONGUS2TIMEVAL@plt>
  jmp    L34
L52:
  nop
L34:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

assistant: ```c
void bfdtx_cal_new_txinterval(unsigned char detectmult, struct timeval *oldtv, struct timeval *newtv) {
  unsigned long long int txint;
  unsigned int percent;
  if (!oldtv || !newtv)
    return;
  BFD_TIMEVAL2LONGUS(oldtv, &txint);
  if (detectmult == 1)
    percent = rand() % 16;
  else
    percent = rand() % 26;
  txint = txint * (75 + percent) / 100;
  BFD_LONGUS2TIMEVAL(&txint, newtv);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<bfdtx_cal_new_txinterval>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L34
  mov    %edi,%ebp
  mov    %rsi,%rdi
  mov    %rdx,%rbx
  test   %rdx,%rdx
  je     L34
  mov    %rsp,%rsi
  call   <BFD_TIMEVAL2LONGUS@plt>
  cmp    $0x1,%bpl
  je     L23
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
L41:
  lea    0x4b(%rdx),%edx
  imul   (%rsp),%rdx
  shr    $0x2,%rdx
  movabs $0x28f5c28f5c28f5c3,%rcx
  mov    %rdx,%rax
  mul    %rcx
  shr    $0x2,%rdx
  mov    %rdx,(%rsp)
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <BFD_LONGUS2TIMEVAL@plt>
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1c,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0xf,%edx
  sub    %ecx,%edx
  jmp    L41
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bfdtx_cal_new_txinterval(unsigned char detectmult, struct timeval *oldtv, struct timeval *newtv) {
  unsigned long long int txint;
  unsigned int percent;
  if (!oldtv || !newtv)
    return;
  BFD_TIMEVAL2LONGUS(oldtv, &txint);
  if (detectmult == 1)
    percent = rand() % 16;
  else
    percent = rand() % 26;
  txint = txint * (75 + percent) / 100;
  BFD_LONGUS2TIMEVAL(&txint, newtv);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bfdtx_cal_new_txinterval>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L10
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L10
  mov    %rsp,%rbp
  mov    %edi,%ebx
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  call   <BFD_TIMEVAL2LONGUS@plt>
  xor    %eax,%eax
  cmp    $0x1,%bl
  je     L41
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
L44:
  add    $0x4b,%edx
  imul   (%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movabs $0x28f5c28f5c28f5c3,%rcx
  shr    $0x2,%rdx
  mov    %rdx,%rax
  mul    %rcx
  shr    $0x2,%rdx
  mov    %rdx,(%rsp)
  call   <BFD_LONGUS2TIMEVAL@plt>
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L41:
  call   <rand@plt>
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1c,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0xf,%edx
  sub    %ecx,%edx
  jmp    L44
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bfdtx_cal_new_txinterval(unsigned char detectmult, struct timeval *oldtv, struct timeval *newtv) {
  unsigned long long int txint;
  unsigned int percent;
  if (!oldtv || !newtv)
    return;
  BFD_TIMEVAL2LONGUS(oldtv, &txint);
  if (detectmult == 1)
    percent = rand() % 16;
  else
    percent = rand() % 26;
  txint = txint * (75 + percent) / 100;
  BFD_LONGUS2TIMEVAL(&txint, newtv);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bfdtx_cal_new_txinterval>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L8
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L8
  mov    %rsp,%rbp
  mov    %edi,%ebx
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  call   <BFD_TIMEVAL2LONGUS@plt>
  xor    %eax,%eax
  cmp    $0x1,%bl
  je     L23
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
L6:
  add    $0x4b,%edx
  imul   (%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movabs $0x28f5c28f5c28f5c3,%rcx
  shr    $0x2,%rdx
  mov    %rdx,%rax
  mul    %rcx
  shr    $0x2,%rdx
  mov    %rdx,(%rsp)
  call   <BFD_LONGUS2TIMEVAL@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L23:
  call   <rand@plt>
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1c,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0xf,%edx
  sub    %ecx,%edx
  jmp    L6
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bfdtx_cal_new_txinterval(unsigned char detectmult, struct timeval *oldtv, struct timeval *newtv) {
  unsigned long long int txint;
  unsigned int percent;
  if (!oldtv || !newtv)
    return;
  BFD_TIMEVAL2LONGUS(oldtv, &txint);
  if (detectmult == 1)
    percent = rand() % 16;
  else
    percent = rand() % 26;
  txint = txint * (75 + percent) / 100;
  BFD_LONGUS2TIMEVAL(&txint, newtv);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ld2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x10,-0x4(%rbp)
  jmp    L19
L14:
  mov    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  ja     L22
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L22:
  shrl   -0x4(%rbp)
L19:
  cmpl   $0x1,-0x14(%rbp)
  jg     L14
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int ld2(u_int32_t v) {
  unsigned int r, bits;
  for (r = 0, bits = 16; v > 1; bits >>= 1) {
    if (v >= (1U << bits)) {
      v >>= bits;
      r += bits;
    }
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ld2>:
  endbr64
  cmp    $0x1,%edi
  jle    L29
  mov    $0x10,%ecx
  mov    $0x0,%edx
  mov    $0x1,%esi
  jmp    L15
L18:
  shr    %ecx
  cmp    $0x1,%edi
  jle    L12
L15:
  mov    %esi,%eax
  shl    %cl,%eax
  cmp    %edi,%eax
  ja     L18
  sar    %cl,%edi
  add    %ecx,%edx
  jmp    L18
L29:
  mov    $0x0,%edx
L12:
  mov    %edx,%eax
  ret
```

assistant: ```c
unsigned int ld2(u_int32_t v) {
  unsigned int r, bits;
  for (r = 0, bits = 16; v > 1; bits >>= 1) {
    if (v >= (1U << bits)) {
      v >>= bits;
      r += bits;
    }
  }
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ld2>:
  endbr64
  cmp    $0x1,%edi
  jle    L38
  mov    $0x10,%ecx
  xor    %r8d,%r8d
  mov    $0x1,%edx
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %edx,%eax
  shl    %cl,%eax
  cmp    %edi,%eax
  ja     L16
  sar    %cl,%edi
  add    %ecx,%r8d
  shr    %ecx
  cmp    $0x1,%edi
  jg     L27
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  shr    %ecx
  jmp    L27
  nopl   0x0(%rax)
L38:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int ld2(u_int32_t v) {
  unsigned int r, bits;
  for (r = 0, bits = 16; v > 1; bits >>= 1) {
    if (v >= (1U << bits)) {
      v >>= bits;
      r += bits;
    }
  }
  return r;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ld2>:
  endbr64
  cmp    $0x1,%edi
  jle    L2
  mov    $0x10,%ecx
  xor    %r8d,%r8d
  mov    $0x1,%edx
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    %edx,%eax
  shl    %cl,%eax
  cmp    %edi,%eax
  ja     L11
  sar    %cl,%edi
  add    %ecx,%r8d
  shr    %ecx
  cmp    $0x1,%edi
  jg     L8
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  shr    %ecx
  jmp    L8
  nopl   0x0(%rax)
L2:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int ld2(u_int32_t v) {
  unsigned int r, bits;
  for (r = 0, bits = 16; v > 1; bits >>= 1) {
    if (v >= (1U << bits)) {
      v >>= bits;
      r += bits;
    }
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<char_to_charhexa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,D1(%rip)
  movl   $0x0,-0x10(%rbp)
  jmp    L2
L11:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  lea    -0xb(%rbp),%rax
  lea    D6(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0xb(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <strcat@plt>
  addl   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L11
  lea    D1(%rip),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D6
datatype: string

tool: D6: %02hx

assistant: ```c
char *char_to_charhexa(unsigned char *data, int l) {
  static char resultat[3000];
  int n;
  char ch[3];
  resultat[0] = '\0';
  for (n = 0; n < l; n++) {
    sprintf(ch, "%02hx", (unsigned char)data[n]);
    strcat(resultat, ch);
  }
  return resultat;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<char_to_charhexa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,D39(%rip)
  test   %esi,%esi
  jle    L10
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r14
  lea    0x5(%rsp),%rbp
  lea    D16(%rip),%r13
  lea    D39(%rip),%r12
L2:
  movzbl (%rbx),%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
  add    $0x1,%rbx
  cmp    %r14,%rbx
  jne    L2
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  lea    D39(%rip),%rax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
type_infered: string

tool: D16: %02hx

assistant: ```c
char *char_to_charhexa(unsigned char *data, int l) {
  static char resultat[3000];
  int n;
  char ch[3];
  resultat[0] = '\0';
  for (n = 0; n < l; n++) {
    sprintf(ch, "%02hx", (unsigned char)data[n]);
    strcat(resultat, ch);
  }
  return resultat;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<char_to_charhexa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D14(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,D14(%rip)
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x5(%rsp),%rbp
  lea    0x1(%rdi,%rax,1),%r14
  lea    D15(%rip),%r13
  nopl   0x0(%rax)
L14:
  movzbl (%rbx),%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
  cmp    %r14,%rbx
  jne    L14
L1:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
datatype: string

tool: D15: %02hx

assistant: ```c
char *char_to_charhexa(unsigned char *data, int l) {
  static char resultat[3000];
  int n;
  char ch[3];
  resultat[0] = '\0';
  for (n = 0; n < l; n++) {
    sprintf(ch, "%02hx", (unsigned char)data[n]);
    strcat(resultat, ch);
  }
  return resultat;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<char_to_charhexa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,D0(%rip)
  test   %esi,%esi
  jle    L16
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x5(%rsp),%rbp
  lea    0x1(%rdi,%rax,1),%r14
  lea    D23(%rip),%r13
  nopl   0x0(%rax)
L1:
  movzbl (%rbx),%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
  cmp    %r14,%rbx
  jne    L1
L16:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D23
type_infered: string

tool: D23: %02hx

assistant: ```c
char *char_to_charhexa(unsigned char *data, int l) {
  static char resultat[3000];
  int n;
  char ch[3];
  resultat[0] = '\0';
  for (n = 0; n < l; n++) {
    sprintf(ch, "%02hx", (unsigned char)data[n]);
    strcat(resultat, ch);
  }
  return resultat;
}
```
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   %rbp
  mov    %rsp,%rbp
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <open@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <open@plt>
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <open@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: string
name: parse_data
arguments: data_tag: D16
data_type: string
name: parse_data
arguments: data_tag: D6
data_type: string

tool: D10: /root/abc

tool: D16: /media/xyz

tool: D6: /mnt/abc

assistant: ```c
int main() {
  open("/root/abc");
  open("/media/xyz");
  open("/mnt/abc");
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <open@plt>
  lea    D23(%rip),%rdi
  call   <open@plt>
  lea    D18(%rip),%rdi
  call   <open@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type: string
name: disassemble_data
arguments: data_tag: D23
type: string
name: disassemble_data
arguments: data_tag: D18
type: string

tool: D9: /root/abc

tool: D23: /media/xyz

tool: D18: /mnt/abc

assistant: ```c
int main() {
  open("/root/abc");
  open("/media/xyz");
  open("/mnt/abc");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D22(%rip),%rdi
  call   <open@plt>
  lea    D21(%rip),%rdi
  call   <open@plt>
  lea    D5(%rip),%rdi
  call   <open@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D22
data_type: string
name: read_data
arguments: data_tag: D21
data_type: string
name: read_data
arguments: data_tag: D5
data_type: string

tool: D22: /root/abc

tool: D21: /media/xyz

tool: D5: /mnt/abc

assistant: ```c
int main() {
  open("/root/abc");
  open("/media/xyz");
  open("/mnt/abc");
}
```