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

```asm
<testVolatile>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  ret
```

assistant: ```c
void testVolatile() {
  volatile int v;
  v = 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
<encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x38(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x9e3779b9,-0x14(%rbp)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L3
L13:
  mov    -0x14(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  xor    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x20(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  add    %ecx,%eax
  xor    %edx,%eax
  add    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  xor    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x24(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  add    %ecx,%eax
  xor    %edx,%eax
  add    %eax,-0x20(%rbp)
  addl   $0x1,-0x18(%rbp)
L3:
  cmpl   $0x1f,-0x18(%rbp)
  jle    L13
  mov    -0x38(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x20(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void encrypt(uint32_t v[2], const uint32_t k[4]) {
  uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
  uint32_t delta = 0x9E3779B9;
  uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
  for (i = 0; i < 32; i++) {
    sum += delta;
    v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
    v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
  }
  v[0] = v0;
  v[1] = v1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encrypt>:
  endbr64
  push   %rbx
  mov    %rdi,%r10
  mov    (%rdi),%ecx
  mov    0x4(%rdi),%edx
  mov    (%rsi),%r11d
  mov    0x4(%rsi),%r9d
  mov    0x8(%rsi),%r8d
  mov    0xc(%rsi),%edi
  mov    $0x0,%esi
L5:
  sub    $0x61c88647,%esi
  mov    %edx,%eax
  shl    $0x4,%eax
  add    %r11d,%eax
  mov    %edx,%ebx
  sar    $0x5,%ebx
  add    %r9d,%ebx
  xor    %ebx,%eax
  lea    (%rsi,%rdx,1),%ebx
  xor    %ebx,%eax
  add    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x4,%eax
  add    %r8d,%eax
  mov    %ecx,%ebx
  sar    $0x5,%ebx
  add    %edi,%ebx
  xor    %ebx,%eax
  lea    (%rsi,%rcx,1),%ebx
  xor    %ebx,%eax
  add    %eax,%edx
  cmp    $0xc6ef3720,%esi
  jne    L5
  mov    %ecx,(%r10)
  mov    %edx,0x4(%r10)
  pop    %rbx
  ret
```

assistant: ```c
void encrypt(uint32_t v[2], const uint32_t k[4]) {
  uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
  uint32_t delta = 0x9E3779B9;
  uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
  for (i = 0; i < 32; i++) {
    sum += delta;
    v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
    v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
  }
  v[0] = v0;
  v[1] = v1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<encrypt>:
  endbr64
  push   %rbx
  mov    %rdi,%r10
  mov    (%rdi),%ecx
  mov    0x4(%rdi),%edx
  mov    (%rsi),%r11d
  mov    0x4(%rsi),%r9d
  mov    0x8(%rsi),%r8d
  mov    0xc(%rsi),%edi
  mov    $0x20,%esi
L9:
  mov    %edx,%eax
  mov    %edx,%ebx
  shl    $0x4,%eax
  sar    $0x5,%ebx
  add    %r9d,%ebx
  add    %r11d,%eax
  xor    %ebx,%eax
  lea    -0x61c88647(%rdx),%ebx
  xor    %ebx,%eax
  add    %eax,%ecx
  mov    %ecx,%eax
  mov    %ecx,%ebx
  shl    $0x4,%eax
  sar    $0x5,%ebx
  add    %edi,%ebx
  add    %r8d,%eax
  xor    %ebx,%eax
  lea    -0x61c88647(%rcx),%ebx
  xor    %ebx,%eax
  add    %eax,%edx
  sub    $0x1,%esi
  jne    L9
  mov    %ecx,(%r10)
  pop    %rbx
  mov    %edx,0x4(%r10)
  ret
```

assistant: ```c
void encrypt(uint32_t v[2], const uint32_t k[4]) {
  uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
  uint32_t delta = 0x9E3779B9;
  uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
  for (i = 0; i < 32; i++) {
    sum += delta;
    v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
    v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
  }
  v[0] = v0;
  v[1] = v1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encrypt>:
  endbr64
  push   %rbx
  mov    0x4(%rdi),%eax
  mov    %rdi,%rcx
  mov    (%rsi),%r8d
  mov    0x4(%rsi),%r9d
  mov    %eax,%edx
  mov    %eax,%r10d
  mov    0x8(%rsi),%edi
  mov    0xc(%rsi),%esi
  sar    $0x5,%edx
  shl    $0x4,%r10d
  add    %r8d,%r10d
  add    %r9d,%edx
  xor    %r10d,%edx
  lea    -0x61c88647(%rax),%r10d
  xor    %r10d,%edx
  add    (%rcx),%edx
  mov    %edx,%r10d
  mov    %edx,%r11d
  sar    $0x5,%r10d
  shl    $0x4,%r11d
  add    %edi,%r11d
  add    %esi,%r10d
  xor    %r11d,%r10d
  lea    -0x61c88647(%rdx),%r11d
  xor    %r11d,%r10d
  add    %eax,%r10d
  mov    %r10d,%eax
  mov    %r10d,%r11d
  sar    $0x5,%eax
  shl    $0x4,%r11d
  add    %r8d,%r11d
  add    %r9d,%eax
  xor    %r11d,%eax
  lea    -0x61c88647(%r10),%r11d
  xor    %r11d,%eax
  add    %edx,%eax
  mov    %eax,%edx
  mov    %eax,%r11d
  sar    $0x5,%edx
  shl    $0x4,%r11d
  add    %edi,%r11d
  add    %esi,%edx
  xor    %r11d,%edx
  lea    -0x61c88647(%rax),%r11d
  xor    %r11d,%edx
  mov    $0x1e,%r11d
  add    %r10d,%edx
L1:
  mov    %edx,%r10d
  mov    %edx,%ebx
  shl    $0x4,%r10d
  sar    $0x5,%ebx
  add    %r9d,%ebx
  add    %r8d,%r10d
  xor    %ebx,%r10d
  lea    -0x61c88647(%rdx),%ebx
  xor    %ebx,%r10d
  add    %r10d,%eax
  mov    %eax,%r10d
  mov    %eax,%ebx
  shl    $0x4,%r10d
  sar    $0x5,%ebx
  add    %esi,%ebx
  add    %edi,%r10d
  xor    %ebx,%r10d
  lea    -0x61c88647(%rax),%ebx
  xor    %ebx,%r10d
  add    %r10d,%edx
  sub    $0x1,%r11d
  jne    L1
  mov    %eax,(%rcx)
  pop    %rbx
  mov    %edx,0x4(%rcx)
  ret
```

assistant: ```c
void encrypt(uint32_t v[2], const uint32_t k[4]) {
  uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
  uint32_t delta = 0x9E3779B9;
  uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
  for (i = 0; i < 32; i++) {
    sum += delta;
    v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
    v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
  }
  v[0] = v0;
  v[1] = v1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<thread_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <print_thread_affinity@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
void *thread_func(void *args) {
  int ID = *(int *)args;
  print_thread_affinity(pthread_self(), ID);
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<thread_func>:
  endbr64
  push   %rbx
  mov    (%rdi),%ebx
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%edi
  mov    %ebx,%esi
  call   <print_thread_affinity@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
void *thread_func(void *args) {
  int ID = *(int *)args;
  print_thread_affinity(pthread_self(), ID);
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<thread_func>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    (%rdi),%ebp
  call   <pthread_self@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <print_thread_affinity@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
void *thread_func(void *args) {
  int ID = *(int *)args;
  print_thread_affinity(pthread_self(), ID);
  return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<thread_func>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    (%rdi),%ebp
  call   <pthread_self@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <print_thread_affinity@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
void *thread_func(void *args) {
  int ID = *(int *)args;
  print_thread_affinity(pthread_self(), ID);
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<pad_callsign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x9,%eax
  jle    L14
  mov    $0xffffffff,%eax
  jmp    L0
L14:
  mov    -0x8(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    $0x0,%eax
L0:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type_infered: string

tool: D4: :         :

assistant: ```c
int pad_callsign(char *target, char *input) {
  if (strlen(input) > 9)
    return -1;
  sprintf(target, ":         :");
  memcpy(target + 1, input, strlen(input));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<pad_callsign>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  cmp    $0x9,%eax
  jg     L15
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    0x1(%rbx),%rdi
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    $0x0,%eax
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L15:
  mov    $0xffffffff,%eax
  jmp    L13
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type: string

tool: D5: :         :

assistant: ```c
int pad_callsign(char *target, char *input) {
  if (strlen(input) > 9)
    return -1;
  sprintf(target, ":         :");
  memcpy(target + 1, input, strlen(input));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<pad_callsign>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x9,%eax
  jg     L19
  lea    D9(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx),%rdi
  mov    %rbp,%rsi
  mov    %eax,%edx
  call   <memcpy@plt>
  xor    %eax,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0xffffffff,%eax
  jmp    L3
```

tool_calls: name: retrieve_data
arguments: data_label: D9
datatype: string

tool: D9: :         :

assistant: ```c
int pad_callsign(char *target, char *input) {
  if (strlen(input) > 9)
    return -1;
  sprintf(target, ":         :");
  memcpy(target + 1, input, strlen(input));
  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
<pad_callsign>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x9,%eax
  jg     L10
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx),%rdi
  mov    %rbp,%rsi
  mov    %eax,%edx
  call   <memcpy@plt>
  xor    %eax,%eax
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    $0xffffffff,%eax
  jmp    L0
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: string

tool: D6: :         :

assistant: ```c
int pad_callsign(char *target, char *input) {
  if (strlen(input) > 9)
    return -1;
  sprintf(target, ":         :");
  memcpy(target + 1, input, strlen(input));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cards_have_rank>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L22:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L26
  mov    $0x1,%eax
  jmp    L30
L26:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L22
  mov    $0x0,%eax
L30:
  pop    %rbp
  ret
```

assistant: ```c
int cards_have_rank(int rank, int x[], int size) {
  int i;
  for (i = 0; i < size; ++i) {
    if (x[i] == rank)
      return 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cards_have_rank>:
  endbr64
  test   %edx,%edx
  jle    L23
  mov    %rsi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rsi,%rdx,4),%rdx
L25:
  cmp    %edi,(%rax)
  je     L20
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L25
  mov    $0x0,%eax
  ret
L23:
  mov    $0x0,%eax
  ret
L20:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int cards_have_rank(int rank, int x[], int size) {
  int i;
  for (i = 0; i < size; ++i) {
    if (x[i] == rank)
      return 1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cards_have_rank>:
  endbr64
  test   %edx,%edx
  jle    L14
  lea    -0x1(%rdx),%eax
  lea    0x4(%rsi,%rax,4),%rax
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L37:
  add    $0x4,%rsi
  cmp    %rax,%rsi
  je     L14
L9:
  cmp    %edi,(%rsi)
  jne    L37
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
int cards_have_rank(int rank, int x[], int size) {
  int i;
  for (i = 0; i < size; ++i) {
    if (x[i] == rank)
      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
<cards_have_rank>:
  endbr64
  test   %edx,%edx
  jle    L13
  lea    -0x1(%rdx),%eax
  lea    0x4(%rsi,%rax,4),%rax
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L32:
  add    $0x4,%rsi
  cmp    %rax,%rsi
  je     L13
L5:
  cmp    %edi,(%rsi)
  jne    L32
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
int cards_have_rank(int rank, int x[], int size) {
  int i;
  for (i = 0; i < size; ++i) {
    if (x[i] == rank)
      return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcspn1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L24
L7:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L33
  mov    -0x10(%rbp),%rax
  jmp    L28
L33:
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L24:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  mov    -0x10(%rbp),%rax
L28:
  pop    %rbp
  ret
```

assistant: ```c
size_t strcspn1(const char *s1, const char *s2) {
  size_t length = 0;
  char *copy = s2;
  while (*s1 != '\0') {
    while (*copy != '\0') {
      if (*s1 == *copy)
        return length;
      copy++;
    }
    copy = s2;
    length++;
    s1++;
  }
  return length;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strcspn1>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L51
  movzbl (%rsi),%r9d
  mov    $0x0,%r8d
L18:
  mov    %rsi,%rdx
  mov    %r9d,%eax
  test   %r9b,%r9b
  je     L17
L28:
  cmp    %cl,%al
  je     L57
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L28
L17:
  add    $0x1,%r8
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  jne    L18
  jmp    L57
L51:
  mov    $0x0,%r8d
L57:
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t strcspn1(const char *s1, const char *s2) {
  size_t length = 0;
  char *copy = s2;
  while (*s1 != '\0') {
    while (*copy != '\0') {
      if (*s1 == *copy)
        return length;
      copy++;
    }
    copy = s2;
    length++;
    s1++;
  }
  return length;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strcspn1>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L44
  movzbl (%rsi),%r9d
  xor    %r8d,%r8d
L36:
  mov    %rsi,%rdx
  mov    %r9d,%eax
  test   %r9b,%r9b
  jne    L6
  jmp    L34
  nop
L63:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L34
L6:
  cmp    %cl,%al
  jne    L63
L66:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L34:
  add    $0x1,%r8
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  jne    L36
  jmp    L66
L44:
  xor    %r8d,%r8d
  jmp    L66
```

assistant: ```c
size_t strcspn1(const char *s1, const char *s2) {
  size_t length = 0;
  char *copy = s2;
  while (*s1 != '\0') {
    while (*copy != '\0') {
      if (*s1 == *copy)
        return length;
      copy++;
    }
    copy = s2;
    length++;
    s1++;
  }
  return length;
}
```
user: What is the c source code of the assembly code below?

```asm
<strcspn1>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L61
  movzbl (%rsi),%r9d
  xor    %r8d,%r8d
L9:
  mov    %rsi,%rdx
  mov    %r9d,%eax
  test   %r9b,%r9b
  jne    L74
  jmp    L21
  nop
L54:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L21
L74:
  cmp    %cl,%al
  jne    L54
L59:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L21:
  add    $0x1,%r8
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  jne    L9
  jmp    L59
L61:
  xor    %r8d,%r8d
  jmp    L59
```

assistant: ```c
size_t strcspn1(const char *s1, const char *s2) {
  size_t length = 0;
  char *copy = s2;
  while (*s1 != '\0') {
    while (*copy != '\0') {
      if (*s1 == *copy)
        return length;
      copy++;
    }
    copy = s2;
    length++;
    s1++;
  }
  return length;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fun4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: what


assistant: ```c
void fun4() {
  printf("what\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<fun4>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D4: what


assistant: ```c
void fun4() {
  printf("what\n");
  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
<fun4>:
  endbr64
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type: string

tool: D9: what


assistant: ```c
void fun4() {
  printf("what\n");
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fun4>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: string

tool: D5: what


assistant: ```c
void fun4() {
  printf("what\n");
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<app_func1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    $0x0,%eax
  call   <boot_func1@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int app_func1(int i1, int i2) {
  boot_func1();
  return (i1 + i2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<app_func1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    $0x0,%eax
  call   <boot_func1@plt>
  lea    (%rbx,%rbp,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int app_func1(int i1, int i2) {
  boot_func1();
  return (i1 + i2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<app_func1>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <boot_func1@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int app_func1(int i1, int i2) {
  boot_func1();
  return (i1 + i2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<app_func1>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <boot_func1@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int app_func1(int i1, int i2) {
  boot_func1();
  return (i1 + i2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D17
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: string
name: unpack_data
arguments: data_label: D18
type_infered: string

tool: D17: Andrei Glingeanu


tool: D28: 201


tool: D18: 06.09.1995


assistant: ```c
int main(void) {
  printf("Andrei Glingeanu\n");
  printf("201\n");
  printf("06.09.1995\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D10
type: string
name: unpack_data
arguments: label: D16
type: string
name: unpack_data
arguments: label: D12
type: string

tool: D10: Andrei Glingeanu


tool: D16: 201


tool: D12: 06.09.1995


assistant: ```c
int main(void) {
  printf("Andrei Glingeanu\n");
  printf("201\n");
  printf("06.09.1995\n");
  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    D19(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D19
datatype: string
name: unpack_data
arguments: data_tag: D2
datatype: string
name: unpack_data
arguments: data_tag: D14
datatype: string

tool: D19: Andrei Glingeanu


tool: D2: 201


tool: D14: 06.09.1995


assistant: ```c
int main(void) {
  printf("Andrei Glingeanu\n");
  printf("201\n");
  printf("06.09.1995\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type: string
name: parse_data
arguments: tag: D25
type: string
name: parse_data
arguments: tag: D8
type: string

tool: D0: Andrei Glingeanu


tool: D25: 201


tool: D8: 06.09.1995


assistant: ```c
int main(void) {
  printf("Andrei Glingeanu\n");
  printf("201\n");
  printf("06.09.1995\n");
  return 0;
}
```