ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L23
  nopl   (%rax)
L50:
  mov    $0x1,%ebx
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L39:
  sub    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    %ebx,%esi
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  cmp    %ebx,%esi
  jl     L64
L15:
  cmp    %ebx,%ebp
  jg     L39
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  cmp    %ebx,%esi
  jge    L15
L64:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  mov    0x4(%rsp),%esi
  cmp    %ebp,%esi
  jge    L50
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
type: string

tool: D4: %d

assistant: ```c
int main() {
  int N;
  int i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i = i + 1) {
    for (j = 1; j <= N; j = j + 1)
      if (i > j) {
        printf("%d", N - i + j);
      } else {
        printf("%d", N);
      }
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L9
  nopl   (%rax)
L21:
  mov    $0x1,%ebx
  nopl   (%rax)
L27:
  cmp    %ebx,%ebp
  jle    L22
  sub    %ebp,%esi
  add    %ebx,%esi
L22:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  cmp    %ebx,%esi
  jge    L27
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  mov    0x4(%rsp),%esi
  cmp    %ebp,%esi
  jge    L21
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D6: %d

assistant: ```c
int main() {
  int N;
  int i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i = i + 1) {
    for (j = 1; j <= N; j = j + 1)
      if (i > j) {
        printf("%d", N - i + j);
      } else {
        printf("%d", N);
      }
    printf("\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x1f,%eax
  pop    %rbp
  ret
```

assistant: ```c
int f2(float __x) {
  union {
    float __f;
    unsigned int __u;
  } __u;
  return (int)(__u.__u >> 31);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<f2>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int f2(float __x) {
  union {
    float __f;
    unsigned int __u;
  } __u;
  return (int)(__u.__u >> 31);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f2>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f2(float __x) {
  union {
    float __f;
    unsigned int __u;
  } __u;
  return (int)(__u.__u >> 31);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<f2>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f2(float __x) {
  union {
    float __f;
    unsigned int __u;
  } __u;
  return (int)(__u.__u >> 31);
}
```
user: Translate the below assembly code into C programming language:

```asm
<tlv_len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x8,%eax
  or     -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int tlv_len(unsigned char *outp) {
  int byte1 = *outp & 1;
  int byte2 = *(outp + 1);
  return byte1 << 8 | byte2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tlv_len>:
  endbr64
  movzbl (%rdi),%eax
  and    $0x1,%eax
  shl    $0x8,%eax
  movzbl 0x1(%rdi),%edx
  or     %edx,%eax
  ret
```

assistant: ```c
int tlv_len(unsigned char *outp) {
  int byte1 = *outp & 1;
  int byte2 = *(outp + 1);
  return byte1 << 8 | byte2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tlv_len>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  and    $0x1,%eax
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
int tlv_len(unsigned char *outp) {
  int byte1 = *outp & 1;
  int byte2 = *(outp + 1);
  return byte1 << 8 | byte2;
}
```
user: What is the c source code of the assembly code below?

```asm
<tlv_len>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  and    $0x1,%eax
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
int tlv_len(unsigned char *outp) {
  int byte1 = *outp & 1;
  int byte2 = *(outp + 1);
  return byte1 << 8 | byte2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tlv_append_empty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <tlv_get_length@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x4,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  add    $0x4,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void tlv_append_empty(uint8_t *p_tlv) {
  uint16_t i_length = tlv_get_length(p_tlv);
  p_tlv[4 + i_length] = 0 >> 8;
  p_tlv[4 + i_length + 1] = 0 & 0xff;
}
```
user: What is the c source code of the assembly code below:

```asm
<tlv_append_empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <tlv_get_length@plt>
  add    $0x4,%eax
  cltq
  movl   $0x0,(%rbx,%rax,4)
  movl   $0x0,0x4(%rbx,%rax,4)
  pop    %rbx
  ret
```

assistant: ```c
void tlv_append_empty(uint8_t *p_tlv) {
  uint16_t i_length = tlv_get_length(p_tlv);
  p_tlv[4 + i_length] = 0 >> 8;
  p_tlv[4 + i_length + 1] = 0 & 0xff;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tlv_append_empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <tlv_get_length@plt>
  add    $0x4,%eax
  cltq
  movl   $0x0,(%rbx,%rax,4)
  movl   $0x0,0x4(%rbx,%rax,4)
  pop    %rbx
  ret
```

assistant: ```c
void tlv_append_empty(uint8_t *p_tlv) {
  uint16_t i_length = tlv_get_length(p_tlv);
  p_tlv[4 + i_length] = 0 >> 8;
  p_tlv[4 + i_length + 1] = 0 & 0xff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<tlv_append_empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <tlv_get_length@plt>
  add    $0x4,%eax
  cltq
  movq   $0x0,(%rbx,%rax,4)
  pop    %rbx
  ret
```

assistant: ```c
void tlv_append_empty(uint8_t *p_tlv) {
  uint16_t i_length = tlv_get_length(p_tlv);
  p_tlv[4 + i_length] = 0 >> 8;
  p_tlv[4 + i_length + 1] = 0 & 0xff;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uint_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    $0x0,%ebx
  mov    $0x0,%r12d
L12:
  mov    -0x24(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  mov    %ebx,%eax
  lea    0x1(%rax),%ebx
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L12
  movslq %ebx,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  sub    $0x1,%ebx
  jmp    L20
L19:
  movslq %ebx,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  movslq %r12d,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %ebx,%edx
  lea    -0x1(%rdx),%ebx
  movslq %edx,%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    %r12d,%eax
  lea    0x1(%rax),%r12d
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x11(%rbp),%eax
  mov    %al,(%rdx)
L20:
  cmp    %r12d,%ebx
  jg     L19
  mov    -0x30(%rbp),%rax
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
char *uint_to_str(unsigned int n, char *s) {
  register int i = 0, j = 0;
  char temp;
  do
    s[i++] = n - (n / 10 * 10) + '0';
  while (n /= 10);
  s[i] = '\0';
  i--;
  while (i > j) {
    temp = s[i];
    s[i--] = s[j];
    s[j++] = temp;
  }
  return s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<uint_to_str>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%r10
  mov    $0x0,%r9d
  mov    $0xcccccccd,%r11d
L0:
  mov    %r9d,%edx
  add    $0x1,%r9d
  mov    %ecx,%eax
  imul   %r11,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r8d
  add    %r8d,%r8d
  mov    %ecx,%edi
  sub    %r8d,%edi
  add    $0x30,%edi
  mov    %dil,(%r10)
  mov    %ecx,%edi
  mov    %eax,%ecx
  add    $0x1,%r10
  cmp    $0x9,%edi
  ja     L0
  movslq %r9d,%r9
  movb   $0x0,(%rsi,%r9,1)
  test   %edx,%edx
  jle    L6
  movslq %edx,%rax
  mov    $0x0,%edx
L9:
  movzbl (%rsi,%rax,1),%ecx
  movzbl (%rsi,%rdx,1),%edi
  mov    %dil,(%rsi,%rax,1)
  mov    %cl,(%rsi,%rdx,1)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L9
L6:
  mov    %rsi,%rax
  ret
```

assistant: ```c
char *uint_to_str(unsigned int n, char *s) {
  register int i = 0, j = 0;
  char temp;
  do
    s[i++] = n - (n / 10 * 10) + '0';
  while (n /= 10);
  s[i] = '\0';
  i--;
  while (i > j) {
    temp = s[i];
    s[i--] = s[j];
    s[j++] = temp;
  }
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uint_to_str>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%r8
  mov    %rsi,%r10
  xor    %r9d,%r9d
  mov    $0xcccccccd,%r11d
  nopl   (%rax)
L13:
  mov    %ecx,%eax
  mov    %ecx,%esi
  mov    %r9d,%edx
  add    $0x1,%r10
  imul   %r11,%rax
  add    $0x1,%r9d
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,-0x1(%r10)
  mov    %ecx,%esi
  mov    %eax,%ecx
  cmp    $0x9,%esi
  ja     L13
  movslq %r9d,%r9
  movb   $0x0,(%r8,%r9,1)
  test   %edx,%edx
  je     L14
  movslq %edx,%rax
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L10:
  movzbl (%r8,%rax,1),%ecx
  movzbl (%r8,%rdx,1),%esi
  mov    %sil,(%r8,%rax,1)
  sub    $0x1,%rax
  mov    %cl,(%r8,%rdx,1)
  add    $0x1,%rdx
  cmp    %eax,%edx
  jl     L10
L14:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *uint_to_str(unsigned int n, char *s) {
  register int i = 0, j = 0;
  char temp;
  do
    s[i++] = n - (n / 10 * 10) + '0';
  while (n /= 10);
  s[i] = '\0';
  i--;
  while (i > j) {
    temp = s[i];
    s[i--] = s[j];
    s[j++] = temp;
  }
  return s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<uint_to_str>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%r8
  mov    %rsi,%r10
  xor    %r9d,%r9d
  mov    $0xcccccccd,%r11d
  nopl   (%rax)
L1:
  mov    %ecx,%eax
  mov    %ecx,%esi
  mov    %r9d,%edx
  add    $0x1,%r10
  imul   %r11,%rax
  add    $0x1,%r9d
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,-0x1(%r10)
  mov    %ecx,%esi
  mov    %eax,%ecx
  cmp    $0x9,%esi
  ja     L1
  movslq %r9d,%r9
  movb   $0x0,(%r8,%r9,1)
  test   %edx,%edx
  je     L5
  movslq %edx,%rax
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L21:
  movzbl (%r8,%rax,1),%ecx
  movzbl (%r8,%rdx,1),%esi
  mov    %sil,(%r8,%rax,1)
  sub    $0x1,%rax
  mov    %cl,(%r8,%rdx,1)
  add    $0x1,%rdx
  cmp    %eax,%edx
  jl     L21
L5:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *uint_to_str(unsigned int n, char *s) {
  register int i = 0, j = 0;
  char temp;
  do
    s[i++] = n - (n / 10 * 10) + '0';
  while (n /= 10);
  s[i] = '\0';
  i--;
  while (i > j) {
    temp = s[i];
    s[i--] = s[j];
    s[j++] = temp;
  }
  return s;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ZeroXmms@plt>
  mov    $0x0,%eax
  call   <DoXmm@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  ZeroXmms();
  DoXmm();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ZeroXmms@plt>
  mov    $0x0,%eax
  call   <DoXmm@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ZeroXmms();
  DoXmm();
  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
  xor    %eax,%eax
  call   <ZeroXmms@plt>
  xor    %eax,%eax
  call   <DoXmm@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ZeroXmms();
  DoXmm();
  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
  xor    %eax,%eax
  call   <ZeroXmms@plt>
  xor    %eax,%eax
  call   <DoXmm@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ZeroXmms();
  DoXmm();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fn5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x30(%rbp),%rax
  add    $0x4,%rax
  lea    -0x30(%rbp),%rdx
  movl   $0xfa1e0ff3,(%rax)
  movw   $0xbb49,0x4(%rax)
  lea    -0x58(%rip),%rcx        # 1139 <should_appear2.0>
  mov    %rcx,0x6(%rax)
  movw   $0xba49,0xe(%rax)
  mov    %rdx,0x10(%rax)
  movl   $0x90e3ff49,0x18(%rax)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  lea    -0x30(%rbp),%rax
  add    $0x4,%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <check@plt>
  mov    -0x30(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int fn5(int x) {
  int i = x;
  inline int should_appear2(void) { return i; }
  check(should_appear2);
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn5>:
  endbr64
  sub    $0x38,%rsp
  mov    %edi,%eax
  mov    %fs:0x28,%rdx
  mov    %rdx,0x28(%rsp)
  xor    %edx,%edx
  lea    0x40(%rsp),%rcx
  mov    %rcx,0x20(%rsp)
  mov    %rsp,%rdx
  lea    0x4(%rsp),%rdi
  movl   $0xfa1e0ff3,0x4(%rsp)
  movw   $0xbb49,0x8(%rsp)
  lea    -0x4a(%rip),%rsi        # 1139 <should_appear2.0>
  mov    %rsi,0xa(%rsp)
  movw   $0xba49,0x12(%rsp)
  mov    %rdx,0x14(%rsp)
  movl   $0x90e3ff49,0x1c(%rsp)
  mov    %eax,(%rsp)
  mov    $0x0,%eax
  call   <check@plt>
  mov    (%rsp),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn5(int x) {
  int i = x;
  inline int should_appear2(void) { return i; }
  check(should_appear2);
  return i;
}
```
user: What is the c source code of the assembly code below:

```asm
<fn5>:
  endbr64
  sub    $0x38,%rsp
  lea    -0x1f(%rip),%rcx        # 1140 <should_appear2.0>
  mov    %edi,%r8d
  mov    $0xffffbb49,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rax
  lea    0x4(%rsp),%rdi
  mov    %r8d,(%rsp)
  mov    %rax,0x20(%rsp)
  mov    %rsp,%rax
  mov    %rcx,0xa(%rsp)
  mov    $0xffffba49,%ecx
  mov    %rax,0x14(%rsp)
  xor    %eax,%eax
  movl   $0xfa1e0ff3,0x4(%rsp)
  mov    %dx,0x8(%rsp)
  mov    %cx,0x12(%rsp)
  movl   $0x90e3ff49,0x1c(%rsp)
  call   <check@plt>
  mov    (%rsp),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x38,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn5(int x) {
  int i = x;
  inline int should_appear2(void) { return i; }
  check(should_appear2);
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn5>:
  endbr64
  sub    $0x38,%rsp
  lea    -0x1f(%rip),%rcx        # 1140 <should_appear2.0>
  mov    %edi,%r8d
  mov    $0xffffbb49,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rax
  lea    0x4(%rsp),%rdi
  mov    %r8d,(%rsp)
  mov    %rax,0x20(%rsp)
  mov    %rsp,%rax
  mov    %rcx,0xa(%rsp)
  mov    $0xffffba49,%ecx
  mov    %rax,0x14(%rsp)
  xor    %eax,%eax
  movl   $0xfa1e0ff3,0x4(%rsp)
  mov    %dx,0x8(%rsp)
  mov    %cx,0x12(%rsp)
  movl   $0x90e3ff49,0x1c(%rsp)
  call   <check@plt>
  mov    (%rsp),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn5(int x) {
  int i = x;
  inline int should_appear2(void) { return i; }
  check(should_appear2);
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rec_inc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  cmpl   $0x0,-0x18(%rbp)
  jne    L11
  mov    -0x14(%rbp),%eax
  jmp    L4
L11:
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <rec_inc>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
L4:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string

tool: D8: nr_of_incs %d


assistant: ```c
int rec_inc(int to_inc, int nr_of_incs) {
  int res;
  printf("nr_of_incs %d\n", nr_of_incs);
  if (nr_of_incs == 0) {
    return to_inc;
  } else {
    res = rec_inc(to_inc + 1, nr_of_incs - 1);
    return res;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rec_inc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  lea    D9(%rip),%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L13
L16:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  lea    -0x1(%rbx),%esi
  lea    0x1(%rbp),%edi
  call   <rec_inc>
  mov    %eax,%ebp
  jmp    L16
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string

tool: D9: nr_of_incs %d


assistant: ```c
int rec_inc(int to_inc, int nr_of_incs) {
  int res;
  printf("nr_of_incs %d\n", nr_of_incs);
  if (nr_of_incs == 0) {
    return to_inc;
  } else {
    res = rec_inc(to_inc + 1, nr_of_incs - 1);
    return res;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<rec_inc>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  jmp    L10
  nop
L14:
  sub    $0x1,%ebx
L10:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L14
  add    $0x8,%rsp
  lea    (%r12,%r13,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: label: D8
datatype: string

tool: D8: nr_of_incs %d


assistant: ```c
int rec_inc(int to_inc, int nr_of_incs) {
  int res;
  printf("nr_of_incs %d\n", nr_of_incs);
  if (nr_of_incs == 0) {
    return to_inc;
  } else {
    res = rec_inc(to_inc + 1, nr_of_incs - 1);
    return res;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rec_inc>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  jmp    L7
  nop
L9:
  sub    $0x1,%ebx
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L9
  add    $0x8,%rsp
  lea    (%r12,%r13,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string

tool: D0: nr_of_incs %d


assistant: ```c
int rec_inc(int to_inc, int nr_of_incs) {
  int res;
  printf("nr_of_incs %d\n", nr_of_incs);
  if (nr_of_incs == 0) {
    return to_inc;
  } else {
    res = rec_inc(to_inc + 1, nr_of_incs - 1);
    return res;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<willEcho>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0xfbff,-0xb(%rbp)
  movb   $0x1,-0x9(%rbp)
  lea    -0xb(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void willEcho(int f) {
  unsigned char e[3] = {255, 251, 1};
  write(f, e, 3);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<willEcho>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movw   $0xfbff,0x5(%rsp)
  movb   $0x1,0x7(%rsp)
  lea    0x5(%rsp),%rsi
  mov    $0x3,%edx
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void willEcho(int f) {
  unsigned char e[3] = {255, 251, 1};
  write(f, e, 3);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<willEcho>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x3,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rsi
  mov    $0xfffffbff,%eax
  movb   $0x1,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void willEcho(int f) {
  unsigned char e[3] = {255, 251, 1};
  write(f, e, 3);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<willEcho>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x3,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rsi
  mov    $0xfffffbff,%eax
  movb   $0x1,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <write@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void willEcho(int f) {
  unsigned char e[3] = {255, 251, 1};
  write(f, e, 3);
}
```