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

```asm
<ldtoa>:
  endbr64
  push   %r12
  lea    D8(%rip),%r12
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <lddec@plt>
  push   0x28(%rsp)
  mov    %r12,%rdi
  mov    $0x2d,%edx
  push   0x28(%rsp)
  sub    %eax,%edx
  lea    D18(%rip),%rsi
  call   <sprintf@plt>
  add    $0x20,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D18
type: string

tool: D18: %.*Lf

assistant: ```c
char *ldtoa(long double x) {
  int dec = lddec(x);
  static char output[(45 + 3) - 1];
  sprintf(output, "%.*Lf", ((45 + 3) - 3 - dec), x);
  return output;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strlcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strnlen@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L31
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rax
  jmp    L14
L31:
  mov    -0x18(%rbp),%rax
  add    %rax,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L6
L30:
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x20(%rbp)
L6:
  mov    -0x38(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jae    L8
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L30
L8:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
L14:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
size_t strlcat(char *dst, char const *src, size_t s) {
  size_t i;
  size_t j = strnlen(dst, s);
  if (!s) {
    return j + strlen(src);
  }
  dst += j;
  for (i = 0; ((i < s - j) && src[i]); i++) {
    dst[i] = src[i];
  }
  dst[i] = 0;
  return j + i + strlen(src + i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlcat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  call   <strnlen@plt>
  mov    %rax,%r12
  test   %rbx,%rbx
  je     L13
  lea    0x0(%r13,%rax,1),%rcx
  mov    $0x0,%eax
  sub    %r12,%rbx
  je     L19
L6:
  movzbl 0x0(%rbp,%rax,1),%edx
  test   %dl,%dl
  je     L3
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L6
  jmp    L19
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    %r12,%rax
  jmp    L54
L3:
  mov    %rax,%rbx
L19:
  movb   $0x0,(%rcx,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    %r12,%rax
L54:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t strlcat(char *dst, char const *src, size_t s) {
  size_t i;
  size_t j = strnlen(dst, s);
  if (!s) {
    return j + strlen(src);
  }
  dst += j;
  for (i = 0; ((i < s - j) && src[i]); i++) {
    dst[i] = src[i];
  }
  dst[i] = 0;
  return j + i + strlen(src + i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strlcat>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strnlen@plt>
  mov    %rax,%r13
  test   %r12,%r12
  je     L15
  mov    %r12,%rdx
  lea    (%rbx,%rax,1),%rdi
  xor    %ebx,%ebx
  mov    %rbp,%r8
  sub    %rax,%rdx
  mov    %rdi,%rax
  jne    L20
  jmp    L58
  nopl   (%rax)
L7:
  mov    %cl,(%rdi,%rbx,1)
  add    $0x1,%rbx
  cmp    %rdx,%rbx
  je     L47
L20:
  movzbl 0x0(%rbp,%rbx,1),%ecx
  lea    0x0(%rbp,%rbx,1),%r8
  lea    (%rdi,%rbx,1),%rax
  test   %cl,%cl
  jne    L7
L58:
  movb   $0x0,(%rax)
  mov    %r8,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  add    %r13,%rax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  pop    %rbx
  add    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L47:
  lea    (%rdi,%rbx,1),%rax
  lea    0x0(%rbp,%rbx,1),%r8
  jmp    L58
```

assistant: ```c
size_t strlcat(char *dst, char const *src, size_t s) {
  size_t i;
  size_t j = strnlen(dst, s);
  if (!s) {
    return j + strlen(src);
  }
  dst += j;
  for (i = 0; ((i < s - j) && src[i]); i++) {
    dst[i] = src[i];
  }
  dst[i] = 0;
  return j + i + strlen(src + i);
}
```
user: What is the c source code of the assembly code below?

```asm
<strlcat>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strnlen@plt>
  mov    %rax,%r13
  test   %r12,%r12
  je     L53
  mov    %r12,%rdx
  lea    (%rbx,%rax,1),%rdi
  xor    %ebx,%ebx
  mov    %rbp,%r8
  sub    %rax,%rdx
  mov    %rdi,%rax
  jne    L58
  jmp    L1
  nopl   (%rax)
L12:
  mov    %cl,(%rdi,%rbx,1)
  add    $0x1,%rbx
  cmp    %rdx,%rbx
  je     L20
L58:
  movzbl 0x0(%rbp,%rbx,1),%ecx
  lea    0x0(%rbp,%rbx,1),%r8
  lea    (%rdi,%rbx,1),%rax
  test   %cl,%cl
  jne    L12
L1:
  movb   $0x0,(%rax)
  mov    %r8,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  add    %r13,%rax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    $0x8,%rsp
  pop    %rbx
  add    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  lea    (%rdi,%rbx,1),%rax
  lea    0x0(%rbp,%rbx,1),%r8
  jmp    L1
```

assistant: ```c
size_t strlcat(char *dst, char const *src, size_t s) {
  size_t i;
  size_t j = strnlen(dst, s);
  if (!s) {
    return j + strlen(src);
  }
  dst += j;
  for (i = 0; ((i < s - j) && src[i]); i++) {
    dst[i] = src[i];
  }
  dst[i] = 0;
  return j + i + strlen(src + i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<s_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L13:
  addl   $0x1,-0x4(%rbp)
L0:
  cmpl   $0x63,-0x4(%rbp)
  jg     L27
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
L27:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int s_length(char s[]) {
  int i;
  for (i = 0; (i < 100) && (s[i] != '\0'); ++i)
    ;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<s_length>:
  endbr64
  mov    $0x0,%eax
L11:
  cmpb   $0x0,(%rdi)
  je     L4
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    $0x64,%eax
  jne    L11
L4:
  ret
```

assistant: ```c
int s_length(char s[]) {
  int i;
  for (i = 0; (i < 100) && (s[i] != '\0'); ++i)
    ;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<s_length>:
  endbr64
  xor    %eax,%eax
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x1,%rax
  cmp    $0x64,%rax
  je     L22
L0:
  cmpb   $0x0,(%rdi,%rax,1)
  mov    %eax,%r8d
  jne    L9
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  mov    $0x64,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int s_length(char s[]) {
  int i;
  for (i = 0; (i < 100) && (s[i] != '\0'); ++i)
    ;
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<s_length>:
  endbr64
  xor    %eax,%eax
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L12:
  add    $0x1,%rax
  cmp    $0x64,%rax
  je     L3
L27:
  cmpb   $0x0,(%rdi,%rax,1)
  mov    %eax,%r8d
  jne    L12
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    $0x64,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int s_length(char s[]) {
  int i;
  for (i = 0; (i < 100) && (s[i] != '\0'); ++i)
    ;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x5,%edi
  call   <f1@plt>
  cmp    $0x5,%eax
  jne    L21
  mov    $0xfffffffb,%edi
  call   <f1@plt>
  cmp    $0x4,%eax
  jne    L21
  mov    $0x0,%edi
  call   <f1@plt>
  test   %eax,%eax
  je     L14
L21:
  mov    $0x0,%eax
  call   <abort@plt>
L14:
  mov    $0x5,%edi
  call   <f2@plt>
  cmp    $0xfffffffa,%eax
  jne    L39
  mov    $0xfffffffb,%edi
  call   <f2@plt>
  cmp    $0xfffffffb,%eax
  jne    L39
  mov    $0x0,%edi
  call   <f2@plt>
  cmp    $0xffffffff,%eax
  je     L59
L39:
  mov    $0x0,%eax
  call   <abort@plt>
L59:
  mov    $0x5,%edi
  call   <f3@plt>
  cmp    $0x5,%eax
  jne    L46
  mov    $0xfffffffb,%edi
  call   <f3@plt>
  cmp    $0x4,%eax
  jne    L46
  mov    $0x0,%edi
  call   <f3@plt>
  cmp    $0xffffffff,%eax
  je     L77
L46:
  mov    $0x0,%eax
  call   <abort@plt>
L77:
  mov    $0x5,%edi
  call   <f4@plt>
  cmp    $0xfffffffa,%eax
  jne    L112
  mov    $0xfffffffb,%edi
  call   <f4@plt>
  cmp    $0xfffffffb,%eax
  jne    L112
  mov    $0x0,%edi
  call   <f4@plt>
  test   %eax,%eax
  je     L16
L112:
  mov    $0x0,%eax
  call   <abort@plt>
L16:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (f1(5) != 5 || f1(-5) != 4 || f1(0) != 0)
    abort();
  if (f2(5) != -6 || f2(-5) != -5 || f2(0) != -1)
    abort();
  if (f3(5) != 5 || f3(-5) != 4 || f3(0) != -1)
    abort();
  if (f4(5) != -6 || f4(-5) != -5 || f4(0) != 0)
    abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <f1@plt>
  cmp    $0x5,%eax
  je     L118
L83:
  mov    $0x0,%eax
  call   <abort@plt>
L88:
  mov    $0x5,%edi
  call   <f2@plt>
  cmp    $0xfffffffa,%eax
  je     L91
L113:
  mov    $0x0,%eax
  call   <abort@plt>
L116:
  mov    $0x5,%edi
  call   <f3@plt>
  cmp    $0x5,%eax
  je     L156
L127:
  mov    $0x0,%eax
  call   <abort@plt>
L87:
  mov    $0x5,%edi
  call   <f4@plt>
  cmp    $0xfffffffa,%eax
  je     L142
L5:
  mov    $0x0,%eax
  call   <abort@plt>
L20:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L118:
  mov    $0xfffffffb,%edi
  call   <f1@plt>
  cmp    $0x4,%eax
  jne    L83
  mov    $0x0,%edi
  call   <f1@plt>
  test   %eax,%eax
  je     L88
  jmp    L83
L91:
  mov    $0xfffffffb,%edi
  call   <f2@plt>
  cmp    $0xfffffffb,%eax
  jne    L113
  mov    $0x0,%edi
  call   <f2@plt>
  cmp    $0xffffffff,%eax
  jne    L113
  jmp    L116
L156:
  mov    $0xfffffffb,%edi
  call   <f3@plt>
  cmp    $0x4,%eax
  jne    L127
  mov    $0x0,%edi
  call   <f3@plt>
  cmp    $0xffffffff,%eax
  jne    L127
  jmp    L87
L142:
  mov    $0xfffffffb,%edi
  call   <f4@plt>
  cmp    $0xfffffffb,%eax
  jne    L5
  mov    $0x0,%edi
  call   <f4@plt>
  test   %eax,%eax
  je     L20
  jmp    L5
```

assistant: ```c
int main() {
  if (f1(5) != 5 || f1(-5) != 4 || f1(0) != 0)
    abort();
  if (f2(5) != -6 || f2(-5) != -5 || f2(0) != -1)
    abort();
  if (f3(5) != 5 || f3(-5) != 4 || f3(0) != -1)
    abort();
  if (f4(5) != -6 || f4(-5) != -5 || f4(0) != 0)
    abort();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <f1@plt>
  cmp    $0x5,%eax
  je     L46
L139:
  xor    %eax,%eax
  call   <abort@plt>
L77:
  mov    $0x5,%edi
  call   <f2@plt>
  cmp    $0xfffffffa,%eax
  je     L9
L150:
  xor    %eax,%eax
  call   <abort@plt>
L89:
  mov    $0x5,%edi
  call   <f3@plt>
  cmp    $0x5,%eax
  je     L156
L141:
  xor    %eax,%eax
  call   <abort@plt>
L138:
  mov    $0x5,%edi
  call   <f4@plt>
  cmp    $0xfffffffa,%eax
  je     L136
L14:
  xor    %eax,%eax
  call   <abort@plt>
L2:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L46:
  mov    $0xfffffffb,%edi
  call   <f1@plt>
  cmp    $0x4,%eax
  jne    L139
  xor    %edi,%edi
  call   <f1@plt>
  test   %eax,%eax
  jne    L139
  jmp    L77
L136:
  mov    $0xfffffffb,%edi
  call   <f4@plt>
  cmp    $0xfffffffb,%eax
  jne    L14
  xor    %edi,%edi
  call   <f4@plt>
  test   %eax,%eax
  jne    L14
  jmp    L2
L156:
  mov    $0xfffffffb,%edi
  call   <f3@plt>
  cmp    $0x4,%eax
  jne    L141
  xor    %edi,%edi
  call   <f3@plt>
  add    $0x1,%eax
  jne    L141
  jmp    L138
L9:
  mov    $0xfffffffb,%edi
  call   <f2@plt>
  cmp    $0xfffffffb,%eax
  jne    L150
  xor    %edi,%edi
  call   <f2@plt>
  add    $0x1,%eax
  jne    L150
  jmp    L89
```

assistant: ```c
int main() {
  if (f1(5) != 5 || f1(-5) != 4 || f1(0) != 0)
    abort();
  if (f2(5) != -6 || f2(-5) != -5 || f2(0) != -1)
    abort();
  if (f3(5) != 5 || f3(-5) != 4 || f3(0) != -1)
    abort();
  if (f4(5) != -6 || f4(-5) != -5 || f4(0) != 0)
    abort();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <f1@plt>
  cmp    $0x5,%eax
  je     L138
L157:
  xor    %eax,%eax
  call   <abort@plt>
L140:
  mov    $0x5,%edi
  call   <f2@plt>
  cmp    $0xfffffffa,%eax
  je     L93
L97:
  xor    %eax,%eax
  call   <abort@plt>
L78:
  mov    $0x5,%edi
  call   <f3@plt>
  cmp    $0x5,%eax
  je     L110
L118:
  xor    %eax,%eax
  call   <abort@plt>
L1:
  mov    $0x5,%edi
  call   <f4@plt>
  cmp    $0xfffffffa,%eax
  je     L52
L37:
  xor    %eax,%eax
  call   <abort@plt>
L58:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L138:
  mov    $0xfffffffb,%edi
  call   <f1@plt>
  cmp    $0x4,%eax
  jne    L157
  xor    %edi,%edi
  call   <f1@plt>
  test   %eax,%eax
  jne    L157
  jmp    L140
L52:
  mov    $0xfffffffb,%edi
  call   <f4@plt>
  cmp    $0xfffffffb,%eax
  jne    L37
  xor    %edi,%edi
  call   <f4@plt>
  test   %eax,%eax
  jne    L37
  jmp    L58
L110:
  mov    $0xfffffffb,%edi
  call   <f3@plt>
  cmp    $0x4,%eax
  jne    L118
  xor    %edi,%edi
  call   <f3@plt>
  add    $0x1,%eax
  jne    L118
  jmp    L1
L93:
  mov    $0xfffffffb,%edi
  call   <f2@plt>
  cmp    $0xfffffffb,%eax
  jne    L97
  xor    %edi,%edi
  call   <f2@plt>
  add    $0x1,%eax
  jne    L97
  jmp    L78
```

assistant: ```c
int main() {
  if (f1(5) != 5 || f1(-5) != 4 || f1(0) != 0)
    abort();
  if (f2(5) != -6 || f2(-5) != -5 || f2(0) != -1)
    abort();
  if (f3(5) != 5 || f3(-5) != 4 || f3(0) != -1)
    abort();
  if (f4(5) != -6 || f4(-5) != -5 || f4(0) != 0)
    abort();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L15
L19:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
  mov    $0x0,%eax
  jmp    L37
L35:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L19
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L37:
  pop    %rbp
  ret
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  while (*str1 == *str2) {
    if (*str1 == '\0') {
      return 0;
    }
    str1++;
    str2++;
  }
  return (*(unsigned char *)str1) - (*(unsigned char *)str2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %cl,%dl
  jne    L28
  mov    $0x1,%eax
L4:
  test   %dl,%dl
  je     L8
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  je     L4
L28:
  movzbl %dl,%eax
  movzbl %cl,%ecx
  sub    %ecx,%eax
  ret
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  while (*str1 == *str2) {
    if (*str1 == '\0') {
      return 0;
    }
    str1++;
    str2++;
  }
  return (*(unsigned char *)str1) - (*(unsigned char *)str2);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %dl,%cl
  jne    L1
  mov    $0x1,%eax
  jmp    L29
  nopl   (%rax)
L20:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L1
L29:
  test   %dl,%dl
  jne    L20
  xor    %eax,%eax
  ret
L1:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  while (*str1 == *str2) {
    if (*str1 == '\0') {
      return 0;
    }
    str1++;
    str2++;
  }
  return (*(unsigned char *)str1) - (*(unsigned char *)str2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %dl,%cl
  jne    L31
  mov    $0x1,%eax
  jmp    L18
  nopl   (%rax)
L38:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L31
L18:
  test   %dl,%dl
  jne    L38
  xor    %eax,%eax
  ret
L31:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  while (*str1 == *str2) {
    if (*str1 == '\0') {
      return 0;
    }
    str1++;
    str2++;
  }
  return (*(unsigned char *)str1) - (*(unsigned char *)str2);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putGawbleBE>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  sar    $0x18,%edx
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl %cl,%edx
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl %cl,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  movzbl %dl,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void putGawbleBE(uint8_t *buffer, uint32_t value) {
  *buffer++ = value >> 24;
  *buffer++ = (value >> 16) & 0xff;
  *buffer++ = (value >> 8) & 0xff;
  *buffer++ = value & 0xff;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<putGawbleBE>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%edx
  sar    $0x18,%edx
  mov    %edx,(%rdi)
  mov    %esi,%edx
  sar    $0x10,%edx
  and    $0xff,%edx
  mov    %edx,0x4(%rdi)
  movzbl %ah,%edx
  mov    %edx,0x8(%rdi)
  and    $0xff,%eax
  mov    %eax,0xc(%rdi)
  ret
```

assistant: ```c
void putGawbleBE(uint8_t *buffer, uint32_t value) {
  *buffer++ = value >> 24;
  *buffer++ = (value >> 16) & 0xff;
  *buffer++ = (value >> 8) & 0xff;
  *buffer++ = value & 0xff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<putGawbleBE>:
  endbr64
  mov    %esi,%edx
  mov    %esi,%eax
  sar    $0x18,%edx
  mov    %edx,(%rdi)
  mov    %esi,%edx
  sar    $0x10,%edx
  and    $0xff,%edx
  mov    %edx,0x4(%rdi)
  movzbl %ah,%edx
  and    $0xff,%eax
  mov    %edx,0x8(%rdi)
  mov    %eax,0xc(%rdi)
  ret
```

assistant: ```c
void putGawbleBE(uint8_t *buffer, uint32_t value) {
  *buffer++ = value >> 24;
  *buffer++ = (value >> 16) & 0xff;
  *buffer++ = (value >> 8) & 0xff;
  *buffer++ = value & 0xff;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putGawbleBE>:
  endbr64
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %esi,%eax
  sar    $0x18,%ecx
  sar    $0x10,%edx
  movd   %ecx,%xmm0
  movzbl %dl,%edx
  movzbl %ah,%ecx
  movzbl %sil,%eax
  movd   %ecx,%xmm1
  movd   %eax,%xmm2
  movd   %edx,%xmm3
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,(%rdi)
  ret
```

assistant: ```c
void putGawbleBE(uint8_t *buffer, uint32_t value) {
  *buffer++ = value >> 24;
  *buffer++ = (value >> 16) & 0xff;
  *buffer++ = (value >> 8) & 0xff;
  *buffer++ = value & 0xff;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x18(%rbp)
  movl   $0x14,-0x14(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x28,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x3c,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D18
type_guessed: string

tool: D18: a=%d,b=%d,*p=%d


assistant: ```c
int main() {
  int a, b;
  int *p;
  a = 10, b = 20;
  p = &a;
  *p = 40;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  *p = b;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  p = &b;
  *p = 60;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x28,%ecx
  mov    $0x14,%edx
  mov    $0x28,%esi
  lea    D9(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x14,%ecx
  mov    $0x14,%edx
  mov    $0x14,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3c,%ecx
  mov    $0x3c,%edx
  mov    $0x14,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_infered: string

tool: D9: a=%d,b=%d,*p=%d


assistant: ```c
int main() {
  int a, b;
  int *p;
  a = 10, b = 20;
  p = &a;
  *p = 40;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  *p = b;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  p = &b;
  *p = 60;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  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
  push   %rbp
  mov    $0x28,%ecx
  mov    $0x14,%edx
  mov    $0x28,%esi
  lea    D2(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x14,%ecx
  mov    $0x14,%edx
  mov    $0x14,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3c,%ecx
  mov    $0x3c,%edx
  mov    $0x14,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_infered: string

tool: D2: a=%d,b=%d,*p=%d


assistant: ```c
int main() {
  int a, b;
  int *p;
  a = 10, b = 20;
  p = &a;
  *p = 40;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  *p = b;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  p = &b;
  *p = 60;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  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
  push   %rbp
  mov    $0x28,%ecx
  mov    $0x14,%edx
  mov    $0x28,%esi
  lea    D5(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x14,%ecx
  mov    $0x14,%edx
  mov    $0x14,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3c,%ecx
  mov    $0x3c,%edx
  mov    $0x14,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
data_type: string

tool: D5: a=%d,b=%d,*p=%d


assistant: ```c
int main() {
  int a, b;
  int *p;
  a = 10, b = 20;
  p = &a;
  *p = 40;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  *p = b;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  p = &b;
  *p = 60;
  printf("a=%d,b=%d,*p=%d\n", a, b, *p);
  return 0;
}
```