ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<get_int_tab>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r14
  test   %ebp,%ebp
  jle    L14
  lea    (%rax,%rbp,8),%r12
  lea    -0x1(%rbx),%ebp
  mov    %rax,%r15
  lea    0x0(,%rbx,4),%r13d
  add    $0x1,%rbp
  nop
L13:
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  test   %ebx,%ebx
  jle    L39
  lea    (%rax,%rbp,4),%rdx
  nopl   0x0(%rax,%rax,1)
L12:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L12
L39:
  add    $0x8,%r15
  cmp    %r12,%r15
  jne    L13
L14:
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **get_int_tab(int x, int y) {
  int i, j;
  int **tab;
  tab = (int **)malloc(x * sizeof(int *));
  for (i = 0; i < x; i++) {
    tab[i] = (int *)malloc(y * sizeof(int));
    for (j = 0; j < y; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_int_tab>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbx,8),%edi
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %ebx,%ebx
  jle    L41
  lea    (%rax,%rbx,8),%r12
  mov    %rax,%r14
  lea    -0x1(%r15),%eax
  mov    %r15d,%ebx
  mov    %eax,0x4(%rsp)
  mov    %r15d,%eax
  lea    0x0(,%r15,4),%r13d
  and    $0xfffffffc,%ebx
  shr    $0x2,%eax
  lea    -0x1(%rax),%ebp
  add    $0x1,%rbp
  shl    $0x4,%rbp
  xchg   %ax,%ax
L11:
  mov    %r13d,%edi
  call   <malloc@plt>
  test   %r15d,%r15d
  pxor   %xmm0,%xmm0
  mov    %rax,(%r14)
  mov    %rax,%rdx
  jle    L14
  cmpl   $0x2,0x4(%rsp)
  jbe    L67
  lea    0x0(%rbp,%rax,1),%rcx
  nopl   0x0(%rax,%rax,1)
L59:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L59
  mov    %ebx,%eax
  cmp    %ebx,%r15d
  je     L14
L70:
  movslq %eax,%rcx
  movl   $0x0,(%rdx,%rcx,4)
  lea    0x0(,%rcx,4),%rdi
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r15d
  jle    L14
  add    $0x2,%eax
  movl   $0x0,0x4(%rdx,%rdi,1)
  cmp    %eax,%r15d
  jle    L14
  movl   $0x0,0x8(%rdx,%rdi,1)
L14:
  add    $0x8,%r14
  cmp    %r12,%r14
  jne    L11
L41:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  xor    %eax,%eax
  jmp    L70
```

assistant: ```c
int **get_int_tab(int x, int y) {
  int i, j;
  int **tab;
  tab = (int **)malloc(x * sizeof(int *));
  for (i = 0; i < x; i++) {
    tab[i] = (int *)malloc(y * sizeof(int));
    for (j = 0; j < y; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <unsorted_mode@plt>
  mov    $0x0,%eax
  call   <sorted_mode@plt>
  mov    $0x0,%eax
  call   <sorted2_mode@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsorted_mode();
  sorted_mode();
  sorted2_mode();
  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
  mov    $0x0,%eax
  call   <unsorted_mode@plt>
  mov    $0x0,%eax
  call   <sorted_mode@plt>
  mov    $0x0,%eax
  call   <sorted2_mode@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsorted_mode();
  sorted_mode();
  sorted2_mode();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <unsorted_mode@plt>
  xor    %eax,%eax
  call   <sorted_mode@plt>
  xor    %eax,%eax
  call   <sorted2_mode@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsorted_mode();
  sorted_mode();
  sorted2_mode();
  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   <unsorted_mode@plt>
  xor    %eax,%eax
  call   <sorted_mode@plt>
  xor    %eax,%eax
  call   <sorted2_mode@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsorted_mode();
  sorted_mode();
  sorted2_mode();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strtabsearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L11
  cmpq   $0x0,-0x20(%rbp)
  jne    L28
L11:
  mov    $0x0,%eax
  jmp    L12
L28:
  movl   $0xffffffff,-0x4(%rbp)
  jmp    L18
L48:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <my_strcmp@plt>
  test   %eax,%eax
  jne    L18
  mov    $0x1,%eax
  jmp    L12
L18:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L48
  mov    $0x0,%eax
L12:
  leave
  ret
```

assistant: ```c
int my_strtabsearch(char **tab, char *str) {
  int i;
  if (tab == NULL || str == NULL)
    return (0);
  i = -1;
  while (tab[++i] != NULL) {
    if (!my_strcmp(tab[i], str))
      return (1);
  }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strtabsearch>:
  endbr64
  test   %rdi,%rdi
  je     L43
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L3
L40:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L15
  mov    %rbp,%rsi
  call   <my_strcmp@plt>
  add    $0x8,%rbx
  test   %eax,%eax
  jne    L40
  mov    $0x1,%eax
  jmp    L2
L15:
  mov    $0x0,%eax
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  mov    $0x0,%eax
  jmp    L2
L43:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int my_strtabsearch(char **tab, char *str) {
  int i;
  if (tab == NULL || str == NULL)
    return (0);
  i = -1;
  while (tab[++i] != NULL) {
    if (!my_strcmp(tab[i], str))
      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
<my_strtabsearch>:
  endbr64
  test   %rdi,%rdi
  je     L39
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  jne    L5
  jmp    L36
  nopl   0x0(%rax)
L3:
  mov    %rbp,%rsi
  add    $0x8,%rbx
  call   <my_strcmp@plt>
  test   %eax,%eax
  je     L27
L5:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  jne    L3
L36:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L27:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L39:
  xor    %eax,%eax
  ret
```

assistant: ```c
int my_strtabsearch(char **tab, char *str) {
  int i;
  if (tab == NULL || str == NULL)
    return (0);
  i = -1;
  while (tab[++i] != NULL) {
    if (!my_strcmp(tab[i], str))
      return (1);
  }
  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
<my_strtabsearch>:
  endbr64
  test   %rdi,%rdi
  je     L3
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  jne    L17
  jmp    L21
  nopl   0x0(%rax)
L20:
  mov    %rbp,%rsi
  add    $0x8,%rbx
  call   <my_strcmp@plt>
  test   %eax,%eax
  je     L8
L17:
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  jne    L20
L21:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L8:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int my_strtabsearch(char **tab, char *str) {
  int i;
  if (tab == NULL || str == NULL)
    return (0);
  i = -1;
  while (tab[++i] != NULL) {
    if (!my_strcmp(tab[i], str))
      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
<suffix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  jge    L17
  mov    $0x0,%eax
  jmp    L34
L17:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L28
  mov    $0x1,%eax
  jmp    L34
L28:
  mov    $0x0,%eax
L34:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int suffix(const char *str, const char *suffix) {
  if (strlen(str) < strlen(suffix))
    return 0;
  if (!strcmp(suffix, str + (strlen(str) - strlen(suffix))))
    return 1;
  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
<suffix>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    %edx,%ebp
  jge    L17
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    %eax,%ebp
  movslq %ebp,%rbp
  lea    (%rbx,%rbp,1),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L1
```

assistant: ```c
int suffix(const char *str, const char *suffix) {
  if (strlen(str) < strlen(suffix))
    return 0;
  if (!strcmp(suffix, str + (strlen(str) - strlen(suffix))))
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<suffix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    %r8d,%r12d
  jl     L4
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %rbp,%rdi
  sub    %eax,%r12d
  movslq %r12d,%r12
  lea    (%rbx,%r12,1),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int suffix(const char *str, const char *suffix) {
  if (strlen(str) < strlen(suffix))
    return 0;
  if (!strcmp(suffix, str + (strlen(str) - strlen(suffix))))
    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
<suffix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    %r8d,%r12d
  jl     L4
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %rbp,%rdi
  sub    %eax,%r12d
  movslq %r12d,%r12
  lea    (%rbx,%r12,1),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int suffix(const char *str, const char *suffix) {
  if (strlen(str) < strlen(suffix))
    return 0;
  if (!strcmp(suffix, str + (strlen(str) - strlen(suffix))))
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t sz = strlen(s) + 1;
  char *d = malloc(sz);
  memcpy(d, s, sz);
  return d;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t sz = strlen(s) + 1;
  char *d = malloc(sz);
  memcpy(d, s, sz);
  return d;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strdup>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t sz = strlen(s) + 1;
  char *d = malloc(sz);
  memcpy(d, s, sz);
  return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L17
L36:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L24
  mov    -0x1c(%rbp),%eax
  jmp    L26
L24:
  addq   $0x1,-0x18(%rbp)
L17:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L36
  mov    $0x0,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
int memcmp(const void *buf1, const void *buf2, const size_t size) {
  const char *buf1c = (const char *)buf1;
  const char *buf2c = (const char *)buf2;
  for (size_t i = 0; i < size; i++) {
    int cmp = buf1c[i] - buf2c[i];
    if (cmp) {
      return cmp;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L23
  mov    $0x0,%ecx
L19:
  movsbl (%rdi,%rcx,1),%eax
  movsbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  jne    L8
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L19
  ret
L23:
  mov    $0x0,%eax
L8:
  ret
```

assistant: ```c
int memcmp(const void *buf1, const void *buf2, const size_t size) {
  const char *buf1c = (const char *)buf1;
  const char *buf2c = (const char *)buf2;
  for (size_t i = 0; i < size; i++) {
    int cmp = buf1c[i] - buf2c[i];
    if (cmp) {
      return cmp;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L27
  xor    %ecx,%ecx
  jmp    L34
  nopl   (%rax)
L37:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L39
L34:
  movsbl (%rdi,%rcx,1),%eax
  movsbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  je     L37
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  ret
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(const void *buf1, const void *buf2, const size_t size) {
  const char *buf1c = (const char *)buf1;
  const char *buf2c = (const char *)buf2;
  for (size_t i = 0; i < size; i++) {
    int cmp = buf1c[i] - buf2c[i];
    if (cmp) {
      return cmp;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L36
  xor    %ecx,%ecx
  jmp    L13
  nopl   (%rax)
L30:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L19
L13:
  movsbl (%rdi,%rcx,1),%eax
  movsbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  je     L30
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  ret
L36:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(const void *buf1, const void *buf2, const size_t size) {
  const char *buf1c = (const char *)buf1;
  const char *buf2c = (const char *)buf2;
  for (size_t i = 0; i < size; i++) {
    int cmp = buf1c[i] - buf2c[i];
    if (cmp) {
      return cmp;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<standt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L4
  mov    $0x0,%eax
  call   <standout@plt>
  jmp    L12
L4:
  mov    $0x0,%eax
  call   <standend@plt>
L12:
  nop
  leave
  ret
```

assistant: ```c
void standt(on) int on;
{
  if (on) {
    standout();
  } else {
    standend();
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<standt>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L1
  mov    $0x0,%eax
  call   <standout@plt>
L3:
  add    $0x8,%rsp
  ret
L1:
  mov    $0x0,%eax
  call   <standend@plt>
  jmp    L3
```

assistant: ```c
void standt(on) int on;
{
  if (on) {
    standout();
  } else {
    standend();
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<standt>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L8
  jmp    1060 <standout@plt>
  nop
L8:
  jmp    1070 <standend@plt>
```

assistant: ```c
void standt(on) int on;
{
  if (on) {
    standout();
  } else {
    standend();
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<standt>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L6
  jmp    1060 <standout@plt>
  nop
L6:
  jmp    1070 <standend@plt>
```

assistant: ```c
void standt(on) int on;
{
  if (on) {
    standout();
  } else {
    standend();
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x17,-0x8(%rbp)
  movl   $0xb2,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D2
datatype: string
name: unpack_data
arguments: data_tag: D0
datatype: string

tool: D2: I am %-10d years old.


tool: D0: And  %10d cm tall.


assistant: ```c
int main() {
  int age = 23;
  int height = 178;
  printf("I am %-10d years old.\n", age);
  printf("And  %10d cm tall.\n", height);
  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
  mov    $0x17,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0xb2,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string
name: unpack_data
arguments: data_tag: D0
type: string

tool: D6: I am %-10d years old.


tool: D0: And  %10d cm tall.


assistant: ```c
int main() {
  int age = 23;
  int height = 178;
  printf("I am %-10d years old.\n", age);
  printf("And  %10d cm tall.\n", height);
  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
  mov    $0x17,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0xb2,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type_guessed: string
name: extract_symbolic_data
arguments: tag: D11
type_guessed: string

tool: D3: I am %-10d years old.


tool: D11: And  %10d cm tall.


assistant: ```c
int main() {
  int age = 23;
  int height = 178;
  printf("I am %-10d years old.\n", age);
  printf("And  %10d cm tall.\n", height);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x17,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0xb2,%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D10
type_guessed: string
name: unpack_data
arguments: label: D19
type_guessed: string

tool: D10: I am %-10d years old.


tool: D19: And  %10d cm tall.


assistant: ```c
int main() {
  int age = 23;
  int height = 178;
  printf("I am %-10d years old.\n", age);
  printf("And  %10d cm tall.\n", height);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jg     L58
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L38
L58:
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L25
L45:
  mov    $0x0,%eax
  call   <fork@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <sleep@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L16:
  nop
L35:
  mov    $0x0,%edi
  call   <wait@plt>
  cmp    %rax,-0x8(%rbp)
  jne    L35
  addl   $0x1,-0x18(%rbp)
L25:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L45
  mov    $0x0,%eax
L38:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D7
type_guessed: string

tool: D7: the number of args is wrong!

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("the number of args is wrong!");
    return -1;
  }
  int n = atoi(argv[1]);
  int *ptr = malloc(sizeof(int) * n);
  pid_t pid;
  for (int i = 0; i != n; ++i) {
    pid = fork();
    if (pid == 0) {
      free(ptr);
      sleep(n - i - 1);
      exit(0);
    }
    while (wait((int *)0) != pid)
      ;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%edi
  jle    L49
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  je     L45
  mov    $0x0,%ebp
  lea    -0x1(%r12),%r14d
  jmp    L34
L49:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%r12d
  jmp    L45
L79:
  mov    $0x0,%edi
  call   <wait@plt>
  cmp    %rbx,%rax
  jne    L79
  add    $0x1,%ebp
  cmp    %ebp,%r12d
  je     L24
L34:
  mov    $0x0,%eax
  call   <fork@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L79
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r14d,%edi
  sub    %ebp,%edi
  call   <sleep@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L79
L24:
  mov    $0x0,%r12d
L45:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type: string

tool: D8: the number of args is wrong!

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("the number of args is wrong!");
    return -1;
  }
  int n = atoi(argv[1]);
  int *ptr = malloc(sizeof(int) * n);
  pid_t pid;
  for (int i = 0; i != n; ++i) {
    pid = fork();
    if (pid == 0) {
      free(ptr);
      sleep(n - i - 1);
      exit(0);
    }
    while (wait((int *)0) != pid)
      ;
  }
  return 0;
}
```