ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <foo@plt>
  cmp    $0xa,%ebx
  jne    L13
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L18:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <foo@plt>
  cmp    $0x64,%ebx
  jne    L18
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int y;
  int z;
  for (y = 0; y < 10; ++y)
    foo(y);
  for (z = 0; z < 100; ++z)
    foo(z);
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <foo@plt>
  cmp    $0xa,%ebx
  jne    L6
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L2:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <foo@plt>
  cmp    $0x64,%ebx
  jne    L2
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int y;
  int z;
  for (y = 0; y < 10; ++y)
    foo(y);
  for (z = 0; z < 100; ++z)
    foo(z);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x8,%esi
  mov    $0x100,%edi
  call   <__mingw_aligned_malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L8
  mov    $0x0,%eax
  jmp    L4
L8:
  mov    $0xfffffffe,%eax
L4:
  leave
  ret
```

assistant: ```c
int main() {
  void *result = __mingw_aligned_malloc(32 * sizeof(double), sizeof(double));
  return result ? 0 : -2;
}
```
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
  sub    $0x8,%rsp
  mov    $0x8,%esi
  mov    $0x100,%edi
  call   <__mingw_aligned_malloc@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xfffffffe,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  void *result = __mingw_aligned_malloc(32 * sizeof(double), sizeof(double));
  return result ? 0 : -2;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%esi
  mov    $0x100,%edi
  call   <__mingw_aligned_malloc@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0xfffffffe,%eax
  ret
```

assistant: ```c
int main() {
  void *result = __mingw_aligned_malloc(32 * sizeof(double), sizeof(double));
  return result ? 0 : -2;
}
```
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    $0x8,%esi
  mov    $0x100,%edi
  call   <__mingw_aligned_malloc@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0xfffffffe,%eax
  ret
```

assistant: ```c
int main() {
  void *result = __mingw_aligned_malloc(32 * sizeof(double), sizeof(double));
  return result ? 0 : -2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  lea    0x70(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0xe0(%rsp),%rdi
  mov    %rbx,%rsi
  call   <strcpy@plt>
  mov    $0x0,%r13d
  lea    0x70(%rsp),%r12
  jmp    L47
L49:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  cltq
  movzbl (%rbx,%rbp,1),%edx
  mov    %dl,(%rsp,%rax,1)
  add    $0x1,%rbx
L39:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L49
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    0xe0(%rsp),%rdi
  mov    %rbx,%rsi
  call   <strncpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
  mov    %rbx,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
L47:
  lea    0xe0(%rsp),%rdi
  mov    %r12,%rsi
  call   <strstr@plt>
  cmp    $0x4,%r13d
  jg     L16
  test   %rax,%rax
  je     L16
  add    $0x1,%r13d
  mov    %rsp,%rdi
  mov    %r12,%rsi
  call   <strstr@plt>
  mov    %rax,%rsi
  lea    0xe0(%rsp),%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  lea    0xe0(%rsp),%rbp
  jmp    L39
L16:
  lea    0x70(%rsp),%rsi
  lea    D37(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  mov    $0x0,%eax
  add    $0x158,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D31
type_infered: string
name: parse_data
arguments: label: D4
type_infered: string
name: parse_data
arguments: label: D19
type_infered: string
name: parse_data
arguments: label: D37
type_infered: string
name: parse_data
arguments: label: D28
type_infered: string

tool: D31: Enter the String: 

tool: D4: %s

tool: D19: Enter The Key: 

tool: D37: No. of Time "%s" appears is: 

tool: D28: %d 


assistant: ```c
int main() {
  printf("Enter the String: \n");
  int i = 0, count = 0;
  ;
  char str1[100], key[100], str2[100];
  scanf("%s", str1);
  printf("Enter The Key: \n");
  scanf("%s", key);
  strcpy(str2, str1);
  while (strstr(str2, key) && count < 5) {
    count++;
    strcpy(str2, strstr(str1, key));
    for (i = strlen(key); i <= strlen(str2); i++)
      str1[i - strlen(key)] = str2[i];
    strncpy(str2, str1, strlen(str1));
    str2[strlen(str1)] = '\0';
    str2[strlen(str1) + 1] = '\0';
  }
  printf("No. of Time \"%s\" appears is: ", key);
  printf("%d \n", count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D6(%rip),%rdi
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  lea    D20(%rip),%rbp
  push   %rbx
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x70(%rsp),%r12
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  lea    0xe0(%rsp),%rbp
  call   <scanf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L22
L56:
  cmp    $0x5,%r14d
  je     L22
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%r14d
  call   <strstr@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  jmp    L1
  xchg   %ax,%ax
L58:
  mov    %r12,%rdi
  call   <strlen@plt>
  movzbl 0x0(%rbp,%rbx,1),%edx
  mov    %eax,%r8d
  mov    %ebx,%eax
  add    $0x1,%rbx
  sub    %r8d,%eax
  cltq
  mov    %dl,(%rsp,%rax,1)
L1:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L58
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  call   <strncpy@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%eax
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
  call   <strstr@plt>
  test   %rax,%rax
  jne    L56
L22:
  mov    %r12,%rsi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r14d,%esi
  lea    D43(%rip),%rdi
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x150,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
type_guessed: string
name: retrieve_data
arguments: label: D20
type_guessed: string
name: retrieve_data
arguments: label: D27
type_guessed: string
name: retrieve_data
arguments: label: D25
type_guessed: string
name: retrieve_data
arguments: label: D43
type_guessed: string

tool: D6: Enter the String: 

tool: D20: %s

tool: D27: Enter The Key: 

tool: D25: No. of Time "%s" appears is: 

tool: D43: %d 


assistant: ```c
int main() {
  printf("Enter the String: \n");
  int i = 0, count = 0;
  ;
  char str1[100], key[100], str2[100];
  scanf("%s", str1);
  printf("Enter The Key: \n");
  scanf("%s", key);
  strcpy(str2, str1);
  while (strstr(str2, key) && count < 5) {
    count++;
    strcpy(str2, strstr(str1, key));
    for (i = strlen(key); i <= strlen(str2); i++)
      str1[i - strlen(key)] = str2[i];
    strncpy(str2, str1, strlen(str1));
    str2[strlen(str1)] = '\0';
    str2[strlen(str1) + 1] = '\0';
  }
  printf("No. of Time \"%s\" appears is: ", key);
  printf("%d \n", count);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D45(%rip),%rdi
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x70(%rsp),%r12
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  lea    0xe0(%rsp),%rbp
  call   <scanf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L21
L30:
  cmp    $0x5,%r14d
  je     L21
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%r14d
  call   <strstr@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  jmp    L54
  xchg   %ax,%ax
L11:
  mov    %r12,%rdi
  call   <strlen@plt>
  movzbl 0x0(%rbp,%rbx,1),%edx
  mov    %eax,%r8d
  mov    %ebx,%eax
  add    $0x1,%rbx
  sub    %r8d,%eax
  cltq
  mov    %dl,(%rsp,%rax,1)
L54:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L11
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  call   <strncpy@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%eax
  cltq
  movb   $0x0,0xe0(%rsp,%rax,1)
  call   <strstr@plt>
  test   %rax,%rax
  jne    L30
L21:
  mov    %r12,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r14d,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x150,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D45
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D44
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D11
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string

tool: D45: Enter the String: 

tool: D10: %s

tool: D44: Enter The Key: 

tool: D11: No. of Time "%s" appears is: 

tool: D5: %d 


assistant: ```c
int main() {
  printf("Enter the String: \n");
  int i = 0, count = 0;
  ;
  char str1[100], key[100], str2[100];
  scanf("%s", str1);
  printf("Enter The Key: \n");
  scanf("%s", key);
  strcpy(str2, str1);
  while (strstr(str2, key) && count < 5) {
    count++;
    strcpy(str2, strstr(str1, key));
    for (i = strlen(key); i <= strlen(str2); i++)
      str1[i - strlen(key)] = str2[i];
    strncpy(str2, str1, strlen(str1));
    str2[strlen(str1)] = '\0';
    str2[strlen(str1) + 1] = '\0';
  }
  printf("No. of Time \"%s\" appears is: ", key);
  printf("%d \n", count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<output_with_tab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  call   <print_tab@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <output_string@plt>
  nop
  leave
  ret
```

assistant: ```c
void output_with_tab(char ptr[]) {
  print_tab();
  output_string(ptr);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<output_with_tab>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x0,%eax
  call   <print_tab@plt>
  mov    %rbx,%rdi
  call   <output_string@plt>
  pop    %rbx
  ret
```

assistant: ```c
void output_with_tab(char ptr[]) {
  print_tab();
  output_string(ptr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<output_with_tab>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <print_tab@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <output_string@plt>
```

assistant: ```c
void output_with_tab(char ptr[]) {
  print_tab();
  output_string(ptr);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<output_with_tab>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <print_tab@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <output_string@plt>
```

assistant: ```c
void output_with_tab(char ptr[]) {
  print_tab();
  output_string(ptr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<compare>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jae    L16
  mov    $0xffffffff,%eax
  jmp    L36
L16:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jbe    L13
  mov    $0x1,%eax
  jmp    L36
L13:
  mov    $0x0,%eax
L36:
  pop    %rbp
  ret
```

assistant: ```c
int compare(const void *d1, const void *d2) {
  unsigned int *data1 = (unsigned int *)d1;
  unsigned int *data2 = (unsigned int *)d2;
  if (*data1 < *data2) {
    return -1;
  }
  if (*data1 > *data2) {
    return 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<compare>:
  endbr64
  mov    (%rdi),%ecx
  mov    (%rsi),%edx
  mov    $0xffffffff,%eax
  cmp    %edx,%ecx
  jb     L1
  seta   %al
  movzbl %al,%eax
L1:
  ret
```

assistant: ```c
int compare(const void *d1, const void *d2) {
  unsigned int *data1 = (unsigned int *)d1;
  unsigned int *data2 = (unsigned int *)d2;
  if (*data1 < *data2) {
    return -1;
  }
  if (*data1 > *data2) {
    return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<compare>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int compare(const void *d1, const void *d2) {
  unsigned int *data1 = (unsigned int *)d1;
  unsigned int *data2 = (unsigned int *)d2;
  if (*data1 < *data2) {
    return -1;
  }
  if (*data1 > *data2) {
    return 1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compare>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int compare(const void *d1, const void *d2) {
  unsigned int *data1 = (unsigned int *)d1;
  unsigned int *data2 = (unsigned int *)d2;
  if (*data1 < *data2) {
    return -1;
  }
  if (*data1 > *data2) {
    return 1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_equal_display_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L7
  mov    $0x1,%eax
  jmp    L19
L7:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <print_error@plt>
  mov    $0x0,%eax
L19:
  leave
  ret
```

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

tool: D9:                 "%s" != "%s"


assistant: ```c
int string_equal_display_error(const char *const left, const char *const right) {
  if (strcmp(left, right) == 0) {
    return 1;
  }
  print_error("                "
              "\"%s\" != \"%s\"\n",
              left, right);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<string_equal_display_error>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L10
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D0(%rip),%rdi
  call   <print_error@plt>
  mov    $0x0,%eax
  jmp    L17
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string

tool: D0:                 "%s" != "%s"


assistant: ```c
int string_equal_display_error(const char *const left, const char *const right) {
  if (strcmp(left, right) == 0) {
    return 1;
  }
  print_error("                "
              "\"%s\" != \"%s\"\n",
              left, right);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_equal_display_error>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L9
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <print_error@plt>
  xor    %eax,%eax
L9:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D2
data_type: string

tool: D2:                 "%s" != "%s"


assistant: ```c
int string_equal_display_error(const char *const left, const char *const right) {
  if (strcmp(left, right) == 0) {
    return 1;
  }
  print_error("                "
              "\"%s\" != \"%s\"\n",
              left, right);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_equal_display_error>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L0
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <print_error@plt>
  xor    %eax,%eax
L0:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_infered: string

tool: D2:                 "%s" != "%s"


assistant: ```c
int string_equal_display_error(const char *const left, const char *const right) {
  if (strcmp(left, right) == 0) {
    return 1;
  }
  print_error("                "
              "\"%s\" != \"%s\"\n",
              left, right);
  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
<gt_ggc_mx_dw_ranges_struct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ggc_test_and_set_mark@plt>
  nop
  leave
  ret
```

assistant: ```c
void gt_ggc_mx_dw_ranges_struct(void *x_p) {
  struct dw_ranges_struct *const x = (struct dw_ranges_struct *)x_p;
  if (ggc_test_and_set_mark(x)) {
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<gt_ggc_mx_dw_ranges_struct>:
  endbr64
  sub    $0x8,%rsp
  call   <ggc_test_and_set_mark@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void gt_ggc_mx_dw_ranges_struct(void *x_p) {
  struct dw_ranges_struct *const x = (struct dw_ranges_struct *)x_p;
  if (ggc_test_and_set_mark(x)) {
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<gt_ggc_mx_dw_ranges_struct>:
  endbr64
  jmp    1050 <ggc_test_and_set_mark@plt>
```

assistant: ```c
void gt_ggc_mx_dw_ranges_struct(void *x_p) {
  struct dw_ranges_struct *const x = (struct dw_ranges_struct *)x_p;
  if (ggc_test_and_set_mark(x)) {
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gt_ggc_mx_dw_ranges_struct>:
  endbr64
  jmp    1050 <ggc_test_and_set_mark@plt>
```

assistant: ```c
void gt_ggc_mx_dw_ranges_struct(void *x_p) {
  struct dw_ranges_struct *const x = (struct dw_ranges_struct *)x_p;
  if (ggc_test_and_set_mark(x)) {
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<m_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L12:
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void m_copy(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    i++;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<m_copy>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L5
  mov    $0x1,%eax
L16:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L16
L5:
  ret
```

assistant: ```c
void m_copy(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    i++;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<m_copy>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L15
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L17
L15:
  ret
```

assistant: ```c
void m_copy(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    i++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<m_copy>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L14
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L8
L14:
  ret
```

assistant: ```c
void m_copy(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    i++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x1,%edi
  call   <f@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <g@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x;
  x = f(1);
  g(x);
  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
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <f@plt>
  mov    %eax,%edi
  call   <g@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x;
  x = f(1);
  g(x);
  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    $0x1,%edi
  call   <f@plt>
  mov    %eax,%edi
  call   <g@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x;
  x = f(1);
  g(x);
  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
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <f@plt>
  mov    %eax,%edi
  call   <g@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x;
  x = f(1);
  g(x);
  return 0;
}
```