ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x65696c754a,%rax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  mov    %edx,%ecx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L3
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  movb   $0x0,0x3(%rsp)
  call   <printf@plt>
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L8:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  mov    %edx,%ecx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L8
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D16
data_type: string
name: retrieve_data
arguments: tag: D10
data_type: string

tool: D16: name is %s.


tool: D10: [%i] %c - %d


assistant: ```c
int main() {
  char name[8] = "Julie";
  printf("name is %s.\n", name);
  int i;
  for (i = 0; i < 8; i++) {
    printf("[%i] %c - %d\n", i, name[i], name[i]);
  }
  name[3] = '\0';
  printf("name is %s.\n", name);
  for (i = 0; i < 8; i++) {
    printf("[%i] %c - %d\n", i, name[i], name[i]);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D7(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x65696c754a,%rax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L29:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  mov    %edx,%ecx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L29
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  movb   $0x0,0x3(%rsp)
  call   <printf@plt>
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L0:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  mov    %edx,%ecx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: string
name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D2: name is %s.


tool: D7: [%i] %c - %d


assistant: ```c
int main() {
  char name[8] = "Julie";
  printf("name is %s.\n", name);
  int i;
  for (i = 0; i < 8; i++) {
    printf("[%i] %c - %d\n", i, name[i], name[i]);
  }
  name[3] = '\0';
  printf("name is %s.\n", name);
  for (i = 0; i < 8; i++) {
    printf("[%i] %c - %d\n", i, name[i], name[i]);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<reb_tools_mod2pi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <fmod@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm2
  addsd  -0x8(%rbp),%xmm2
  movq   %xmm2,%rax
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <fmod@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double reb_tools_mod2pi(double f) {
  const double pi2 = 2. * M_PI;
  return fmod(pi2 + fmod(f, pi2), pi2);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reb_tools_mod2pi>:
  endbr64
  sub    $0x18,%rsp
  mov    D4(%rip),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sdl (%rax),%xmm1
  addsd  %xmm1,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
  call   <fmod@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
double reb_tools_mod2pi(double f) {
  const double pi2 = 2. * M_PI;
  return fmod(pi2 + fmod(f, pi2), pi2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reb_tools_mod2pi>:
  endbr64
  sub    $0x18,%rsp
  mov    D4(%rip),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sdl (%rax),%xmm1
  addsd  %xmm1,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  movsd  0x8(%rsp),%xmm1
  add    $0x18,%rsp
  addsd  %xmm1,%xmm0
  jmp    1050 <fmod@plt>
```

assistant: ```c
double reb_tools_mod2pi(double f) {
  const double pi2 = 2. * M_PI;
  return fmod(pi2 + fmod(f, pi2), pi2);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<reb_tools_mod2pi>:
  endbr64
  sub    $0x18,%rsp
  mov    D5(%rip),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sdl (%rax),%xmm1
  addsd  %xmm1,%xmm1
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  movsd  0x8(%rsp),%xmm1
  add    $0x18,%rsp
  addsd  %xmm1,%xmm0
  jmp    1050 <fmod@plt>
```

assistant: ```c
double reb_tools_mod2pi(double f) {
  const double pi2 = 2. * M_PI;
  return fmod(pi2 + fmod(f, pi2), pi2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void error(const char *msg) {
  perror(msg);
  exit(EXIT_FAILURE);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void error(const char *msg) {
  perror(msg);
  exit(EXIT_FAILURE);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void error(const char *msg) {
  perror(msg);
  exit(EXIT_FAILURE);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<error>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void error(const char *msg) {
  perror(msg);
  exit(EXIT_FAILURE);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<palindrome>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L31
  mov    %rbx,%rbp
L94:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %al,0x0(%rbp)
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L94
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L31
  lea    0x1(%rbx),%r12
  mov    $0x0,%ebp
L37:
  movsbl %dil,%edi
  call   <islower@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebp
  add    $0x1,%r12
  movzbl -0x1(%r12),%edi
  test   %dil,%dil
  jne    L37
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L41
L148:
  mov    %ebp,%r13d
  jmp    L8
L79:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L41
L8:
  movsbl %dil,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L79
  movzbl (%rbx),%edx
  movslq %r13d,%rax
  mov    %dl,-0x1(%r12,%rax,1)
  sub    $0x1,%r13d
  jmp    L79
L41:
  mov    %ebp,%ecx
  shr    $0x1f,%ecx
  add    %ebp,%ecx
  sar    %ecx
  cmp    $0x1,%ebp
  jle    L88
  movslq %ebp,%rbp
  lea    -0x1(%r12,%rbp,1),%rdx
  mov    $0x0,%eax
L0:
  movzbl (%rdx),%esi
  cmp    %sil,(%r12,%rax,1)
  jne    L89
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%ecx
  jg     L0
L88:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x1,%eax
L140:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L89:
  mov    $0x0,%eax
  jmp    L140
L31:
  mov    $0x0,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%edi
  mov    $0x0,%ebp
  test   %dil,%dil
  jne    L148
  jmp    L88
```

assistant: ```c
int palindrome(char *String) {
  int temp = 0;
  int label = 0;
  int chars = 0;
  for (temp = 0; String[temp]; temp++) {
    String[temp] = tolower(String[temp]);
  }
  for (temp = 0; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      chars++;
    }
  }
  char *str = (char *)malloc(chars * sizeof(char));
  for (temp = 0, label = chars; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      str[label - 1] = String[temp];
      label--;
    }
  }
  for (temp = 0; temp < (chars / 2); temp++) {
    if (str[temp] != str[chars - 1 - temp]) {
      return 0;
    }
  }
  free(str);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<palindrome>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L106
  mov    %r12,%rbx
  nopl   0x0(%rax)
L96:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L96
  movsbl (%r12),%edi
  test   %dil,%dil
  je     L106
  lea    0x1(%r12),%rbp
  xor    %ebx,%ebx
  mov    %rbp,%r13
  nopl   0x0(%rax)
L4:
  call   <islower@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebx
  movsbl 0x0(%r13),%edi
  add    $0x1,%r13
  test   %dil,%dil
  jne    L4
  mov    %ebx,%edi
  mov    %ebx,%r15d
  call   <malloc@plt>
  movsbl (%r12),%edi
  sar    %r15d
  mov    %rax,%r14
  test   %dil,%dil
  je     L122
L83:
  mov    %ebx,%r13d
  jmp    L129
  nopl   0x0(%rax)
L95:
  add    $0x1,%rbp
L129:
  call   <islower@plt>
  test   %rax,%rax
  je     L71
  movzbl (%r12),%edx
  movslq %r13d,%rax
  sub    $0x1,%r13d
  mov    %dl,-0x1(%r14,%rax,1)
L71:
  movsbl 0x0(%rbp),%edi
  mov    %rbp,%r12
  test   %dil,%dil
  jne    L95
L122:
  test   %r15d,%r15d
  je     L94
  sub    $0x1,%ebx
  lea    -0x1(%r15),%edx
  mov    %r14,%rax
  movslq %ebx,%rbx
  lea    0x1(%r14,%rdx,1),%rdx
  add    %r14,%rbx
  jmp    L86
  nop
L34:
  add    $0x1,%rax
  sub    $0x1,%rbx
  cmp    %rax,%rdx
  je     L94
L86:
  movzbl (%rbx),%ecx
  cmp    %cl,(%rax)
  je     L34
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L106:
  xor    %edi,%edi
  xor    %r15d,%r15d
  lea    0x1(%r12),%rbp
  xor    %ebx,%ebx
  call   <malloc@plt>
  movsbl (%r12),%edi
  mov    %rax,%r14
  test   %dil,%dil
  jne    L83
L94:
  mov    %r14,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int palindrome(char *String) {
  int temp = 0;
  int label = 0;
  int chars = 0;
  for (temp = 0; String[temp]; temp++) {
    String[temp] = tolower(String[temp]);
  }
  for (temp = 0; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      chars++;
    }
  }
  char *str = (char *)malloc(chars * sizeof(char));
  for (temp = 0, label = chars; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      str[label - 1] = String[temp];
      label--;
    }
  }
  for (temp = 0; temp < (chars / 2); temp++) {
    if (str[temp] != str[chars - 1 - temp]) {
      return 0;
    }
  }
  free(str);
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<palindrome>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L41
  mov    %r12,%rbx
  nopl   0x0(%rax)
L14:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L14
  movsbl (%r12),%edi
  test   %dil,%dil
  je     L41
  lea    0x1(%r12),%rbp
  xor    %ebx,%ebx
  mov    %rbp,%r13
  nopl   0x0(%rax)
L81:
  call   <islower@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebx
  movsbl 0x0(%r13),%edi
  add    $0x1,%r13
  test   %dil,%dil
  jne    L81
  mov    %ebx,%edi
  mov    %ebx,%r15d
  call   <malloc@plt>
  movsbl (%r12),%edi
  sar    %r15d
  mov    %rax,%r14
  test   %dil,%dil
  je     L111
L43:
  mov    %ebx,%r13d
  jmp    L34
  nopl   0x0(%rax)
L54:
  movzbl (%r12),%edx
  movslq %r13d,%rax
  mov    %rbp,%r12
  sub    $0x1,%r13d
  mov    %dl,-0x1(%r14,%rax,1)
  movsbl 0x0(%rbp),%edi
  test   %dil,%dil
  je     L111
L129:
  add    $0x1,%rbp
L34:
  call   <islower@plt>
  test   %rax,%rax
  jne    L54
  movsbl 0x0(%rbp),%edi
  mov    %rbp,%r12
  test   %dil,%dil
  jne    L129
L111:
  test   %r15d,%r15d
  je     L114
  sub    $0x1,%ebx
  lea    -0x1(%r15),%edx
  mov    %r14,%rax
  movslq %ebx,%rbx
  lea    0x1(%r14,%rdx,1),%rdx
  add    %r14,%rbx
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x1,%rax
  sub    $0x1,%rbx
  cmp    %rax,%rdx
  je     L114
L50:
  movzbl (%rbx),%ecx
  cmp    %cl,(%rax)
  je     L13
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L41:
  xor    %edi,%edi
  xor    %r15d,%r15d
  lea    0x1(%r12),%rbp
  xor    %ebx,%ebx
  call   <malloc@plt>
  movsbl (%r12),%edi
  mov    %rax,%r14
  test   %dil,%dil
  jne    L43
L114:
  mov    %r14,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int palindrome(char *String) {
  int temp = 0;
  int label = 0;
  int chars = 0;
  for (temp = 0; String[temp]; temp++) {
    String[temp] = tolower(String[temp]);
  }
  for (temp = 0; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      chars++;
    }
  }
  char *str = (char *)malloc(chars * sizeof(char));
  for (temp = 0, label = chars; String[temp] != '\0'; temp++) {
    if (islower(String[temp])) {
      str[label - 1] = String[temp];
      label--;
    }
  }
  for (temp = 0; temp < (chars / 2); temp++) {
    if (str[temp] != str[chars - 1 - temp]) {
      return 0;
    }
  }
  free(str);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<open_output_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %rdx,-0x438(%rbp)
  mov    %rcx,-0x440(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x440(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L18
  mov    -0x440(%rbp),%rax
  mov    %rax,%rdi
  call   <open_explicit_file@plt>
  mov    %rax,-0x418(%rbp)
  mov    -0x418(%rbp),%rax
  jmp    L18
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
FILE *open_output_file(char *in_name, char *prompt, char *file_extension, char *out_name) {
  char temp[256 + 1];
  char path[256 + 1];
  char local_prompt[1024];
  FILE *file_handle;
  if (out_name[0] != '\0') {
    file_handle = open_explicit_file(out_name);
    return file_handle;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<open_output_file>:
  endbr64
  cmpb   $0x0,(%rcx)
  jne    L7
  ret
L7:
  sub    $0x8,%rsp
  mov    %rcx,%rdi
  call   <open_explicit_file@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
FILE *open_output_file(char *in_name, char *prompt, char *file_extension, char *out_name) {
  char temp[256 + 1];
  char path[256 + 1];
  char local_prompt[1024];
  FILE *file_handle;
  if (out_name[0] != '\0') {
    file_handle = open_explicit_file(out_name);
    return file_handle;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<open_output_file>:
  endbr64
  cmpb   $0x0,(%rcx)
  jne    L0
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %rcx,%rdi
  jmp    1050 <open_explicit_file@plt>
```

assistant: ```c
FILE *open_output_file(char *in_name, char *prompt, char *file_extension, char *out_name) {
  char temp[256 + 1];
  char path[256 + 1];
  char local_prompt[1024];
  FILE *file_handle;
  if (out_name[0] != '\0') {
    file_handle = open_explicit_file(out_name);
    return file_handle;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<open_output_file>:
  endbr64
  cmpb   $0x0,(%rcx)
  jne    L7
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rcx,%rdi
  jmp    1050 <open_explicit_file@plt>
```

assistant: ```c
FILE *open_output_file(char *in_name, char *prompt, char *file_extension, char *out_name) {
  char temp[256 + 1];
  char path[256 + 1];
  char local_prompt[1024];
  FILE *file_handle;
  if (out_name[0] != '\0') {
    file_handle = open_explicit_file(out_name);
    return file_handle;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int test6(int x) {
  int y = INT_MIN;
  return x - y;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test6>:
  endbr64
  mov    D5(%rip),%rax
  sub    (%rax),%edi
  mov    %edi,%eax
  ret
```

assistant: ```c
int test6(int x) {
  int y = INT_MIN;
  return x - y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test6>:
  endbr64
  mov    D8(%rip),%rax
  sub    (%rax),%edi
  mov    %edi,%eax
  ret
```

assistant: ```c
int test6(int x) {
  int y = INT_MIN;
  return x - y;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test6>:
  endbr64
  mov    D5(%rip),%rax
  sub    (%rax),%edi
  mov    %edi,%eax
  ret
```

assistant: ```c
int test6(int x) {
  int y = INT_MIN;
  return x - y;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<solve_with_double_doubles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    %rsi,-0x28(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x111,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int solve_with_double_doubles(int max, double eps, int *nit) {
  int fail = _ada_use_c2phc(273, &max, nit, &eps);
  return fail;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<solve_with_double_doubles>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  movsd  %xmm0,(%rsp)
  mov    %rsi,%rdx
  mov    %rsp,%rcx
  lea    0xc(%rsp),%rsi
  mov    $0x111,%edi
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_with_double_doubles(int max, double eps, int *nit) {
  int fail = _ada_use_c2phc(273, &max, nit, &eps);
  return fail;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<solve_with_double_doubles>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rdx
  mov    %edi,0xc(%rsp)
  mov    %rsp,%rcx
  lea    0xc(%rsp),%rsi
  mov    $0x111,%edi
  movsd  %xmm0,(%rsp)
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_with_double_doubles(int max, double eps, int *nit) {
  int fail = _ada_use_c2phc(273, &max, nit, &eps);
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<solve_with_double_doubles>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rdx
  mov    %edi,0xc(%rsp)
  mov    %rsp,%rcx
  lea    0xc(%rsp),%rsi
  mov    $0x111,%edi
  movsd  %xmm0,(%rsp)
  call   <_ada_use_c2phc@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_with_double_doubles(int max, double eps, int *nit) {
  int fail = _ada_use_c2phc(273, &max, nit, &eps);
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<u_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <u_strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L23
  movl   $0x0,-0x10(%rbp)
  jmp    L10
L19:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L10:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L19
L23:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
UChar *u_strdup(UChar *ustr) {
  int i, len = u_strlen(ustr);
  UChar *dup = calloc(len + 1, sizeof(UChar));
  if (dup != NULL)
    for (i = 0; i < len; i++)
      dup[i] = ustr[i];
  return dup;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<u_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <u_strlen@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%edi
  mov    $0x4,%esi
  call   <calloc@plt>
  test   %rax,%rax
  je     L17
  test   %ebx,%ebx
  jle    L17
  mov    %ebx,%ecx
  mov    $0x0,%edx
L26:
  mov    0x0(%rbp,%rdx,4),%esi
  mov    %esi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L26
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
UChar *u_strdup(UChar *ustr) {
  int i, len = u_strlen(ustr);
  UChar *dup = calloc(len + 1, sizeof(UChar));
  if (dup != NULL)
    for (i = 0; i < len; i++)
      dup[i] = ustr[i];
  return dup;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<u_strdup>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <u_strlen@plt>
  mov    $0x4,%esi
  lea    0x1(%rax),%edi
  mov    %eax,%ebx
  call   <calloc@plt>
  test   %rax,%rax
  je     L7
  test   %ebx,%ebx
  jle    L7
  movslq %ebx,%rcx
  xor    %edx,%edx
  nop
L17:
  mov    0x0(%rbp,%rdx,4),%esi
  mov    %esi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L17
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
UChar *u_strdup(UChar *ustr) {
  int i, len = u_strlen(ustr);
  UChar *dup = calloc(len + 1, sizeof(UChar));
  if (dup != NULL)
    for (i = 0; i < len; i++)
      dup[i] = ustr[i];
  return dup;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<u_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <u_strlen@plt>
  mov    $0x4,%esi
  lea    0x1(%rax),%edi
  mov    %eax,%ebp
  call   <calloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L82
  test   %ebp,%ebp
  jle    L82
  lea    0x4(%rbx),%rdx
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L68
  lea    -0x1(%rbp),%eax
  cmp    $0x2,%eax
  jbe    L68
  mov    %ebp,%edx
  xor    %eax,%eax
  shr    $0x2,%edx
  shl    $0x4,%rdx
  nopl   (%rax)
L53:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L53
  mov    %ebp,%edx
  and    $0xfffffffc,%edx
  test   $0x3,%bpl
  je     L82
  mov    %edx,%ecx
  mov    (%rbx,%rcx,4),%esi
  lea    0x0(,%rcx,4),%rax
  mov    %esi,(%r8,%rcx,4)
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%ebp
  jle    L82
  mov    0x4(%rbx,%rax,1),%ecx
  add    $0x2,%edx
  mov    %ecx,0x4(%r8,%rax,1)
  cmp    %edx,%ebp
  jle    L82
  mov    0x8(%rbx,%rax,1),%edx
  mov    %edx,0x8(%r8,%rax,1)
L82:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L68:
  mov    %ebp,%eax
  xor    %edx,%edx
  nopl   0x0(%rax)
L61:
  mov    (%rbx,%rdx,4),%ecx
  mov    %ecx,(%r8,%rdx,4)
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L61
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
UChar *u_strdup(UChar *ustr) {
  int i, len = u_strlen(ustr);
  UChar *dup = calloc(len + 1, sizeof(UChar));
  if (dup != NULL)
    for (i = 0; i < len; i++)
      dup[i] = ustr[i];
  return dup;
}
```