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

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D81(%rip),%rax
  mov    %rax,(%rsp)
  mov    D84(%rip),%rax
  mov    %rax,0x8(%rsp)
  mov    D55(%rip),%rax
  mov    %rax,0x10(%rsp)
  mov    D3(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    D20(%rip),%rax
  mov    %rax,0x20(%rsp)
  mov    D89(%rip),%rax
  mov    %rax,0x28(%rsp)
  mov    D77(%rip),%rax
  mov    %rax,0x30(%rsp)
  mov    D28(%rip),%rax
  mov    %rax,0x38(%rsp)
  mov    %rsp,%rdx
  mov    $0x2,%ecx
  mov    $0x1,%esi
  mov    $0x4,%edi
  call   <stdlib_strided_dstdev@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D81
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D84
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D55
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D3
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D20
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D89
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D77
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D28
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D9
data_type: string

tool: D81: 1

tool: D84: 2

tool: D55: 3

tool: D3: 4

tool: D20: 5

tool: D89: 6

tool: D77: 7

tool: D28: 8

tool: D9: sample standard deviation: %lf


assistant: ```c
int main() {
  double x[] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0};
  int64_t N = 4;
  int64_t stride = 2;
  double v = stdlib_strided_dstdev(N, 1, x, stride);
  printf("sample standard deviation: %lf\n", v);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x4,%edi
  mov    $0x2,%ecx
  mov    $0x1,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D52(%rip),%rax
  mov    %rsp,%rdx
  mov    %rax,(%rsp)
  mov    D33(%rip),%rax
  mov    %rax,0x8(%rsp)
  mov    D28(%rip),%rax
  mov    %rax,0x10(%rsp)
  mov    D27(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    D73(%rip),%rax
  mov    %rax,0x20(%rsp)
  mov    D63(%rip),%rax
  mov    %rax,0x28(%rsp)
  mov    D21(%rip),%rax
  mov    %rax,0x30(%rsp)
  mov    D83(%rip),%rax
  mov    %rax,0x38(%rsp)
  call   <stdlib_strided_dstdev@plt>
  lea    D68(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D52
type_infered: f64
name: parse_data
arguments: label: D33
type_infered: f64
name: parse_data
arguments: label: D28
type_infered: f64
name: parse_data
arguments: label: D27
type_infered: f64
name: parse_data
arguments: label: D73
type_infered: f64
name: parse_data
arguments: label: D63
type_infered: f64
name: parse_data
arguments: label: D21
type_infered: f64
name: parse_data
arguments: label: D83
type_infered: f64
name: parse_data
arguments: label: D68
type_infered: string

tool: D52: 1

tool: D33: 2

tool: D28: 3

tool: D27: 4

tool: D73: 5

tool: D63: 6

tool: D21: 7

tool: D83: 8

tool: D68: sample standard deviation: %lf


assistant: ```c
int main() {
  double x[] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0};
  int64_t N = 4;
  int64_t stride = 2;
  double v = stdlib_strided_dstdev(N, 1, x, stride);
  printf("sample standard deviation: %lf\n", v);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x4,%edi
  mov    $0x2,%ecx
  movapd D38(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    $0x1,%esi
  movaps %xmm0,(%rsp)
  movapd D20(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movapd D21(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movapd D13(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  call   <stdlib_strided_dstdev@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D38
type_infered: f64[2]
name: unpack_data
arguments: tag: D20
type_infered: f64[2]
name: unpack_data
arguments: tag: D21
type_infered: f64[2]
name: unpack_data
arguments: tag: D13
type_infered: f64[2]
name: unpack_data
arguments: tag: D47
type_infered: string

tool: {"D38": [1.0,2.0]}

tool: {"D20": [3.0,4.0]}

tool: {"D21": [5.0,6.0]}

tool: {"D13": [7.0,8.0]}

tool: D47: sample standard deviation: %lf


assistant: ```c
int main() {
  double x[] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0};
  int64_t N = 4;
  int64_t stride = 2;
  double v = stdlib_strided_dstdev(N, 1, x, stride);
  printf("sample standard deviation: %lf\n", v);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<parse_reply>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  call   <strlen@plt>
  mov    %eax,0xc(%rsp)
  mov    %r15,%rsi
  lea    D78(%rip),%rdi
  call   <printf@plt>
  mov    %r15,%rdi
  mov    $0x0,%ebp
  lea    D42(%rip),%r12
  lea    D51(%rip),%r14
  mov    $0x0,%r13d
L22:
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L17
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rbp,%rax,1),%ebp
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %r13,%rdi
  test   %rax,%rax
  jne    L22
  lea    D42(%rip),%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rbp,%rax,1),%ebp
  lea    D44(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  cmp    $0x1,%rax
  sbb    %ebx,%ebx
  and    $0x5,%ebx
  sub    $0x1,%ebx
  lea    D15(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  mov    $0x1,%eax
  cmove  %eax,%ebx
  lea    D31(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  mov    $0x3,%eax
  cmove  %eax,%ebx
  lea    D3(%rip),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  mov    $0x2,%eax
  cmove  %eax,%ebx
  mov    0xc(%rsp),%edx
  sub    %ebp,%edx
  add    $0x1,%edx
  movslq %ebp,%rsi
  add    %r15,%rsi
  mov    %r15,%rdi
  call   <strncpy@plt>
  jmp    L15
L17:
  movb   $0x0,(%r15)
L15:
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: label: D78
datatype: string
name: read_data
arguments: label: D42
datatype: string
name: read_data
arguments: label: D51
datatype: string
name: read_data
arguments: label: D44
datatype: string
name: read_data
arguments: label: D15
datatype: string
name: read_data
arguments: label: D31
datatype: string
name: read_data
arguments: label: D3
datatype: string

tool: D78: Eingabe: %s


tool: D42:  


tool: D51: key

tool: D44: Escape

tool: D15: Down

tool: D31: Right

tool: D3: Enter

assistant: ```c
int parse_reply(char *input) {
  char *next = input;
  char *retstr;
  int retint = -1;
  int inputlen = strlen(input);
  int len = 0;
  printf("Eingabe: %s\n", input);
  while (1) {
    retstr = strtok(next, " \n");
    next = NULL;
    if (retstr == NULL) {
      input[0] = 0;
      break;
    }
    len += strlen(retstr) + 1;
    if (strcmp(retstr, "key") == 0) {
      retstr = strtok(next, " \n");
      len += strlen(retstr) + 1;
      if (strcmp(retstr, "Escape") == 0)
        retint = 4;
      if (strcmp(retstr, "Down") == 0)
        retint = 1;
      if (strcmp(retstr, "Right") == 0)
        retint = 3;
      if (strcmp(retstr, "Enter") == 0)
        retint = 2;
      input = strncpy(input, &input[len], inputlen - len + 1);
      return retint;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_reply>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  lea    D69(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    D22(%rip),%rdi
  mov    %r14,%rsi
  mov    %eax,%r15d
  call   <printf@plt>
  mov    %r14,%rdi
  jmp    L12
  nopl   0x0(%rax)
L11:
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %r13,%rsi
  lea    0x1(%rbx,%rax,1),%ebx
  call   <strcmp@plt>
  xor    %edi,%edi
  test   %rax,%rax
  je     L18
L12:
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L11
  movb   $0x0,(%r14)
  xor    %r12d,%r12d
L33:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L18:
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  lea    D36(%rip),%rsi
  lea    0x1(%rbx,%rax,1),%ebx
  call   <strcmp@plt>
  mov    %rbp,%rdi
  lea    D16(%rip),%rsi
  cmp    $0x1,%rax
  sbb    %r12d,%r12d
  call   <strcmp@plt>
  and    $0x5,%r12d
  mov    %rbp,%rdi
  lea    D8(%rip),%rsi
  sub    $0x1,%r12d
  test   %rax,%rax
  mov    $0x1,%eax
  cmove  %eax,%r12d
  call   <strcmp@plt>
  mov    %rbp,%rdi
  lea    D53(%rip),%rsi
  test   %rax,%rax
  mov    $0x3,%eax
  cmove  %eax,%r12d
  call   <strcmp@plt>
  movslq %ebx,%rsi
  mov    %r14,%rdi
  test   %rax,%rax
  mov    $0x2,%eax
  cmove  %eax,%r12d
  sub    %ebx,%r15d
  add    %r14,%rsi
  lea    0x1(%r15),%edx
  call   <strncpy@plt>
  jmp    L33
```

tool_calls: name: unpack_data
arguments: data_label: D6
datatype: string
name: unpack_data
arguments: data_label: D69
datatype: string
name: unpack_data
arguments: data_label: D22
datatype: string
name: unpack_data
arguments: data_label: D36
datatype: string
name: unpack_data
arguments: data_label: D16
datatype: string
name: unpack_data
arguments: data_label: D8
datatype: string
name: unpack_data
arguments: data_label: D53
datatype: string

tool: D6: key

tool: D69:  


tool: D22: Eingabe: %s


tool: D36: Escape

tool: D16: Down

tool: D8: Right

tool: D53: Enter

assistant: ```c
int parse_reply(char *input) {
  char *next = input;
  char *retstr;
  int retint = -1;
  int inputlen = strlen(input);
  int len = 0;
  printf("Eingabe: %s\n", input);
  while (1) {
    retstr = strtok(next, " \n");
    next = NULL;
    if (retstr == NULL) {
      input[0] = 0;
      break;
    }
    len += strlen(retstr) + 1;
    if (strcmp(retstr, "key") == 0) {
      retstr = strtok(next, " \n");
      len += strlen(retstr) + 1;
      if (strcmp(retstr, "Escape") == 0)
        retint = 4;
      if (strcmp(retstr, "Down") == 0)
        retint = 1;
      if (strcmp(retstr, "Right") == 0)
        retint = 3;
      if (strcmp(retstr, "Enter") == 0)
        retint = 2;
      input = strncpy(input, &input[len], inputlen - len + 1);
      return retint;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<parse_reply>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  lea    D64(%rip),%r13
  push   %r12
  lea    D66(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    D1(%rip),%rdi
  mov    %r14,%rsi
  mov    %eax,%r15d
  call   <printf@plt>
  mov    %r14,%rdi
  jmp    L10
  nopl   0x0(%rax)
L13:
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %r13,%rsi
  lea    0x1(%rbx,%rax,1),%ebx
  call   <strcmp@plt>
  xor    %edi,%edi
  test   %rax,%rax
  je     L33
L10:
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L13
  movb   $0x0,(%r14)
  xor    %r12d,%r12d
L36:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L33:
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  lea    D52(%rip),%rsi
  lea    0x1(%rbx,%rax,1),%ebx
  call   <strcmp@plt>
  mov    %rbp,%rdi
  lea    D14(%rip),%rsi
  cmp    $0x1,%rax
  sbb    %r12d,%r12d
  call   <strcmp@plt>
  and    $0x5,%r12d
  mov    %rbp,%rdi
  lea    D8(%rip),%rsi
  sub    $0x1,%r12d
  test   %rax,%rax
  mov    $0x1,%eax
  cmove  %eax,%r12d
  call   <strcmp@plt>
  mov    %rbp,%rdi
  lea    D59(%rip),%rsi
  test   %rax,%rax
  mov    $0x3,%eax
  cmove  %eax,%r12d
  call   <strcmp@plt>
  movslq %ebx,%rsi
  mov    %r14,%rdi
  test   %rax,%rax
  mov    $0x2,%eax
  cmove  %eax,%r12d
  sub    %ebx,%r15d
  add    %r14,%rsi
  lea    0x1(%r15),%edx
  call   <strncpy@plt>
  jmp    L36
```

tool_calls: name: disassemble_data
arguments: data_label: D64
type_infered: string
name: disassemble_data
arguments: data_label: D66
type_infered: string
name: disassemble_data
arguments: data_label: D1
type_infered: string
name: disassemble_data
arguments: data_label: D52
type_infered: string
name: disassemble_data
arguments: data_label: D14
type_infered: string
name: disassemble_data
arguments: data_label: D8
type_infered: string
name: disassemble_data
arguments: data_label: D59
type_infered: string

tool: D64: key

tool: D66:  


tool: D1: Eingabe: %s


tool: D52: Escape

tool: D14: Down

tool: D8: Right

tool: D59: Enter

assistant: ```c
int parse_reply(char *input) {
  char *next = input;
  char *retstr;
  int retint = -1;
  int inputlen = strlen(input);
  int len = 0;
  printf("Eingabe: %s\n", input);
  while (1) {
    retstr = strtok(next, " \n");
    next = NULL;
    if (retstr == NULL) {
      input[0] = 0;
      break;
    }
    len += strlen(retstr) + 1;
    if (strcmp(retstr, "key") == 0) {
      retstr = strtok(next, " \n");
      len += strlen(retstr) + 1;
      if (strcmp(retstr, "Escape") == 0)
        retint = 4;
      if (strcmp(retstr, "Down") == 0)
        retint = 1;
      if (strcmp(retstr, "Right") == 0)
        retint = 3;
      if (strcmp(retstr, "Enter") == 0)
        retint = 2;
      input = strncpy(input, &input[len], inputlen - len + 1);
      return retint;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<mod_add>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  test   %rax,%rax
  jg     L1
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jl     L5
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  sub    %rdx,%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,%rax
  jmp    L4
L5:
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  sub    %rdx,%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,%rax
  jmp    L4
L1:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,%rax
L4:
  pop    %rbp
  ret
```

assistant: ```c
long mod_add(long a, long b, long m) {
  a %= m;
  b %= m;
  if (a + b <= 0l) {
    if (a >= b)
      return (b - (m - a)) % m;
    else
      return (a - (m - b)) % m;
  } else
    return (a + b) % m;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mod_add>:
  endbr64
  mov    %rdi,%rax
  mov    %rdx,%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %rsi,%rax
  cqto
  idiv   %rcx
  lea    (%r8,%rdx,1),%rax
  test   %rax,%rax
  jg     L19
  cmp    %rdx,%r8
  jl     L7
  mov    %rcx,%rsi
  sub    %r8,%rsi
  mov    %rdx,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  jmp    L14
L7:
  mov    %rcx,%rsi
  sub    %rdx,%rsi
  mov    %r8,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  jmp    L14
L19:
  cqto
  idiv   %rcx
L14:
  mov    %rdx,%rax
  ret
```

assistant: ```c
long mod_add(long a, long b, long m) {
  a %= m;
  b %= m;
  if (a + b <= 0l) {
    if (a >= b)
      return (b - (m - a)) % m;
    else
      return (a - (m - b)) % m;
  } else
    return (a + b) % m;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mod_add>:
  endbr64
  mov    %rdi,%rax
  mov    %rdx,%rcx
  cqto
  idiv   %rcx
  mov    %rsi,%rax
  mov    %rdx,%r8
  cqto
  idiv   %rcx
  lea    (%r8,%rdx,1),%rax
  test   %rax,%rax
  jg     L11
  mov    %rcx,%rsi
  cmp    %rdx,%r8
  jge    L12
  sub    %rdx,%rsi
  mov    %r8,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
L12:
  sub    %r8,%rsi
  mov    %rdx,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
  nopl   (%rax)
L11:
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
long mod_add(long a, long b, long m) {
  a %= m;
  b %= m;
  if (a + b <= 0l) {
    if (a >= b)
      return (b - (m - a)) % m;
    else
      return (a - (m - b)) % m;
  } else
    return (a + b) % m;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<mod_add>:
  endbr64
  mov    %rdi,%rax
  mov    %rdx,%rcx
  cqto
  idiv   %rcx
  mov    %rsi,%rax
  mov    %rdx,%r8
  cqto
  idiv   %rcx
  lea    (%r8,%rdx,1),%rax
  test   %rax,%rax
  jg     L18
  mov    %rcx,%rsi
  cmp    %rdx,%r8
  jge    L4
  sub    %rdx,%rsi
  mov    %r8,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
L4:
  sub    %r8,%rsi
  mov    %rdx,%rax
  sub    %rsi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
  nopl   (%rax)
L18:
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
long mod_add(long a, long b, long m) {
  a %= m;
  b %= m;
  if (a + b <= 0l) {
    if (a >= b)
      return (b - (m - a)) % m;
    else
      return (a - (m - b)) % m;
  } else
    return (a + b) % m;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dctrl_get_current_dir_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L23
  mov    $0x0,%eax
  jmp    L31
L23:
  mov    -0x18(%rbp),%rax
  mov    $0x1000,%esi
  mov    %rax,%rdi
  call   <getcwd@plt>
  test   %rax,%rax
  jne    L36
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L31
L36:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L31:
  leave
  ret
```

assistant: ```c
char *dctrl_get_current_dir_name() {
  char *buf;
  char *path;
  size_t len;
  buf = malloc(4096);
  if (NULL == buf)
    return NULL;
  if (NULL == getcwd(buf, 4096)) {
    free(buf);
    return NULL;
  }
  len = strlen(buf);
  path = realloc(buf, len + 1);
  if (NULL == path)
    ;
  path = buf;
  return path;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dctrl_get_current_dir_name>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L3
  mov    $0x1000,%esi
  mov    %rax,%rdi
  call   <getcwd@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L1
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rsi
  mov    %rbx,%rdi
  call   <realloc@plt>
L3:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rbx
  jmp    L3
```

assistant: ```c
char *dctrl_get_current_dir_name() {
  char *buf;
  char *path;
  size_t len;
  buf = malloc(4096);
  if (NULL == buf)
    return NULL;
  if (NULL == getcwd(buf, 4096)) {
    free(buf);
    return NULL;
  }
  len = strlen(buf);
  path = realloc(buf, len + 1);
  if (NULL == path)
    ;
  path = buf;
  return path;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dctrl_get_current_dir_name>:
  endbr64
  push   %r12
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L13
  mov    %rax,%rdi
  mov    $0x1000,%esi
  call   <getcwd@plt>
  mov    %r12,%rdi
  test   %rax,%rax
  je     L15
  call   <strlen@plt>
  mov    %r12,%rdi
  lea    0x1(%rax),%rsi
  call   <realloc@plt>
L13:
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L15:
  call   <free@plt>
  xor    %r12d,%r12d
  jmp    L13
```

assistant: ```c
char *dctrl_get_current_dir_name() {
  char *buf;
  char *path;
  size_t len;
  buf = malloc(4096);
  if (NULL == buf)
    return NULL;
  if (NULL == getcwd(buf, 4096)) {
    free(buf);
    return NULL;
  }
  len = strlen(buf);
  path = realloc(buf, len + 1);
  if (NULL == path)
    ;
  path = buf;
  return path;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dctrl_get_current_dir_name>:
  endbr64
  push   %r12
  mov    $0x1000,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L11
  mov    %rax,%rdi
  mov    $0x1000,%esi
  call   <getcwd@plt>
  mov    %r12,%rdi
  test   %rax,%rax
  je     L2
  call   <strlen@plt>
  mov    %r12,%rdi
  lea    0x1(%rax),%rsi
  call   <realloc@plt>
L11:
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  call   <free@plt>
  xor    %r12d,%r12d
  jmp    L11
```

assistant: ```c
char *dctrl_get_current_dir_name() {
  char *buf;
  char *path;
  size_t len;
  buf = malloc(4096);
  if (NULL == buf)
    return NULL;
  if (NULL == getcwd(buf, 4096)) {
    free(buf);
    return NULL;
  }
  len = strlen(buf);
  path = realloc(buf, len + 1);
  if (NULL == path)
    ;
  path = buf;
  return path;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L8
L27:
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    $0x0,%eax
  jmp    L10
L8:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%ecx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  cmp    %al,%cl
  je     L27
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    -0x10(%rbp),%rdx
  sub    $0x1,%rdx
  movzbl (%rdx),%edx
  movzbl %dl,%edx
  sub    %edx,%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int strcmp(const char *s1, const char *s2) {
  while (*s1++ == *s2++)
    if (s1[-1] == '\0')
      return (0);
  return ((int)((unsigned char)s1[-1] - (unsigned char)s2[-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
<strcmp>:
  endbr64
  mov    $0x0,%eax
L11:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L18
  add    $0x1,%rax
  test   %dl,%dl
  jne    L11
  mov    $0x0,%eax
  ret
L18:
  movzbl %dl,%eax
  movzbl %cl,%ecx
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(const char *s1, const char *s2) {
  while (*s1++ == *s2++)
    if (s1[-1] == '\0')
      return (0);
  return ((int)((unsigned char)s1[-1] - (unsigned char)s2[-1]));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcmp>:
  endbr64
  xor    %eax,%eax
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L14:
  add    $0x1,%rax
  test   %dl,%dl
  je     L10
L21:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  je     L14
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcmp(const char *s1, const char *s2) {
  while (*s1++ == *s2++)
    if (s1[-1] == '\0')
      return (0);
  return ((int)((unsigned char)s1[-1] - (unsigned char)s2[-1]));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strcmp>:
  endbr64
  xor    %eax,%eax
  jmp    L12
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x1,%rax
  test   %dl,%dl
  je     L3
L12:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  je     L0
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcmp(const char *s1, const char *s2) {
  while (*s1++ == *s2++)
    if (s1[-1] == '\0')
      return (0);
  return ((int)((unsigned char)s1[-1] - (unsigned char)s2[-1]));
}
```