ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readlong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L5:
  mov    -0x88(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <gets@plt>
  lea    -0x78(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  lea    D1(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L16
  jmp    L5
L16:
  nop
  mov    -0x78(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_guessed: string
name: retrieve_data
arguments: label: D1
type_guessed: string

tool: D5: %s

tool: D1: %ld

assistant: ```c
long readlong(prompt)
char *prompt;
{
  int err;
  long res;
  char string[100];
  do {
    printf("%s", prompt);
    gets(string);
    if (sscanf(string, "%ld", &res) == 1)
      break;
  } while (1);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readlong>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%r14
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%r13
  lea    D5(%rip),%r12
L6:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <gets@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L6
  mov    0x8(%rsp),%rax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D18
type: string
name: disassemble_data
arguments: tag: D5
type: string

tool: D18: %s

tool: D5: %ld

assistant: ```c
long readlong(prompt)
char *prompt;
{
  int err;
  long res;
  char string[100];
  do {
    printf("%s", prompt);
    gets(string);
    if (sscanf(string, "%ld", &res) == 1)
      break;
  } while (1);
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readlong>:
  endbr64
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <gets@plt>
  xor    %eax,%eax
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L13
  mov    0x8(%rsp),%rax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D13
type_infered: string

tool: D2: %s

tool: D13: %ld

assistant: ```c
long readlong(prompt)
char *prompt;
{
  int err;
  long res;
  char string[100];
  do {
    printf("%s", prompt);
    gets(string);
    if (sscanf(string, "%ld", &res) == 1)
      break;
  } while (1);
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readlong>:
  endbr64
  push   %r14
  lea    D16(%rip),%r14
  push   %r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <gets@plt>
  xor    %eax,%eax
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L17
  mov    0x8(%rsp),%rax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D16
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string

tool: D16: %s

tool: D15: %ld

assistant: ```c
long readlong(prompt)
char *prompt;
{
  int err;
  long res;
  char string[100];
  do {
    printf("%s", prompt);
    gets(string);
    if (sscanf(string, "%ld", &res) == 1)
      break;
  } while (1);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<timestring2Int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  movb   $0x0,-0x9(%rbp)
  cmpl   $0x9,-0x10(%rbp)
  jg     L9
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0x1(%rax),%eax
  mov    %al,-0xa(%rbp)
  jmp    L54
L9:
  mov    -0x28(%rbp),%rax
  movzbl 0x3(%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0x4(%rax),%eax
  mov    %al,-0xa(%rbp)
L54:
  lea    -0xb(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  imul   $0x3c,%eax,%eax
  mov    %eax,-0x20(%rbp)
  cmpl   $0x9,-0x10(%rbp)
  jg     L47
  mov    -0x28(%rbp),%rax
  movzbl 0x3(%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0x4(%rax),%eax
  mov    %al,-0xa(%rbp)
  jmp    L60
L47:
  mov    -0x28(%rbp),%rax
  movzbl 0x6(%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0x7(%rax),%eax
  mov    %al,-0xa(%rbp)
L60:
  lea    -0xb(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  add    %eax,-0x20(%rbp)
  cmpl   $0x9,-0x10(%rbp)
  jg     L48
  mov    -0x28(%rbp),%rax
  movzbl 0x6(%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0x7(%rax),%eax
  mov    %al,-0xa(%rbp)
  jmp    L29
L48:
  mov    -0x28(%rbp),%rax
  movzbl 0x9(%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl 0xa(%rax),%eax
  mov    %al,-0xa(%rbp)
L29:
  lea    -0xb(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L12
  addl   $0x1,-0x20(%rbp)
L12:
  mov    -0x20(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L68
  call   <__stack_chk_fail@plt>
L68:
  leave
  ret
```

assistant: ```c
int timestring2Int(char *timestring) {
  int result = 0;
  int minutes = 0, seconds = 0, ticks = 0;
  int stringLength = strlen(timestring);
  char subString[3];
  subString[2] = '\0';
  if (stringLength < 10) {
    subString[0] = timestring[0];
    subString[1] = timestring[1];
  } else {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  }
  minutes = atoi(subString);
  result = minutes * 60;
  if (stringLength < 10) {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  } else {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  }
  seconds = atoi(subString);
  result += seconds;
  if (stringLength < 10) {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  } else {
    subString[0] = timestring[9];
    subString[1] = timestring[10];
  }
  ticks = atoi(subString);
  if (ticks > 0) {
    ++result;
  }
  return (result);
}
```
user: What is the c source code of the assembly code below:

```asm
<timestring2Int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  movb   $0x0,0x7(%rsp)
  cmp    $0x9,%eax
  jg     L5
  movzbl (%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x1(%rbx),%eax
  mov    %al,0x6(%rsp)
  lea    0x5(%rsp),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  imul   $0x3c,%eax,%r12d
  movzbl 0x3(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x4(%rbx),%eax
  mov    %al,0x6(%rsp)
  mov    %rbp,%rdi
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x6(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x7(%rbx),%eax
L6:
  mov    %al,0x6(%rsp)
  lea    0x5(%rsp),%rdi
  call   <atoi@plt>
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r12d
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    %r12d,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  movzbl 0x3(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x4(%rbx),%eax
  mov    %al,0x6(%rsp)
  lea    0x5(%rsp),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  imul   $0x3c,%eax,%r12d
  movzbl 0x6(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x7(%rbx),%eax
  mov    %al,0x6(%rsp)
  mov    %rbp,%rdi
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x9(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0xa(%rbx),%eax
  jmp    L6
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int timestring2Int(char *timestring) {
  int result = 0;
  int minutes = 0, seconds = 0, ticks = 0;
  int stringLength = strlen(timestring);
  char subString[3];
  subString[2] = '\0';
  if (stringLength < 10) {
    subString[0] = timestring[0];
    subString[1] = timestring[1];
  } else {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  }
  minutes = atoi(subString);
  result = minutes * 60;
  if (stringLength < 10) {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  } else {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  }
  seconds = atoi(subString);
  result += seconds;
  if (stringLength < 10) {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  } else {
    subString[0] = timestring[9];
    subString[1] = timestring[10];
  }
  ticks = atoi(subString);
  if (ticks > 0) {
    ++result;
  }
  return (result);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timestring2Int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  movb   $0x0,0x7(%rsp)
  cmp    $0x9,%eax
  jg     L28
  movzwl (%rbx),%eax
  lea    0x5(%rsp),%rbp
  mov    %rbp,%rdi
  mov    %ax,0x5(%rsp)
  call   <atoi@plt>
  mov    %rbp,%rdi
  imul   $0x3c,%eax,%r12d
  movzwl 0x3(%rbx),%eax
  mov    %ax,0x5(%rsp)
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x6(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x7(%rbx),%eax
L5:
  mov    %rbp,%rdi
  mov    %al,0x6(%rsp)
  call   <atoi@plt>
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r12d
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  movzwl 0x3(%rbx),%eax
  lea    0x5(%rsp),%rbp
  mov    %rbp,%rdi
  mov    %ax,0x5(%rsp)
  call   <atoi@plt>
  mov    %rbp,%rdi
  imul   $0x3c,%eax,%r12d
  movzwl 0x6(%rbx),%eax
  mov    %ax,0x5(%rsp)
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x9(%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0xa(%rbx),%eax
  jmp    L5
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int timestring2Int(char *timestring) {
  int result = 0;
  int minutes = 0, seconds = 0, ticks = 0;
  int stringLength = strlen(timestring);
  char subString[3];
  subString[2] = '\0';
  if (stringLength < 10) {
    subString[0] = timestring[0];
    subString[1] = timestring[1];
  } else {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  }
  minutes = atoi(subString);
  result = minutes * 60;
  if (stringLength < 10) {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  } else {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  }
  seconds = atoi(subString);
  result += seconds;
  if (stringLength < 10) {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  } else {
    subString[0] = timestring[9];
    subString[1] = timestring[10];
  }
  ticks = atoi(subString);
  if (ticks > 0) {
    ++result;
  }
  return (result);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<timestring2Int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  movb   $0x0,0x6(%rsp)
  cmp    $0x9,%eax
  jg     L27
  movzwl (%rbx),%eax
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  mov    %ax,0x4(%rsp)
  call   <atoi@plt>
  mov    %rbp,%rdi
  imul   $0x3c,%eax,%r12d
  movzwl 0x3(%rbx),%eax
  mov    %ax,0x4(%rsp)
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x6(%rbx),%eax
  mov    %al,0x4(%rsp)
  movzbl 0x7(%rbx),%eax
L3:
  mov    %rbp,%rdi
  mov    %al,0x5(%rsp)
  call   <atoi@plt>
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r12d
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  movzwl 0x3(%rbx),%eax
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  mov    %ax,0x4(%rsp)
  call   <atoi@plt>
  mov    %rbp,%rdi
  imul   $0x3c,%eax,%r12d
  movzwl 0x6(%rbx),%eax
  mov    %ax,0x4(%rsp)
  call   <atoi@plt>
  add    %eax,%r12d
  movzbl 0x9(%rbx),%eax
  mov    %al,0x4(%rsp)
  movzbl 0xa(%rbx),%eax
  jmp    L3
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int timestring2Int(char *timestring) {
  int result = 0;
  int minutes = 0, seconds = 0, ticks = 0;
  int stringLength = strlen(timestring);
  char subString[3];
  subString[2] = '\0';
  if (stringLength < 10) {
    subString[0] = timestring[0];
    subString[1] = timestring[1];
  } else {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  }
  minutes = atoi(subString);
  result = minutes * 60;
  if (stringLength < 10) {
    subString[0] = timestring[3];
    subString[1] = timestring[4];
  } else {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  }
  seconds = atoi(subString);
  result += seconds;
  if (stringLength < 10) {
    subString[0] = timestring[6];
    subString[1] = timestring[7];
  } else {
    subString[0] = timestring[9];
    subString[1] = timestring[10];
  }
  ticks = atoi(subString);
  if (ticks > 0) {
    ++result;
  }
  return (result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tolower>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x40,-0x4(%rbp)
  jle    L10
  cmpl   $0x5a,-0x4(%rbp)
  jg     L10
  mov    -0x4(%rbp),%eax
  add    $0x20,%eax
  jmp    L12
L10:
  mov    -0x4(%rbp),%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int tolower(int c) {
  if ((c >= 'A') && (c <= 'Z'))
    return c + ('a' - 'A');
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<tolower>:
  endbr64
  mov    %edi,%eax
  lea    -0x41(%rdi),%ecx
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolower(int c) {
  if ((c >= 'A') && (c <= 'Z'))
    return c + ('a' - 'A');
  return c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tolower>:
  endbr64
  lea    -0x41(%rdi),%ecx
  mov    %edi,%eax
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolower(int c) {
  if ((c >= 'A') && (c <= 'Z'))
    return c + ('a' - 'A');
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tolower>:
  endbr64
  lea    -0x41(%rdi),%ecx
  mov    %edi,%eax
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolower(int c) {
  if ((c >= 'A') && (c <= 'Z'))
    return c + ('a' - 'A');
  return c;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<speed_bigint_mul131>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x84(%rbp)
  jmp    L11
L28:
  mov    $0x0,%eax
  call   <cpucycles@plt>
  mov    -0x84(%rbp),%edx
  movslq %edx,%rdx
  mov    %rax,-0x80(%rbp,%rdx,8)
  lea    -0x30(%rbp),%rdx
  lea    -0x30(%rbp),%rcx
  lea    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bigint_mul131@plt>
  addl   $0x1,-0x84(%rbp)
L11:
  cmpl   $0x9,-0x84(%rbp)
  jle    L28
  lea    -0x80(%rbp),%rax
  mov    $0xa,%edx
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <print_bench@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
type_guessed: string

tool: D4: bigint_mul131

assistant: ```c
void speed_bigint_mul131() {
  unsigned long long t[10];
  int i;
  unsigned char r[34];
  for (i = 0; i < 10; i++) {
    t[i] = cpucycles();
    bigint_mul131(r, r, r);
  }
  print_bench("bigint_mul131", t, 10);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<speed_bigint_mul131>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  mov    %r12,%rbp
L9:
  mov    $0x0,%eax
  call   <cpucycles@plt>
  mov    %rax,(%rbx)
  mov    %rbp,%rdx
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  call   <bigint_mul131@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L9
  mov    %rsp,%rsi
  mov    $0xa,%edx
  lea    D3(%rip),%rdi
  call   <print_bench@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D3
datatype: string

tool: D3: bigint_mul131

assistant: ```c
void speed_bigint_mul131() {
  unsigned long long t[10];
  int i;
  unsigned char r[34];
  for (i = 0; i < 10; i++) {
    t[i] = cpucycles();
    bigint_mul131(r, r, r);
  }
  print_bench("bigint_mul131", t, 10);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<speed_bigint_mul131>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x50(%rsp),%rbx
  mov    %r12,%rbp
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  add    $0x8,%rbp
  call   <cpucycles@plt>
  mov    %rbx,%rdx
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  mov    %rax,-0x8(%rbp)
  call   <bigint_mul131@plt>
  cmp    %rbx,%rbp
  jne    L17
  mov    $0xa,%edx
  mov    %r12,%rsi
  lea    D9(%rip),%rdi
  call   <print_bench@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
data_type: string

tool: D9: bigint_mul131

assistant: ```c
void speed_bigint_mul131() {
  unsigned long long t[10];
  int i;
  unsigned char r[34];
  for (i = 0; i < 10; i++) {
    t[i] = cpucycles();
    bigint_mul131(r, r, r);
  }
  print_bench("bigint_mul131", t, 10);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<speed_bigint_mul131>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x50(%rsp),%rbx
  mov    %r12,%rbp
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  add    $0x8,%rbp
  call   <cpucycles@plt>
  mov    %rbx,%rdx
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  mov    %rax,-0x8(%rbp)
  call   <bigint_mul131@plt>
  cmp    %rbx,%rbp
  jne    L5
  mov    $0xa,%edx
  mov    %r12,%rsi
  lea    D3(%rip),%rdi
  call   <print_bench@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_guessed: string

tool: D3: bigint_mul131

assistant: ```c
void speed_bigint_mul131() {
  unsigned long long t[10];
  int i;
  unsigned char r[34];
  for (i = 0; i < 10; i++) {
    t[i] = cpucycles();
    bigint_mul131(r, r, r);
  }
  print_bench("bigint_mul131", t, 10);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str_into_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L26
L20:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  je     L9
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x8(%rbp)
L9:
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L20
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int str_into_int(char *input) {
  int i, number, length;
  length = strlen(input) - 1;
  for (i = 0, number = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  return number;
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_into_int>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    $0x1,%eax
  jle    L16
  mov    %eax,%esi
  lea    -0x2(%rax),%edi
  mov    $0x0,%edx
  mov    $0x0,%eax
  sub    $0x2,%esi
  jmp    L10
L47:
  lea    0x1(%rdx),%rcx
  cmp    %rdi,%rdx
  je     L27
  mov    %rcx,%rdx
L10:
  movsbl (%rbx,%rdx,1),%ecx
  lea    -0x30(%rax,%rcx,1),%eax
  cmp    %edx,%esi
  je     L47
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  jmp    L47
L16:
  mov    $0x0,%eax
L27:
  pop    %rbx
  ret
```

assistant: ```c
int str_into_int(char *input) {
  int i, number, length;
  length = strlen(input) - 1;
  for (i = 0, number = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  return number;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<str_into_int>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    $0x1,%eax
  jle    L21
  lea    -0x2(%rax),%edi
  xor    %edx,%edx
  xor    %eax,%eax
  mov    %rdi,%rsi
  jmp    L6
  xchg   %ax,%ax
L19:
  mov    %rcx,%rdx
L6:
  movsbl (%rbx,%rdx,1),%ecx
  lea    -0x30(%rax,%rcx,1),%eax
  cmp    %edx,%esi
  je     L7
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
L7:
  lea    0x1(%rdx),%rcx
  cmp    %rdx,%rdi
  jne    L19
  pop    %rbx
  ret
  nop
L21:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int str_into_int(char *input) {
  int i, number, length;
  length = strlen(input) - 1;
  for (i = 0, number = 0; i < length; i++) {
    number += input[i] - '0';
    if (i != length - 1) {
      number *= 10;
    }
  }
  return number;
}
```