ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<GetWord>:
  endbr64
  mov    %rsi,%rax
  movzbl (%rsi),%esi
  lea    -0x1(%rsi),%ecx
  cmp    $0x1f,%cl
  ja     L5
  nopw   0x0(%rax,%rax,1)
L55:
  movzbl 0x1(%rax),%esi
  add    $0x1,%rax
  lea    -0x1(%rsi),%ecx
  cmp    $0x1f,%cl
  jbe    L55
L5:
  lea    (%rdi,%rdx,1),%rcx
  test   %rdx,%rdx
  jne    L28
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L17:
  add    $0x1,%rdi
  add    $0x1,%rax
  mov    %dl,-0x1(%rdi)
  cmp    %rcx,%rdi
  je     L18
L28:
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  ja     L17
L18:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
char *GetWord(char *toWordPtr, char *fromStrPtr, long limit) {
  while ((unsigned char)*fromStrPtr <= 0x20 && *fromStrPtr)
    fromStrPtr++;
  while (limit && (unsigned char)*fromStrPtr > 0x20 && *fromStrPtr) {
    *toWordPtr++ = *fromStrPtr++;
    limit--;
  }
  *toWordPtr = 0x00;
  return (char *)fromStrPtr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  jmp    L11
L23:
  cmpl   $0x2f,-0x4(%rbp)
  jne    L34
  mov    -0x18(%rbp),%rax
  jmp    L1
L34:
  addq   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L23
  mov    $0x0,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
const char *sep(const char *p) {
  int c;
  while ((c = *p)) {
    if (c == '/')
      return p;
    p++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<sep>:
  endbr64
  movsbl (%rdi),%eax
  test   %eax,%eax
  je     L5
L23:
  cmp    $0x2f,%eax
  je     L1
  add    $0x1,%rdi
  movsbl (%rdi),%eax
  test   %eax,%eax
  jne    L23
  mov    $0x0,%eax
  ret
L5:
  mov    $0x0,%eax
  ret
L1:
  mov    %rdi,%rax
  ret
```

assistant: ```c
const char *sep(const char *p) {
  int c;
  while ((c = *p)) {
    if (c == '/')
      return p;
    p++;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sep>:
  endbr64
  jmp    L27
  cs nopw 0x0(%rax,%rax,1)
L7:
  cmp    $0x2f,%eax
  je     L9
  add    $0x1,%rdi
L27:
  movsbl (%rdi),%eax
  test   %eax,%eax
  jne    L7
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %rdi,%rax
  ret
```

assistant: ```c
const char *sep(const char *p) {
  int c;
  while ((c = *p)) {
    if (c == '/')
      return p;
    p++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sep>:
  endbr64
  jmp    L9
  cs nopw 0x0(%rax,%rax,1)
L5:
  cmp    $0x2f,%eax
  je     L4
  add    $0x1,%rdi
L9:
  movsbl (%rdi),%eax
  test   %eax,%eax
  jne    L5
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %rdi,%rax
  ret
```

assistant: ```c
const char *sep(const char *p) {
  int c;
  while ((c = *p)) {
    if (c == '/')
      return p;
    p++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<bitcount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  shr    %eax
  and    $0x55555555,%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x33333333,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x4(%rbp),%eax
  shr    $0x2,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x4,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0x1010101,%eax,%eax
  shr    $0x18,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int bitcount(unsigned int n) {
  unsigned int m;
  n -= (n >> 1) & 0x55555555;
  m = n & 0x33333333;
  n = m + ((n ^ m) >> 2);
  n = (n + (n >> 4)) & 0x0F0F0F0F;
  n = (n * 0x01010101) >> 24;
  return (n);
}
```
user: What is the c source code of the assembly code below:

```asm
<bitcount>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  shr    %edx
  and    $0x55555555,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  and    $0x33333333,%edx
  shr    $0x2,%eax
  and    $0x33333333,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x4,%edx
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  imul   $0x1010101,%eax,%eax
  shr    $0x18,%eax
  ret
```

assistant: ```c
unsigned int bitcount(unsigned int n) {
  unsigned int m;
  n -= (n >> 1) & 0x55555555;
  m = n & 0x33333333;
  n = m + ((n ^ m) >> 2);
  n = (n + (n >> 4)) & 0x0F0F0F0F;
  n = (n * 0x01010101) >> 24;
  return (n);
}
```
user: What is the c source code of the assembly code below:

```asm
<bitcount>:
  endbr64
  mov    %edi,%edx
  mov    %edi,%eax
  shr    %edx
  and    $0x55555555,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  shr    $0x2,%eax
  and    $0x33333333,%edx
  and    $0x33333333,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x4,%edx
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  imul   $0x1010101,%eax,%eax
  shr    $0x18,%eax
  ret
```

assistant: ```c
unsigned int bitcount(unsigned int n) {
  unsigned int m;
  n -= (n >> 1) & 0x55555555;
  m = n & 0x33333333;
  n = m + ((n ^ m) >> 2);
  n = (n + (n >> 4)) & 0x0F0F0F0F;
  n = (n * 0x01010101) >> 24;
  return (n);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<str_remove_nulls>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <assert@plt>
  mov    -0x1c(%rbp),%eax
  not    %eax
  shr    $0x1f,%eax
  movzbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  mov    -0x18(%rbp),%rbx
  mov    $0x0,%r12d
  jmp    L16
L21:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L25
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rbx),%edx
  mov    %dl,(%rax)
L25:
  add    $0x1,%rbx
  add    $0x1,%r12d
L16:
  cmp    -0x1c(%rbp),%r12d
  jl     L21
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void str_remove_nulls(char *str, int const length) {
  register char *p;
  register int slot;
  assert(str);
  assert(length >= 0);
  p = str;
  for (slot = 0; slot < length; slot++) {
    if (*p != '\0')
      *str++ = *p;
    p++;
  }
  *str = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_remove_nulls>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    $0x0,%eax
  call   <assert@plt>
  mov    %ebp,%edi
  not    %edi
  shr    $0x1f,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  test   %ebp,%ebp
  jle    L10
  mov    %ebp,%ecx
  add    %rbx,%rcx
  mov    %rbx,%rax
  jmp    L13
L21:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L10
L13:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L21
  mov    %dl,(%rbx)
  lea    0x1(%rbx),%rbx
  jmp    L21
L10:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void str_remove_nulls(char *str, int const length) {
  register char *p;
  register int slot;
  assert(str);
  assert(length >= 0);
  p = str;
  for (slot = 0; slot < length; slot++) {
    if (*p != '\0')
      *str++ = *p;
    p++;
  }
  *str = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<str_remove_nulls>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  test   %ebp,%ebp
  jle    L4
  movslq %ebp,%rcx
  mov    %rbx,%rax
  add    %rbx,%rcx
  nopl   0x0(%rax)
L27:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L2
  mov    %dl,(%rbx)
  add    $0x1,%rbx
L2:
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L27
L4:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void str_remove_nulls(char *str, int const length) {
  register char *p;
  register int slot;
  assert(str);
  assert(length >= 0);
  p = str;
  for (slot = 0; slot < length; slot++) {
    if (*p != '\0')
      *str++ = *p;
    p++;
  }
  *str = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<str_remove_nulls>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  test   %ebp,%ebp
  jle    L15
  movslq %ebp,%rcx
  mov    %rbx,%rax
  add    %rbx,%rcx
  nopl   0x0(%rax)
L4:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L9
  mov    %dl,(%rbx)
  add    $0x1,%rbx
L9:
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L4
L15:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void str_remove_nulls(char *str, int const length) {
  register char *p;
  register int slot;
  assert(str);
  assert(length >= 0);
  p = str;
  for (slot = 0; slot < length; slot++) {
    if (*p != '\0')
      *str++ = *p;
    p++;
  }
  *str = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fpeek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int fpeek(FILE *stream) {
  int c;
  c = fgetc(stream);
  ungetc(c, stream);
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fpeek>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <fgetc@plt>
  mov    %eax,%ebx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fpeek(FILE *stream) {
  int c;
  c = fgetc(stream);
  ungetc(c, stream);
  return c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fpeek>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <fgetc@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  mov    %eax,%r12d
  call   <ungetc@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int fpeek(FILE *stream) {
  int c;
  c = fgetc(stream);
  ungetc(c, stream);
  return c;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fpeek>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <fgetc@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  mov    %eax,%r12d
  call   <ungetc@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int fpeek(FILE *stream) {
  int c;
  c = fgetc(stream);
  ungetc(c, stream);
  return c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<leMatriz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L0
L5:
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L8:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L8
  addl   $0x1,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L5
  nop
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D13
type_infered: string
name: retrieve_data
arguments: tag: D14
type_infered: string

tool: D13: 
Digite os valores: 

tool: D14: %d

assistant: ```c
void leMatriz(int **matriz, int lin, int col) {
  printf("\nDigite os valores: ");
  for (int i = 0; i < lin; i++)
    for (int j = 0; j < col; j++)
      scanf("%d", &matriz[i][j]);
}
```
user: What is the c source code of the assembly code below?

```asm
<leMatriz>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%r14d
  lea    D3(%rip),%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L8
  mov    %r12,%rbp
  lea    -0x1(%rbx),%eax
  lea    0x8(%r12,%rax,8),%r15
  lea    -0x1(%r14),%eax
  lea    0x4(,%rax,4),%r13
  lea    D6(%rip),%r12
  jmp    L35
L54:
  mov    %rbx,%rsi
  add    0x0(%rbp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L54
L21:
  add    $0x8,%rbp
  cmp    %r15,%rbp
  je     L8
L35:
  mov    $0x0,%ebx
  test   %r14d,%r14d
  jg     L54
  jmp    L21
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D3
type_infered: string
name: parse_data
arguments: label: D6
type_infered: string

tool: D3: 
Digite os valores: 

tool: D6: %d

assistant: ```c
void leMatriz(int **matriz, int lin, int col) {
  printf("\nDigite os valores: ");
  for (int i = 0; i < lin; i++)
    for (int j = 0; j < col; j++)
      scanf("%d", &matriz[i][j]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<leMatriz>:
  endbr64
  push   %r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D4(%rip),%rdi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L9
  lea    -0x1(%rbx),%eax
  lea    D11(%rip),%r12
  lea    0x8(%rbp,%rax,8),%r15
  lea    -0x1(%r14),%eax
  lea    0x4(,%rax,4),%r13
  nopl   (%rax)
L30:
  xor    %ebx,%ebx
  test   %r14d,%r14d
  jle    L16
  nop
L23:
  mov    0x0(%rbp),%rsi
  mov    %r12,%rdi
  add    %rbx,%rsi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbx,%r13
  jne    L23
L16:
  add    $0x8,%rbp
  cmp    %r15,%rbp
  jne    L30
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string
name: read_data
arguments: data_tag: D11
type_infered: string

tool: D4: 
Digite os valores: 

tool: D11: %d

assistant: ```c
void leMatriz(int **matriz, int lin, int col) {
  printf("\nDigite os valores: ");
  for (int i = 0; i < lin; i++)
    for (int j = 0; j < col; j++)
      scanf("%d", &matriz[i][j]);
}
```
user: What is the c source code of the assembly code below:

```asm
<leMatriz>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    D14(%rip),%rdi
  push   %rbx
  movslq %edx,%rbx
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L17
  test   %ebx,%ebx
  jle    L17
  lea    -0x1(%r12),%eax
  lea    D10(%rip),%r14
  lea    0x8(%rbp,%rax,8),%r13
  lea    0x0(,%rbx,4),%r12
  nopl   0x0(%rax,%rax,1)
L31:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L7:
  mov    0x0(%rbp),%rsi
  mov    %r14,%rdi
  add    %rbx,%rsi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbx,%r12
  jne    L7
  add    $0x8,%rbp
  cmp    %r13,%rbp
  jne    L31
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D10
type_guessed: string

tool: D14: 
Digite os valores: 

tool: D10: %d

assistant: ```c
void leMatriz(int **matriz, int lin, int col) {
  printf("\nDigite os valores: ");
  for (int i = 0; i < lin; i++)
    for (int j = 0; j < col; j++)
      scanf("%d", &matriz[i][j]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_check_times>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  imul   %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    %rax,-0x8(%rbp)
  ja     L5
  cmpl   $0x10000000,-0xc(%rbp)
  jbe    L0
L5:
  mov    $0x180,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <_throw_badalloc_fn@plt>
L0:
  mov    -0xc(%rbp),%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
data_type: string

tool: D3: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c

assistant: ```c
unsigned int _check_times(unsigned x, unsigned y) {
  unsigned long long whole_ans = ((unsigned long long)x) * ((unsigned long long)y);
  unsigned word_ans = (unsigned)whole_ans;
  if (word_ans < whole_ans || word_ans > (1 << 28))
    (_throw_badalloc_fn("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c", 384));
  return word_ans;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_check_times>:
  endbr64
  push   %rbx
  mov    %edi,%edi
  mov    %esi,%esi
  imul   %rsi,%rdi
  mov    %edi,%ebx
  mov    %edi,%eax
  cmp    %rax,%rdi
  ja     L12
  cmp    $0x10000000,%edi
  jbe    L17
L12:
  mov    $0x180,%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <_throw_badalloc_fn@plt>
L17:
  mov    %ebx,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
type_infered: string

tool: D8: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c

assistant: ```c
unsigned int _check_times(unsigned x, unsigned y) {
  unsigned long long whole_ans = ((unsigned long long)x) * ((unsigned long long)y);
  unsigned word_ans = (unsigned)whole_ans;
  if (word_ans < whole_ans || word_ans > (1 << 28))
    (_throw_badalloc_fn("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c", 384));
  return word_ans;
}
```
user: What is the c source code of the assembly code below?

```asm
<_check_times>:
  endbr64
  mov    %edi,%edi
  mov    %esi,%esi
  push   %r12
  imul   %rsi,%rdi
  mov    %edi,%eax
  mov    %edi,%r12d
  cmp    %rax,%rdi
  ja     L6
  cmp    $0x10000000,%edi
  ja     L6
  mov    %r12d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    $0x180,%esi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <_throw_badalloc_fn@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: string

tool: D3: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c

assistant: ```c
unsigned int _check_times(unsigned x, unsigned y) {
  unsigned long long whole_ans = ((unsigned long long)x) * ((unsigned long long)y);
  unsigned word_ans = (unsigned)whole_ans;
  if (word_ans < whole_ans || word_ans > (1 << 28))
    (_throw_badalloc_fn("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c", 384));
  return word_ans;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_check_times>:
  endbr64
  mov    %edi,%edi
  mov    %esi,%esi
  push   %r12
  imul   %rsi,%rdi
  mov    %edi,%eax
  mov    %edi,%r12d
  cmp    %rax,%rdi
  ja     L9
  cmp    $0x10000000,%edi
  ja     L9
  mov    %r12d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    $0x180,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <_throw_badalloc_fn@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

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

tool: D0: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c

assistant: ```c
unsigned int _check_times(unsigned x, unsigned y) {
  unsigned long long whole_ans = ((unsigned long long)x) * ((unsigned long long)y);
  unsigned word_ans = (unsigned)whole_ans;
  if (word_ans < whole_ans || word_ans > (1 << 28))
    (_throw_badalloc_fn("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/pgerakios/Concurrent-Cyclone-with-inference/refs/heads/master/bin/genfiles/jump_analysis.c", 384));
  return word_ans;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L42
L2:
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L2
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L6
L42:
  mov    -0x10(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L42
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L6:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D29
type: string
name: unpack_data
arguments: data_tag: D1
type: string
name: unpack_data
arguments: data_tag: D22
type: string
name: unpack_data
arguments: data_tag: D41
type: string

tool: D29: a = 

tool: D1: %d

tool: D22: b = 

tool: D41: NWD = %d


assistant: ```c
int main(void) {
  int a, b, r;
  printf("a = ");
  scanf("%d", &a);
  printf("b = ");
  scanf("%d", &b);
  if (a > b) {
    do {
      r = a % b;
      a = b;
      b = r;
    } while (b);
    printf("NWD = %d\n", a);
  } else {
    do {
      r = b % a;
      b = a;
      a = r;
    } while (a);
    printf("NWD = %d\n", b);
  }
  return 0;
}
```