ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_and_change_bit>:
  endbr64
  btc    %edi,(%rsi)
  sbb    %eax,%eax
  ret
```

assistant: ```c
int test_and_change_bit(int nr, volatile void *addr) {
  int oldbit;
  asm volatile(""
               "btcl %2,%1\n\tsbbl %0,%0"
               : "=r"(oldbit), "=m"((*(volatile long *)addr))
               : "Ir"(nr), "m"((*(volatile long *)addr))
               : "memory");
  return oldbit;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_mySub_large>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,(%rax)
  mov    $0x3e8,%esi
  mov    $0x7d0,%edi
  mov    $0x0,%eax
  call   <mySub@plt>
  cmp    $0x3e8,%eax
  je     L18
  lea    D9(%rip),%rax
  jmp    L13
L18:
  mov    $0x0,%eax
L13:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string
name: parse_data
arguments: label: D9
datatype: string

tool: D2: test_mySub_large

tool: D9: 2000 - 1000 == 1000

assistant: ```c
char *test_mySub_large(const char **_ja_name) {
  do {
    *_ja_name = __func__;
    if (!(mySub(2000, 1000) == 1000))
      return "2000 - 1000 == 1000";
  } while (0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_mySub_large>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rax
  mov    %rax,(%rdi)
  mov    $0x3e8,%esi
  mov    $0x7d0,%edi
  mov    $0x0,%eax
  call   <mySub@plt>
  cmp    $0x3e8,%eax
  lea    D16(%rip),%rax
  mov    $0x0,%edx
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D16
type: string

tool: D14: test_mySub_large

tool: D16: 2000 - 1000 == 1000

assistant: ```c
char *test_mySub_large(const char **_ja_name) {
  do {
    *_ja_name = __func__;
    if (!(mySub(2000, 1000) == 1000))
      return "2000 - 1000 == 1000";
  } while (0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_mySub_large>:
  endbr64
  lea    D19(%rip),%rax
  sub    $0x8,%rsp
  mov    $0x3e8,%esi
  mov    %rax,(%rdi)
  xor    %eax,%eax
  mov    $0x7d0,%edi
  call   <mySub@plt>
  mov    $0x0,%edx
  cmp    $0x3e8,%eax
  lea    D2(%rip),%rax
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
datatype: string
name: extract_symbolic_data
arguments: tag: D2
datatype: string

tool: D19: test_mySub_large

tool: D2: 2000 - 1000 == 1000

assistant: ```c
char *test_mySub_large(const char **_ja_name) {
  do {
    *_ja_name = __func__;
    if (!(mySub(2000, 1000) == 1000))
      return "2000 - 1000 == 1000";
  } while (0);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_mySub_large>:
  endbr64
  lea    D8(%rip),%rax
  sub    $0x8,%rsp
  mov    $0x3e8,%esi
  mov    %rax,(%rdi)
  xor    %eax,%eax
  mov    $0x7d0,%edi
  call   <mySub@plt>
  mov    $0x0,%edx
  cmp    $0x3e8,%eax
  lea    D9(%rip),%rax
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type: string
name: parse_data
arguments: tag: D9
type: string

tool: D8: test_mySub_large

tool: D9: 2000 - 1000 == 1000

assistant: ```c
char *test_mySub_large(const char **_ja_name) {
  do {
    *_ja_name = __func__;
    if (!(mySub(2000, 1000) == 1000))
      return "2000 - 1000 == 1000";
  } while (0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<call_mul_digit_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  lea    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x1c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <mul_digit_string@plt>
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x40(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L13
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rsi
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  mov    %rsi,%r8
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L13:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string

tool: D2: %d * %s is %s and should be %s


assistant: ```c
void call_mul_digit_string(const int index, const int *scalars, const char **numeric_input, const char **numeric_output) {
  int a = scalars[index];
  char *b = (char *)calloc(strlen(numeric_input[index]) + 1, sizeof(char));
  strncpy(b, numeric_input[index], strlen(numeric_input[index]));
  char *c = (char *)calloc(strlen(b) + 1, sizeof(char));
  mul_digit_string(a, b, &c);
  if (strncmp(c, numeric_output[index], strlen(numeric_output[index])))
    printf("%d * %s is %s and should be %s\n", a, b, c, numeric_output[index]);
  free(c);
  free(b);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<call_mul_digit_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rcx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %edi,%rbx
  mov    (%rsi,%rbx,4),%r14d
  shl    $0x3,%rbx
  lea    (%rdx,%rbx,1),%rbp
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    0x0(%rbp),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,(%rsp)
  mov    %rsp,%rdx
  mov    %r12,%rsi
  mov    %r14d,%edi
  call   <mul_digit_string@plt>
  add    %r13,%rbx
  mov    (%rbx),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    (%rsp),%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L15
L2:
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  mov    (%rbx),%r8
  mov    (%rsp),%rcx
  mov    %r12,%rdx
  mov    %r14d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L2
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D3
type_infered: string

tool: D3: %d * %s is %s and should be %s


assistant: ```c
void call_mul_digit_string(const int index, const int *scalars, const char **numeric_input, const char **numeric_output) {
  int a = scalars[index];
  char *b = (char *)calloc(strlen(numeric_input[index]) + 1, sizeof(char));
  strncpy(b, numeric_input[index], strlen(numeric_input[index]));
  char *c = (char *)calloc(strlen(b) + 1, sizeof(char));
  mul_digit_string(a, b, &c);
  if (strncmp(c, numeric_output[index], strlen(numeric_output[index])))
    printf("%d * %s is %s and should be %s\n", a, b, c, numeric_output[index]);
  free(c);
  free(b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<call_mul_digit_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x10,%rsp
  mov    (%rsi,%rbx,4),%r13d
  shl    $0x3,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    (%rdx,%rbx,1),%rbp
  add    %rbx,%r12
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    0x0(%rbp),%rbp
  mov    %rax,%r14
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    %rax,%rdx
  call   <strncpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    %rsp,%rdx
  mov    %r14,%rsi
  mov    %r13d,%edi
  mov    %rax,(%rsp)
  call   <mul_digit_string@plt>
  mov    (%r12),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    (%rsp),%rdi
  mov    %rbp,%rsi
  mov    %rax,%rdx
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L28
L5:
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L28:
  mov    (%r12),%r8
  mov    (%rsp),%rcx
  mov    %r14,%rdx
  mov    %r13d,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L5
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D1
data_type: string

tool: D1: %d * %s is %s and should be %s


assistant: ```c
void call_mul_digit_string(const int index, const int *scalars, const char **numeric_input, const char **numeric_output) {
  int a = scalars[index];
  char *b = (char *)calloc(strlen(numeric_input[index]) + 1, sizeof(char));
  strncpy(b, numeric_input[index], strlen(numeric_input[index]));
  char *c = (char *)calloc(strlen(b) + 1, sizeof(char));
  mul_digit_string(a, b, &c);
  if (strncmp(c, numeric_output[index], strlen(numeric_output[index])))
    printf("%d * %s is %s and should be %s\n", a, b, c, numeric_output[index]);
  free(c);
  free(b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<call_mul_digit_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x10,%rsp
  mov    (%rsi,%rbx,4),%r13d
  shl    $0x3,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    (%rdx,%rbx,1),%rbp
  add    %rbx,%r12
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    0x0(%rbp),%rbp
  mov    %rax,%r14
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    %rax,%rdx
  call   <strncpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    %rsp,%rdx
  mov    %r14,%rsi
  mov    %r13d,%edi
  mov    %rax,(%rsp)
  call   <mul_digit_string@plt>
  mov    (%r12),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    (%rsp),%rdi
  mov    %rbp,%rsi
  mov    %rax,%rdx
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L2
L26:
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L2:
  mov    (%r12),%r8
  mov    (%rsp),%rcx
  mov    %r14,%rdx
  mov    %r13d,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  jmp    L26
L18:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %d * %s is %s and should be %s


assistant: ```c
void call_mul_digit_string(const int index, const int *scalars, const char **numeric_input, const char **numeric_output) {
  int a = scalars[index];
  char *b = (char *)calloc(strlen(numeric_input[index]) + 1, sizeof(char));
  strncpy(b, numeric_input[index], strlen(numeric_input[index]));
  char *c = (char *)calloc(strlen(b) + 1, sizeof(char));
  mul_digit_string(a, b, &c);
  if (strncmp(c, numeric_output[index], strlen(numeric_output[index])))
    printf("%d * %s is %s and should be %s\n", a, b, c, numeric_output[index]);
  free(c);
  free(b);
}
```
user: Translate the below assembly code into C programming language:

```asm
<munmap_reserve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <munmap@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    %rax,%rdi
  call   <mmap@plt>
  nop
  leave
  ret
```

assistant: ```c
void munmap_reserve(void *page, int size) {
  munmap(page, size);
  mmap(page, size, PROT_NONE, MAP_PRIVATE | MAP_FIXED, 0, 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<munmap_reserve>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <munmap@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%ecx
  mov    D6(%rip),%rax
  or     (%rax),%ecx
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <mmap@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void munmap_reserve(void *page, int size) {
  munmap(page, size);
  mmap(page, size, PROT_NONE, MAP_PRIVATE | MAP_FIXED, 0, 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<munmap_reserve>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <munmap@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  mov    D18(%rip),%rax
  xor    %r8d,%r8d
  mov    (%rax),%ecx
  mov    D10(%rip),%rax
  or     (%rax),%ecx
  mov    D29(%rip),%rax
  mov    (%rax),%edx
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1060 <mmap@plt>
```

assistant: ```c
void munmap_reserve(void *page, int size) {
  munmap(page, size);
  mmap(page, size, PROT_NONE, MAP_PRIVATE | MAP_FIXED, 0, 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<munmap_reserve>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <munmap@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  mov    D15(%rip),%rax
  xor    %r8d,%r8d
  mov    (%rax),%ecx
  mov    D5(%rip),%rax
  or     (%rax),%ecx
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1060 <mmap@plt>
```

assistant: ```c
void munmap_reserve(void *page, int size) {
  munmap(page, size);
  mmap(page, size, PROT_NONE, MAP_PRIVATE | MAP_FIXED, 0, 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sq_diagonal_area>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    $0x2,%edi
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm0
  mulss  -0x14(%rbp),%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float sq_diagonal_area(float area) {
  float square_root_2 = sqrt(2);
  float diagonal = square_root_2 * area;
  return diagonal;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sq_diagonal_area>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  mov    $0x2,%edi
  call   <sqrt@plt>
  mulss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
float sq_diagonal_area(float area) {
  float square_root_2 = sqrt(2);
  float diagonal = square_root_2 * area;
  return diagonal;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sq_diagonal_area>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edi
  movss  %xmm0,0xc(%rsp)
  call   <sqrt@plt>
  mulss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
float sq_diagonal_area(float area) {
  float square_root_2 = sqrt(2);
  float diagonal = square_root_2 * area;
  return diagonal;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sq_diagonal_area>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edi
  movss  %xmm0,0xc(%rsp)
  call   <sqrt@plt>
  mulss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
float sq_diagonal_area(float area) {
  float square_root_2 = sqrt(2);
  float diagonal = square_root_2 * area;
  return diagonal;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  movzbl (%rdi),%esi
  mov    $0x0,%edi
  test   %sil,%sil
  je     L89
  lea    -0x1(%rsi),%eax
  cmp    $0x1f,%al
  ja     L63
  mov    $0x1,%edx
L88:
  mov    %edx,%eax
  movzbl (%rbx,%rdx,1),%esi
  add    $0x1,%rdx
  lea    -0x1(%rsi),%ecx
  cmp    $0x1f,%cl
  jbe    L88
  mov    %eax,0xc(%rsp)
L94:
  movq   $0x1,0x10(%rsp)
  lea    0x1(%rax),%edx
  movslq %edx,%rdx
  add    %rbx,%rdx
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L26
  mov    $0x1,%eax
L25:
  lea    (%rax,%rax,4),%rax
  add    %rax,%rax
  movzbl 0x1(%rdx),%edi
  lea    -0x30(%rdi),%ecx
  add    $0x1,%rdx
  cmp    $0x9,%cl
  jbe    L25
  mov    %rax,0x10(%rsp)
L26:
  lea    0xc(%rsp),%rdx
  movsbl %sil,%esi
  lea    0x10(%rsp),%rdi
  call   <is_negative@plt>
  movslq 0xc(%rsp),%rdx
  movzbl (%rbx,%rdx,1),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L57
  mov    0x10(%rsp),%rcx
  lea    0x1(%rbx,%rdx,1),%rsi
  mov    $0x0,%edi
  movabs $0x6666666666666667,%r8
L53:
  movsbl %al,%eax
  sub    $0x30,%eax
  imul   %ecx,%eax
  lea    (%rax,%rdi,1),%edi
  mov    %rcx,%rax
  imul   %r8
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  mov    %rdx,%rcx
  sub    %rax,%rcx
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L53
L89:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    %edi,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L63:
  mov    $0x0,%eax
  jmp    L94
L57:
  mov    $0x0,%edi
  jmp    L89
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_atoi(char *str) {
  int num_start;
  int mult_it;
  long int mult;
  int resault;
  num_start = 0;
  if (str[num_start] == 0)
    return (0);
  while (str[num_start] > 0 && str[num_start] <= 32 && str[num_start] != '-')
    num_start++;
  mult = 1;
  mult_it = num_start;
  while (str[++mult_it] >= '0' && str[mult_it] <= '9')
    mult *= 10;
  is_negative(&mult, str[num_start], &num_start);
  resault = 0;
  while (str[num_start] >= '0' && str[num_start] <= '9') {
    resault += (str[num_start++] - 48) * mult;
    mult /= 10;
  }
  return (resault);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  movsbl (%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  test   %sil,%sil
  je     L47
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  cmp    $0x1f,%al
  mov    $0x1,%eax
  ja     L89
L25:
  movsbl (%rbx,%rax,1),%esi
  mov    %eax,%ecx
  add    $0x1,%rax
  lea    -0x1(%rsi),%edx
  cmp    $0x1f,%dl
  jbe    L25
  mov    %ecx,0xc(%rsp)
  add    $0x1,%ecx
  movslq %ecx,%rax
L75:
  movq   $0x1,0x10(%rsp)
  movzbl (%rbx,%rax,1),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L22
  movslq %ecx,%rcx
  mov    $0x1,%eax
  lea    0x1(%rbx,%rcx,1),%rcx
  nopl   (%rax)
L76:
  movzbl (%rcx),%edi
  lea    (%rax,%rax,4),%rax
  add    $0x1,%rcx
  add    %rax,%rax
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  jbe    L76
  mov    %rax,0x10(%rsp)
L22:
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  call   <is_negative@plt>
  movslq 0xc(%rsp),%rdx
  movzbl (%rbx,%rdx,1),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L47
  mov    0x10(%rsp),%rcx
  lea    0x1(%rbx,%rdx,1),%rsi
  xor    %r8d,%r8d
  movabs $0x6666666666666667,%rdi
  nopw   0x0(%rax,%rax,1)
L17:
  sub    $0x30,%eax
  add    $0x1,%rsi
  movsbq %al,%rax
  imul   %rcx,%rax
  add    %eax,%r8d
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  mov    %rdx,%rcx
  sub    %rax,%rcx
  movzbl -0x1(%rsi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L17
L79:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x20,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L47:
  xor    %r8d,%r8d
  jmp    L79
  nopl   (%rax)
L89:
  mov    $0x1,%ecx
  jmp    L75
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_atoi(char *str) {
  int num_start;
  int mult_it;
  long int mult;
  int resault;
  num_start = 0;
  if (str[num_start] == 0)
    return (0);
  while (str[num_start] > 0 && str[num_start] <= 32 && str[num_start] != '-')
    num_start++;
  mult = 1;
  mult_it = num_start;
  while (str[++mult_it] >= '0' && str[mult_it] <= '9')
    mult *= 10;
  is_negative(&mult, str[num_start], &num_start);
  resault = 0;
  while (str[num_start] >= '0' && str[num_start] <= '9') {
    resault += (str[num_start++] - 48) * mult;
    mult /= 10;
  }
  return (resault);
}
```