ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<more_code>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <marker1@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int more_code() {
  char C, *pC, **ppC, ***pppC, ****ppppC, *****pppppC, ******ppppppC;
  unsigned char UC, *pUC;
  short S, *pS;
  unsigned short US, *pUS;
  int I, *pI;
  unsigned int UI, *pUI;
  long L, *pL;
  unsigned long UL, *pUL;
  float F, *pF;
  double D, *pD;
  C = 'A';
  UC = 21;
  S = -14;
  US = 7;
  I = 102;
  UI = 1002;
  L = -234;
  UL = 234;
  F = 1.25E10;
  D = -1.25E-37;
  pC = &C;
  ppC = &pC;
  pppC = &ppC;
  ppppC = &pppC;
  pppppC = &ppppC;
  ppppppC = &pppppC;
  pUC = &UC;
  pS = &S;
  pUS = &US;
  pI = &I;
  pUI = &UI;
  pL = &L;
  pUL = &UL;
  pF = &F;
  pD = &D;
  marker1();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<parse_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L19
  mov    D12(%rip),%rax
  mov    (%rax),%rdx
  mov    D5(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
L19:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int parse_int(char *str) {
  char *buf;
  int val = (int)strtol(str, &buf, 10);
  if (val == 0)
    assert(errno != ERANGE);
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_int>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L28
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L28:
  mov    D8(%rip),%rdx
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  jmp    L24
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(char *str) {
  char *buf;
  int val = (int)strtol(str, &buf, 10);
  if (val == 0)
    assert(errno != ERANGE);
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_int>:
  endbr64
  push   %r12
  mov    $0xa,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L20
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  mov    D15(%rip),%rdx
  mov    D0(%rip),%rax
  xor    %edi,%edi
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  setne  %dil
  call   <assert@plt>
  jmp    L28
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(char *str) {
  char *buf;
  int val = (int)strtol(str, &buf, 10);
  if (val == 0)
    assert(errno != ERANGE);
  return val;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parse_int>:
  endbr64
  push   %r12
  mov    $0xa,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L1
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    D15(%rip),%rdx
  mov    D14(%rip),%rax
  xor    %edi,%edi
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  setne  %dil
  call   <assert@plt>
  jmp    L4
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(char *str) {
  char *buf;
  int val = (int)strtol(str, &buf, 10);
  if (val == 0)
    assert(errno != ERANGE);
  return val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getnum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L8
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x42,%eax
  cmp    $0x35,%eax
  ja     L8
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D14(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D14(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  shll   $0x9,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L8
  shll   $0xa,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L8
  shll   $0x14,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L8
  shll   $0x2,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  nop
L8:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getnum(char *s, char **e) {
  int ret = -1;
  *e = (char *)0;
  ret = strtol(s, e, 0);
  if (*e)
    switch (**e) {
    case 'b':
    case 'B':
      ret *= 512;
      *e = *e + 1;
      break;
    case 'k':
    case 'K':
      ret *= 1024;
      *e = *e + 1;
      break;
    case 'm':
    case 'M':
      ret *= 1024 * 1024;
      *e = *e + 1;
      break;
    case 'w':
    case 'W':
      ret *= 4;
      *e = *e + 1;
      break;
    }
  return (ret);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getnum>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  movq   $0x0,(%rsi)
  mov    $0x0,%edx
  call   <strtol@plt>
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L24
  movzbl (%rdx),%ecx
  sub    $0x42,%ecx
  cmp    $0x35,%cl
  ja     L24
  movzbl %cl,%ecx
  lea    D4(%rip),%rsi
  movslq (%rsi,%rcx,4),%rcx
  add    %rsi,%rcx
  notrack jmp *%rcx
  shl    $0x9,%eax
  add    $0x1,%rdx
  mov    %rdx,(%rbx)
L24:
  pop    %rbx
  ret
  shl    $0xa,%eax
  add    $0x1,%rdx
  mov    %rdx,(%rbx)
  jmp    L24
  shl    $0x14,%eax
  add    $0x1,%rdx
  mov    %rdx,(%rbx)
  jmp    L24
  shl    $0x2,%eax
  add    $0x1,%rdx
  mov    %rdx,(%rbx)
  jmp    L24
```

assistant: ```c
int getnum(char *s, char **e) {
  int ret = -1;
  *e = (char *)0;
  ret = strtol(s, e, 0);
  if (*e)
    switch (**e) {
    case 'b':
    case 'B':
      ret *= 512;
      *e = *e + 1;
      break;
    case 'k':
    case 'K':
      ret *= 1024;
      *e = *e + 1;
      break;
    case 'm':
    case 'M':
      ret *= 1024 * 1024;
      *e = *e + 1;
      break;
    case 'w':
    case 'W':
      ret *= 4;
      *e = *e + 1;
      break;
    }
  return (ret);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getnum>:
  endbr64
  push   %rbx
  xor    %edx,%edx
  mov    %rsi,%rbx
  movq   $0x0,(%rsi)
  call   <strtol@plt>
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L19
  movzbl (%rdx),%ecx
  sub    $0x42,%ecx
  cmp    $0x35,%cl
  ja     L19
  lea    D5(%rip),%rsi
  movzbl %cl,%ecx
  movslq (%rsi,%rcx,4),%rcx
  add    %rsi,%rcx
  notrack jmp *%rcx
  nopl   (%rax)
  add    $0x1,%rdx
  shl    $0x2,%eax
  mov    %rdx,(%rbx)
L19:
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0x14,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0xa,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0x9,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int getnum(char *s, char **e) {
  int ret = -1;
  *e = (char *)0;
  ret = strtol(s, e, 0);
  if (*e)
    switch (**e) {
    case 'b':
    case 'B':
      ret *= 512;
      *e = *e + 1;
      break;
    case 'k':
    case 'K':
      ret *= 1024;
      *e = *e + 1;
      break;
    case 'm':
    case 'M':
      ret *= 1024 * 1024;
      *e = *e + 1;
      break;
    case 'w':
    case 'W':
      ret *= 4;
      *e = *e + 1;
      break;
    }
  return (ret);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getnum>:
  endbr64
  push   %rbx
  xor    %edx,%edx
  mov    %rsi,%rbx
  movq   $0x0,(%rsi)
  call   <strtol@plt>
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L13
  movzbl (%rdx),%ecx
  sub    $0x42,%ecx
  cmp    $0x35,%cl
  ja     L13
  lea    D7(%rip),%rsi
  movzbl %cl,%ecx
  movslq (%rsi,%rcx,4),%rcx
  add    %rsi,%rcx
  notrack jmp *%rcx
  nopl   (%rax)
  add    $0x1,%rdx
  shl    $0x2,%eax
  mov    %rdx,(%rbx)
L13:
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0x14,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0xa,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
  add    $0x1,%rdx
  shl    $0x9,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int getnum(char *s, char **e) {
  int ret = -1;
  *e = (char *)0;
  ret = strtol(s, e, 0);
  if (*e)
    switch (**e) {
    case 'b':
    case 'B':
      ret *= 512;
      *e = *e + 1;
      break;
    case 'k':
    case 'K':
      ret *= 1024;
      *e = *e + 1;
      break;
    case 'm':
    case 'M':
      ret *= 1024 * 1024;
      *e = *e + 1;
      break;
    case 'w':
    case 'W':
      ret *= 4;
      *e = *e + 1;
      break;
    }
  return (ret);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  jmp    L0
L9:
  lea    -0xd(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  je     L27
  cmpl   $0x4,-0x14(%rbp)
  jne    L57
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
  jmp    L0
L57:
  cmpl   $0x5,-0x14(%rbp)
  jne    L0
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L0:
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L9
  jmp    L28
L27:
  nop
L28:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D25
type_guessed: string
name: parse_data
arguments: label: D9
type_guessed: string
name: parse_data
arguments: label: D24
type_guessed: string

tool: D25: Case %d: Hajj-e-Akbar


tool: D9: Case %d: Hajj-e-Asghar


tool: D24: %s

assistant: ```c
int main() {
  char a[5];
  int sz, i = 1;
  while (scanf("%s", a) == 1) {
    sz = strlen(a);
    if (sz == 1) {
      break;
    } else if (sz == 4) {
      printf("Case %d: Hajj-e-Akbar\n", i);
      i++;
    } else if (sz == 5) {
      printf("Case %d: Hajj-e-Asghar\n", i);
      i++;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x1,%r12d
  lea    0x3(%rsp),%rbx
  lea    D28(%rip),%rbp
  lea    D19(%rip),%r14
  lea    D21(%rip),%r13
  jmp    L13
L45:
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%r12d
L13:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L61
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  je     L61
  cmp    $0x4,%eax
  je     L45
  cmp    $0x5,%eax
  jne    L13
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%r12d
  jmp    L13
L61:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D28
type_guessed: string
name: parse_data
arguments: data_tag: D19
type_guessed: string
name: parse_data
arguments: data_tag: D21
type_guessed: string

tool: D28: %s

tool: D19: Case %d: Hajj-e-Asghar


tool: D21: Case %d: Hajj-e-Akbar


assistant: ```c
int main() {
  char a[5];
  int sz, i = 1;
  while (scanf("%s", a) == 1) {
    sz = strlen(a);
    if (sz == 1) {
      break;
    } else if (sz == 4) {
      printf("Case %d: Hajj-e-Akbar\n", i);
      i++;
    } else if (sz == 5) {
      printf("Case %d: Hajj-e-Asghar\n", i);
      i++;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%r14
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbx
L3:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L26
L28:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  je     L26
  cmp    $0x4,%eax
  je     L48
  cmp    $0x5,%eax
  jne    L3
  mov    %r12d,%esi
  mov    %r14,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L28
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L48:
  mov    %r12d,%esi
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  jmp    L3
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D18
data_type: string
name: parse_data
arguments: label: D14
data_type: string
name: parse_data
arguments: label: D8
data_type: string

tool: D18: Case %d: Hajj-e-Asghar


tool: D14: Case %d: Hajj-e-Akbar


tool: D8: %s

assistant: ```c
int main() {
  char a[5];
  int sz, i = 1;
  while (scanf("%s", a) == 1) {
    sz = strlen(a);
    if (sz == 1) {
      break;
    } else if (sz == 4) {
      printf("Case %d: Hajj-e-Akbar\n", i);
      i++;
    } else if (sz == 5) {
      printf("Case %d: Hajj-e-Asghar\n", i);
      i++;
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D29(%rip),%r14
  push   %r13
  lea    D20(%rip),%r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbx
L46:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L26
L53:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  je     L26
  cmp    $0x4,%eax
  je     L39
  cmp    $0x5,%eax
  jne    L46
  mov    %r12d,%esi
  mov    %r14,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L53
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L39:
  mov    %r12d,%esi
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <printf@plt>
  jmp    L46
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D29
type: string
name: parse_data
arguments: tag: D20
type: string
name: parse_data
arguments: tag: D18
type: string

tool: D29: Case %d: Hajj-e-Asghar


tool: D20: Case %d: Hajj-e-Akbar


tool: D18: %s

assistant: ```c
int main() {
  char a[5];
  int sz, i = 1;
  while (scanf("%s", a) == 1) {
    sz = strlen(a);
    if (sz == 1) {
      break;
    } else if (sz == 4) {
      printf("Case %d: Hajj-e-Akbar\n", i);
      i++;
    } else if (sz == 5) {
      printf("Case %d: Hajj-e-Asghar\n", i);
      i++;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<findmid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L51
L27:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x28(%rbp)
  jle    L7
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L51
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x28(%rbp)
  jge    L5
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L51:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L27
  jmp    L20
L5:
  nop
L20:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L19
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    $0x2,%rax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  jmp    L12
L19:
  mov    $0xffffffff,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int findmid(int *nums, int *x, int size, int target) {
  int left = 0, right = size - 1, mid;
  while (left <= right) {
    mid = (left + right) / 2;
    if (x[mid] < target)
      left = mid + 1;
    else if (x[mid] > target)
      right = mid - 1;
    else
      break;
  }
  return left > right ? -1 : x - nums + mid;
}
```
user: Translate the below assembly code into C programming language:

```asm
<findmid>:
  endbr64
  mov    %rdi,%r9
  sub    $0x1,%edx
  js     L31
  mov    $0x0,%r8d
  jmp    L12
L51:
  jle    L26
  lea    -0x1(%rax),%edx
L7:
  cmp    %edx,%r8d
  jg     L53
L12:
  lea    (%r8,%rdx,1),%edi
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  movslq %eax,%rdi
  mov    (%rsi,%rdi,4),%edi
  cmp    %ecx,%edi
  jge    L51
  lea    0x1(%rax),%r8d
  jmp    L7
L53:
  mov    $0xffffffff,%eax
  ret
L26:
  cmp    %edx,%r8d
  jg     L23
  sub    %r9,%rsi
  sar    $0x2,%rsi
  add    %esi,%eax
  ret
L31:
  mov    $0xffffffff,%eax
  ret
L23:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int findmid(int *nums, int *x, int size, int target) {
  int left = 0, right = size - 1, mid;
  while (left <= right) {
    mid = (left + right) / 2;
    if (x[mid] < target)
      left = mid + 1;
    else if (x[mid] > target)
      right = mid - 1;
    else
      break;
  }
  return left > right ? -1 : x - nums + mid;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<findmid>:
  endbr64
  sub    $0x1,%edx
  js     L50
  xor    %r8d,%r8d
  jmp    L59
  xchg   %ax,%ax
L46:
  lea    0x1(%rax),%r8d
  cmp    %r8d,%edx
  jl     L50
L59:
  lea    (%rdx,%r8,1),%eax
  sar    %eax
  movslq %eax,%r9
  cmp    %ecx,(%rsi,%r9,4)
  jl     L46
  jle    L19
  lea    -0x1(%rax),%edx
  cmp    %r8d,%edx
  jge    L59
L50:
  mov    $0xffffffff,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  sub    %rdi,%rsi
  sar    $0x2,%rsi
  add    %esi,%eax
  ret
```

assistant: ```c
int findmid(int *nums, int *x, int size, int target) {
  int left = 0, right = size - 1, mid;
  while (left <= right) {
    mid = (left + right) / 2;
    if (x[mid] < target)
      left = mid + 1;
    else if (x[mid] > target)
      right = mid - 1;
    else
      break;
  }
  return left > right ? -1 : x - nums + mid;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<findmid>:
  endbr64
  sub    $0x1,%edx
  js     L33
  xor    %r8d,%r8d
  jmp    L49
  xchg   %ax,%ax
L37:
  lea    0x1(%rax),%r8d
  cmp    %r8d,%edx
  jl     L33
L49:
  lea    (%rdx,%r8,1),%eax
  sar    %eax
  movslq %eax,%r9
  cmp    %ecx,(%rsi,%r9,4)
  jl     L37
  jle    L0
  lea    -0x1(%rax),%edx
  cmp    %r8d,%edx
  jge    L49
L33:
  mov    $0xffffffff,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  sub    %rdi,%rsi
  sar    $0x2,%rsi
  add    %esi,%eax
  ret
```

assistant: ```c
int findmid(int *nums, int *x, int size, int target) {
  int left = 0, right = size - 1, mid;
  while (left <= right) {
    mid = (left + right) / 2;
    if (x[mid] < target)
      left = mid + 1;
    else if (x[mid] > target)
      right = mid - 1;
    else
      break;
  }
  return left > right ? -1 : x - nums + mid;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<SHA256_VST_KAT_4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x141cfc9842c4b0e3,%rax
  movabs $0x24b96f99c8f4fb9a,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x4c939b64e441ae27,%rax
  movabs $0x55b852781b9995a4,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  lea    -0x50(%rbp),%rcx
  lea    -0x51(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <SHA256_VST@plt>
  lea    -0x30(%rbp),%rsi
  lea    -0x50(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rcx
  mov    $0x20,%edx
  mov    %rax,%rdi
  call   <check_KAT@plt>
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string

tool: D5: SHA256_VST_KAT_4

assistant: ```c
SHA256_VST_KAT_4() {
  unsigned char input[0] = {};
  unsigned char result[32];
  unsigned char expected_result[32] = {0xe3, 0xb0, 0xc4, 0x42, 0x98, 0xfc, 0x1c, 0x14, 0x9a, 0xfb, 0xf4, 0xc8, 0x99, 0x6f, 0xb9, 0x24, 0x27, 0xae, 0x41, 0xe4, 0x64, 0x9b, 0x93, 0x4c, 0xa4, 0x95, 0x99, 0x1b, 0x78, 0x52, 0xb8, 0x55};
  SHA256_VST(input, result, 0);
  check_KAT(result, expected_result, 32, "SHA256_VST_KAT_4");
}
```
user: Translate the below assembly code into C programming language:

```asm
<SHA256_VST_KAT_4>:
  endbr64
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movabs $0x141cfc9842c4b0e3,%rax
  movabs $0x24b96f99c8f4fb9a,%rdx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movabs $0x4c939b64e441ae27,%rax
  movabs $0x55b852781b9995a4,%rdx
  mov    %rax,0x40(%rsp)
  mov    %rdx,0x48(%rsp)
  lea    0x10(%rsp),%rbx
  lea    0xf(%rsp),%rdi
  mov    $0x0,%edx
  mov    %rbx,%rsi
  call   <SHA256_VST@plt>
  lea    0x30(%rsp),%rsi
  lea    D1(%rip),%rcx
  mov    $0x20,%edx
  mov    %rbx,%rdi
  call   <check_KAT@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
datatype: string

tool: D1: SHA256_VST_KAT_4

assistant: ```c
SHA256_VST_KAT_4() {
  unsigned char input[0] = {};
  unsigned char result[32];
  unsigned char expected_result[32] = {0xe3, 0xb0, 0xc4, 0x42, 0x98, 0xfc, 0x1c, 0x14, 0x9a, 0xfb, 0xf4, 0xc8, 0x99, 0x6f, 0xb9, 0x24, 0x27, 0xae, 0x41, 0xe4, 0x64, 0x9b, 0x93, 0x4c, 0xa4, 0x95, 0x99, 0x1b, 0x78, 0x52, 0xb8, 0x55};
  SHA256_VST(input, result, 0);
  check_KAT(result, expected_result, 32, "SHA256_VST_KAT_4");
}
```