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
<reverse>:
  endbr64
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,%r8d
  shr    $0x1f,%r8d
  add    %eax,%r8d
  sar    %r8d
  cmp    %r8d,%edx
  jg     L13
  movslq %edx,%rcx
  add    %rdi,%rcx
  movslq %esi,%rsi
  lea    (%rdi,%rsi,1),%rax
  lea    -0x1(%rdi,%rsi,1),%rdi
  sub    %edx,%r8d
  sub    %r8,%rdi
L16:
  movzbl (%rcx),%edx
  movzbl (%rax),%esi
  mov    %sil,(%rcx)
  mov    %dl,(%rax)
  add    $0x1,%rcx
  sub    $0x1,%rax
  cmp    %rdi,%rax
  jne    L16
L13:
  ret
```

assistant: ```c
void reverse(char s[], int end, int pos) {
  char t;
  int middle;
  for (middle = (end + pos) / 2; pos <= middle; pos++, end--) {
    t = s[pos];
    s[pos] = s[end];
    s[end] = t;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverse>:
  endbr64
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,%r8d
  shr    $0x1f,%r8d
  add    %eax,%r8d
  sar    %r8d
  cmp    %r8d,%edx
  jg     L11
  movslq %esi,%rsi
  movslq %edx,%rcx
  sub    %edx,%r8d
  add    %rdi,%rcx
  lea    (%rdi,%rsi,1),%rax
  lea    -0x1(%rdi,%rsi,1),%rdi
  sub    %r8,%rdi
  nopl   0x0(%rax)
L2:
  movzbl (%rcx),%edx
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  mov    %dl,0x1(%rax)
  cmp    %rdi,%rax
  jne    L2
L11:
  ret
```

assistant: ```c
void reverse(char s[], int end, int pos) {
  char t;
  int middle;
  for (middle = (end + pos) / 2; pos <= middle; pos++, end--) {
    t = s[pos];
    s[pos] = s[end];
    s[end] = t;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverse>:
  endbr64
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,%r8d
  shr    $0x1f,%r8d
  add    %eax,%r8d
  sar    %r8d
  cmp    %r8d,%edx
  jg     L11
  movslq %esi,%rsi
  movslq %edx,%rcx
  sub    %edx,%r8d
  add    %rdi,%rcx
  lea    (%rdi,%rsi,1),%rax
  lea    -0x1(%rdi,%rsi,1),%rdi
  sub    %r8,%rdi
  nopl   0x0(%rax)
L8:
  movzbl (%rcx),%edx
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  mov    %dl,0x1(%rax)
  cmp    %rdi,%rax
  jne    L8
L11:
  ret
```

assistant: ```c
void reverse(char s[], int end, int pos) {
  char t;
  int middle;
  for (middle = (end + pos) / 2; pos <= middle; pos++, end--) {
    t = s[pos];
    s[pos] = s[end];
    s[end] = t;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lgetline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L47
L10:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L47:
  subl   $0x1,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L24
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L24
  cmpl   $0xa,-0x8(%rbp)
  jne    L10
L24:
  cmpl   $0xa,-0x8(%rbp)
  jne    L12
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int lgetline(char s[], int lim) {
  int c, i;
  i = 0;
  while (--lim > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lgetline>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L47
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L44:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D6(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L19
  cmp    $0xa,%eax
  je     L19
  add    $0x1,%ebx
  mov    %al,0x0(%rbp)
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L44
  mov    %r12d,%ebx
  jmp    L30
L47:
  mov    $0x0,%ebx
L19:
  cmp    $0xa,%eax
  je     L6
L30:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  lea    0x1(%rbx),%ebx
  jmp    L30
```

assistant: ```c
int lgetline(char s[], int lim) {
  int c, i;
  i = 0;
  while (--lim > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lgetline>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L21
  mov    D3(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L58:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L35
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %ebp,%edx
  je     L54
  mov    %edx,%r14d
L30:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L58
  cmp    $0xa,%eax
  je     L24
L21:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  lea    0x1(%r14),%edx
L35:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L54:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int lgetline(char s[], int lim) {
  int c, i;
  i = 0;
  while (--lim > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lgetline>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L18
  mov    D0(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L41
  nopl   0x0(%rax,%rax,1)
L59:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L30
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %ebp,%edx
  je     L50
  mov    %edx,%r14d
L41:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L59
  cmp    $0xa,%eax
  je     L51
L18:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L51:
  lea    0x1(%r14),%edx
L30:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L50:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int lgetline(char s[], int lim) {
  int c, i;
  i = 0;
  while (--lim > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<BooyerMoor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L34
L3:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L20
  cmpl   $0x0,-0xc(%rbp)
  jne    L28
  mov    -0x10(%rbp),%eax
  jmp    L5
L28:
  subl   $0x1,-0x10(%rbp)
  subl   $0x1,-0xc(%rbp)
  jmp    L34
L20:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <Last@plt>
  cmp    %eax,-0xc(%rbp)
  jle    L10
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <Last@plt>
  add    $0x1,%eax
  jmp    L78
L10:
  mov    -0xc(%rbp),%eax
L78:
  mov    -0x8(%rbp),%edx
  sub    %eax,%edx
  add    %edx,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L3
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L5:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type: string

tool: D9: No Match

assistant: ```c
int BooyerMoor(char *T, char *P) {
  int m = strlen(P);
  int n = strlen(T);
  int i = m - 1;
  int j = m - 1;
  while (i < n - m) {
    if (P[i] == T[i]) {
      if (j == 0) {
        return i;
      } else {
        i = i - 1;
        j = j - 1;
      }
    } else {
      i += m - (((j) < (1 + Last(T[i]))) ? (j) : (1 + Last(T[i])));
      j = m - 1;
    }
  }
  printf("No Match");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<BooyerMoor>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rdi,(%rsp)
  mov    %rsi,%r13
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,0xc(%rsp)
  mov    %r15,%rdi
  call   <strlen@plt>
  lea    -0x1(%rbx),%r15d
  sub    %ebx,%eax
  cmp    %eax,%r15d
  jge    L48
  mov    %eax,%r14d
  mov    %r15d,%ebp
  mov    %r15d,%ebx
  jmp    L29
L43:
  movsbl %dil,%edi
  call   <Last@plt>
  cmp    %ebp,%eax
  jl     L27
L23:
  mov    0xc(%rsp),%eax
  sub    %ebp,%eax
  add    %eax,%ebx
  mov    %r15d,%ebp
L45:
  cmp    %r14d,%ebx
  jge    L48
L29:
  movslq %ebx,%rax
  mov    (%rsp),%rdx
  lea    (%rdx,%rax,1),%r12
  movzbl (%r12),%edi
  cmp    %dil,0x0(%r13,%rax,1)
  jne    L43
  test   %ebp,%ebp
  je     L36
  sub    $0x1,%ebx
  sub    $0x1,%ebp
  jmp    L45
L27:
  movsbl (%r12),%edi
  call   <Last@plt>
  lea    0x1(%rax),%ebp
  jmp    L23
L48:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%ebx
L36:
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
datatype: string

tool: D3: No Match

assistant: ```c
int BooyerMoor(char *T, char *P) {
  int m = strlen(P);
  int n = strlen(T);
  int i = m - 1;
  int j = m - 1;
  while (i < n - m) {
    if (P[i] == T[i]) {
      if (j == 0) {
        return i;
      } else {
        i = i - 1;
        j = j - 1;
      }
    } else {
      i += m - (((j) < (1 + Last(T[i]))) ? (j) : (1 + Last(T[i])));
      j = m - 1;
    }
  }
  printf("No Match");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<BooyerMoor>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%r15d
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  lea    -0x1(%r15),%edx
  sub    %r15d,%eax
  mov    %edx,0xc(%rsp)
  cmp    %eax,%edx
  jge    L59
  mov    %eax,%r13d
  mov    %edx,%r14d
  mov    %edx,%r12d
  jmp    L29
  nopl   0x0(%rax)
L19:
  test   %r14d,%r14d
  je     L72
  sub    $0x1,%r12d
  sub    $0x1,%r14d
  cmp    %r13d,%r12d
  jge    L59
L29:
  movslq %r12d,%rax
  lea    (%rbx,%rax,1),%r15
  movsbl (%r15),%edi
  cmp    %dil,0x0(%rbp,%rax,1)
  je     L19
  call   <Last@plt>
  cmp    %r14d,%eax
  jl     L18
L66:
  mov    0x8(%rsp),%eax
  sub    %r14d,%eax
  mov    0xc(%rsp),%r14d
  add    %eax,%r12d
  cmp    %r13d,%r12d
  jl     L29
L59:
  lea    D1(%rip),%rdi
  xor    %r12d,%r12d
  call   <printf@plt>
L72:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  movsbl (%r15),%edi
  call   <Last@plt>
  lea    0x1(%rax),%r14d
  jmp    L66
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D1: No Match

assistant: ```c
int BooyerMoor(char *T, char *P) {
  int m = strlen(P);
  int n = strlen(T);
  int i = m - 1;
  int j = m - 1;
  while (i < n - m) {
    if (P[i] == T[i]) {
      if (j == 0) {
        return i;
      } else {
        i = i - 1;
        j = j - 1;
      }
    } else {
      i += m - (((j) < (1 + Last(T[i]))) ? (j) : (1 + Last(T[i])));
      j = m - 1;
    }
  }
  printf("No Match");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<BooyerMoor>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%r15d
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  lea    -0x1(%r15),%edx
  sub    %r15d,%eax
  mov    %edx,0xc(%rsp)
  cmp    %eax,%edx
  jge    L39
  mov    %eax,%r13d
  mov    %edx,%r14d
  mov    %edx,%r12d
  jmp    L34
  nopl   0x0(%rax)
L16:
  test   %r14d,%r14d
  je     L26
  sub    $0x1,%r12d
  sub    $0x1,%r14d
  cmp    %r13d,%r12d
  jge    L39
L34:
  movslq %r12d,%rax
  lea    (%rbx,%rax,1),%r15
  movsbl (%r15),%edi
  cmp    %dil,0x0(%rbp,%rax,1)
  je     L16
  call   <Last@plt>
  cmp    %r14d,%eax
  jl     L54
L38:
  mov    0x8(%rsp),%eax
  sub    %r14d,%eax
  mov    0xc(%rsp),%r14d
  add    %eax,%r12d
  cmp    %r13d,%r12d
  jl     L34
L39:
  lea    D2(%rip),%rdi
  xor    %r12d,%r12d
  call   <printf@plt>
L26:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L54:
  movsbl (%r15),%edi
  call   <Last@plt>
  lea    0x1(%rax),%r14d
  jmp    L38
```

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

tool: D2: No Match

assistant: ```c
int BooyerMoor(char *T, char *P) {
  int m = strlen(P);
  int n = strlen(T);
  int i = m - 1;
  int j = m - 1;
  while (i < n - m) {
    if (P[i] == T[i]) {
      if (j == 0) {
        return i;
      } else {
        i = i - 1;
        j = j - 1;
      }
    } else {
      i += m - (((j) < (1 + Last(T[i]))) ? (j) : (1 + Last(T[i])));
      j = m - 1;
    }
  }
  printf("No Match");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x1(%rbp)
  cmpb   $0x60,-0x1(%rbp)
  jle    L48
  cmpb   $0x7a,-0x1(%rbp)
  jg     L48
  movzbl -0x1(%rbp),%eax
  sub    $0x20,%eax
  mov    %al,-0x1(%rbp)
  jmp    L23
L48:
  cmpb   $0x40,-0x1(%rbp)
  jle    L23
  cmpb   $0x5a,-0x1(%rbp)
  jg     L23
  movzbl -0x1(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x1(%rbp)
L23:
  movsbl -0x1(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D3: 


assistant: ```c
main(void) {
  char ch;
  ch = getchar();
  if (ch >= 'a' && ch <= 'z')
    ch -= 32;
  else if (ch >= 'A' && ch <= 'Z')
    ch += 32;
  putchar(ch);
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getchar@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L7
  sub    $0x20,%eax
L17:
  movsbl %al,%edi
  call   <putchar@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L7:
  mov    %eax,%ecx
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L17
  lea    0x20(%rcx),%eax
  jmp    L17
```

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

tool: D0: 


assistant: ```c
main(void) {
  char ch;
  ch = getchar();
  if (ch >= 'a' && ch <= 'z')
    ch -= 32;
  else if (ch >= 'A' && ch <= 'Z')
    ch += 32;
  putchar(ch);
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L17
  sub    $0x20,%eax
L2:
  movsbl %al,%edi
  call   <putchar@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L17:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L2
  add    $0x20,%eax
  jmp    L2
```

tool_calls: name: read_data
arguments: tag: D8
type: string

tool: D8: 


assistant: ```c
main(void) {
  char ch;
  ch = getchar();
  if (ch >= 'a' && ch <= 'z')
    ch -= 32;
  else if (ch >= 'A' && ch <= 'Z')
    ch += 32;
  putchar(ch);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L6
  sub    $0x20,%eax
L13:
  movsbl %al,%edi
  call   <putchar@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L6:
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  ja     L13
  add    $0x20,%eax
  jmp    L13
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D0: 


assistant: ```c
main(void) {
  char ch;
  ch = getchar();
  if (ch >= 'a' && ch <= 'z')
    ch -= 32;
  else if (ch >= 'A' && ch <= 'Z')
    ch += 32;
  putchar(ch);
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_strpair>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L104
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L104
  jmp    L110
L99:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  je     L40
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x20(%rbp)
  jne    L37
L40:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L144
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  test   %eax,%eax
  jne    L25
  addq   $0x1,-0x18(%rbp)
  jmp    L110
L25:
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jne    L144
  cmpq   $0x0,-0x28(%rbp)
  je     L156
  mov    -0x28(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
L156:
  mov    -0x18(%rbp),%rax
  jmp    L97
L144:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x20(%rbp)
  jne    L37
  subl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L37
  cmpq   $0x0,-0x28(%rbp)
  je     L108
  mov    -0x28(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
L108:
  mov    -0x18(%rbp),%rax
  jmp    L97
L37:
  addl   $0x1,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L110:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L99
L104:
  cmpq   $0x0,-0x28(%rbp)
  je     L151
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
L151:
  mov    $0x0,%eax
L97:
  pop    %rbp
  ret
```

assistant: ```c
char *_strpair(char *src, int iLvalue, int iRvalue, int *piCount) {
  int iLbr = 0, iCnt = 0;
  if (src && *src == iLvalue) {
    while (*src) {
      if (*src == iLvalue || *src == iRvalue) {
        if (*src == iLvalue) {
          if (!iLbr++) {
            src++;
            continue;
          }
          if (iLvalue == iRvalue) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
        if (*src == iRvalue) {
          if (--iLbr == 0) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
      }
      iCnt++;
      src++;
    }
  }
  if (piCount)
    *piCount = 0;
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_strpair>:
  endbr64
  mov    %esi,%r8d
  test   %rdi,%rdi
  je     L4
  movzbl (%rdi),%eax
  movsbl %al,%esi
  cmp    %r8d,%esi
  je     L90
L4:
  mov    %rcx,%r9
  test   %rcx,%rcx
  je     L135
  movl   $0x0,(%rcx)
  mov    $0x0,%r9d
L135:
  mov    %r9,%rax
  ret
L90:
  test   %al,%al
  je     L4
  lea    0x1(%rdi),%rsi
  mov    $0x0,%edi
  mov    $0x0,%r10d
  jmp    L131
L74:
  lea    0x1(%r10),%r11d
  test   %r10d,%r10d
  je     L23
  cmp    %edx,%r8d
  je     L116
  cmp    %edx,%eax
  je     L117
  mov    %r11d,%r10d
L175:
  add    $0x1,%edi
  mov    %rsi,%rax
L172:
  movzbl (%rax),%eax
  add    $0x1,%rsi
  test   %al,%al
  je     L4
L131:
  lea    -0x1(%rsi),%r9
  movsbl %al,%eax
  cmp    %edx,%eax
  je     L50
  cmp    %r8d,%eax
  jne    L175
L50:
  cmp    %r8d,%eax
  je     L74
L75:
  sub    $0x1,%r10d
  jne    L175
  test   %rcx,%rcx
  je     L135
  mov    %edi,(%rcx)
  jmp    L135
L23:
  mov    %rsi,%rax
  mov    $0x1,%r10d
  jmp    L172
L116:
  test   %rcx,%rcx
  je     L135
  mov    %edi,(%rcx)
  jmp    L135
L117:
  mov    %r11d,%r10d
  jmp    L75
```

assistant: ```c
char *_strpair(char *src, int iLvalue, int iRvalue, int *piCount) {
  int iLbr = 0, iCnt = 0;
  if (src && *src == iLvalue) {
    while (*src) {
      if (*src == iLvalue || *src == iRvalue) {
        if (*src == iLvalue) {
          if (!iLbr++) {
            src++;
            continue;
          }
          if (iLvalue == iRvalue) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
        if (*src == iRvalue) {
          if (--iLbr == 0) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
      }
      iCnt++;
      src++;
    }
  }
  if (piCount)
    *piCount = 0;
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_strpair>:
  endbr64
  test   %rdi,%rdi
  je     L6
  movsbl (%rdi),%r9d
  cmp    %esi,%r9d
  je     L75
L6:
  xor    %r8d,%r8d
  test   %rcx,%rcx
  je     L0
  movl   $0x0,(%rcx)
L0:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L75:
  test   %r9b,%r9b
  je     L6
  lea    0x1(%rdi),%rax
  xor    %r10d,%r10d
  xor    %edi,%edi
  jmp    L115
  nopl   0x0(%rax,%rax,1)
L43:
  cmp    %edx,%r9d
  je     L128
L112:
  add    $0x1,%edi
L29:
  movzbl (%rax),%r8d
  add    $0x1,%rax
  test   %r8b,%r8b
  je     L6
  movsbl %r8b,%r9d
L115:
  lea    -0x1(%rax),%r8
  cmp    %esi,%r9d
  jne    L43
  lea    0x1(%r10),%r11d
  test   %r10d,%r10d
  je     L25
  cmp    %edx,%r9d
  je     L76
  mov    %r11d,%r10d
  jmp    L112
  nopl   (%rax)
L128:
  sub    $0x1,%r10d
  jne    L112
L76:
  test   %rcx,%rcx
  je     L0
  mov    %edi,(%rcx)
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L25:
  mov    $0x1,%r10d
  jmp    L29
```

assistant: ```c
char *_strpair(char *src, int iLvalue, int iRvalue, int *piCount) {
  int iLbr = 0, iCnt = 0;
  if (src && *src == iLvalue) {
    while (*src) {
      if (*src == iLvalue || *src == iRvalue) {
        if (*src == iLvalue) {
          if (!iLbr++) {
            src++;
            continue;
          }
          if (iLvalue == iRvalue) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
        if (*src == iRvalue) {
          if (--iLbr == 0) {
            if (piCount)
              *piCount = iCnt;
            return src;
          }
        }
      }
      iCnt++;
      src++;
    }
  }
  if (piCount)
    *piCount = 0;
  return NULL;
}
```