ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  mov    %rax,%r10
  mov    $0x0,%r11d
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L0:
  cmp    %rdx,%rsp
  je     L39
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L0
L39:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L22
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L22:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <foo@plt>
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int bar(unsigned long x) {
  char a[x];
  return foo(a);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  add    $0xf,%rdi
  mov    %rdi,%rdx
  and    $0xfffffffffffffff0,%rdx
  and    $0xfffffffffffff000,%rdi
  mov    %rsp,%rax
  sub    %rdi,%rax
L11:
  cmp    %rax,%rsp
  je     L35
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L11
L35:
  mov    %rdx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L25
  orq    $0x0,-0x8(%rsp,%rax,1)
L25:
  mov    %rsp,%rdi
  call   <foo@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  leave
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(unsigned long x) {
  char a[x];
  return foo(a);
}
```
user: What is the c source code of the assembly code below:

```asm
<bar>:
  endbr64
  push   %rbp
  add    $0xf,%rdi
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    %rdi,%rax
  and    $0xfffffffffffff000,%rdi
  sub    %rdi,%rdx
  and    $0xfffffffffffffff0,%rax
  cmp    %rdx,%rsp
  je     L11
L18:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdx,%rsp
  jne    L18
L11:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L42
L23:
  mov    %rsp,%rdi
  call   <foo@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  leave
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L23
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(unsigned long x) {
  char a[x];
  return foo(a);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bar>:
  endbr64
  push   %rbp
  add    $0xf,%rdi
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    %rdi,%rax
  and    $0xfffffffffffff000,%rdi
  sub    %rdi,%rdx
  and    $0xfffffffffffffff0,%rax
  cmp    %rdx,%rsp
  je     L36
L1:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdx,%rsp
  jne    L1
L36:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L42
L38:
  mov    %rsp,%rdi
  call   <foo@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L34
  leave
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L38
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(unsigned long x) {
  char a[x];
  return foo(a);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L17
  addq   $0x1,-0x10(%rbp)
L17:
  cmpq   $0x0,-0x20(%rbp)
  jns    L35
  negq   -0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L35
L11:
  addq   $0x1,-0x10(%rbp)
L35:
  mov    -0x18(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L11
  addq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <itoa_cpy@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  char *s;
  long nb;
  long tmp;
  size_t len;
  nb = n;
  tmp = n;
  len = 0;
  if (n < 0)
    ++len;
  if (nb < 0) {
    nb = nb * -1;
    tmp = nb;
  }
  while (tmp /= 10)
    ++len;
  ++len;
  s = itoa_cpy(n, nb, len);
  return (s);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_itoa>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%r9d
  movslq %edi,%rdi
  shr    $0x3f,%rdi
  mov    %r9d,%esi
  neg    %esi
  cmovs  %r9d,%esi
  mov    %esi,%esi
  movabs $0x6666666666666667,%rdx
  mov    %rsi,%rax
  imul   %rdx
  mov    %rdx,%rcx
  sar    $0x2,%rcx
  je     L8
  movabs $0x6666666666666667,%r8
L5:
  add    $0x1,%rdi
  mov    %rcx,%rax
  imul   %r8
  sar    $0x2,%rdx
  sar    $0x3f,%rcx
  sub    %rcx,%rdx
  mov    %rdx,%rcx
  jne    L5
L8:
  lea    0x1(%rdi),%rdx
  mov    %r9d,%edi
  call   <itoa_cpy@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  char *s;
  long nb;
  long tmp;
  size_t len;
  nb = n;
  tmp = n;
  len = 0;
  if (n < 0)
    ++len;
  if (nb < 0) {
    nb = nb * -1;
    tmp = nb;
  }
  while (tmp /= 10)
    ++len;
  ++len;
  s = itoa_cpy(n, nb, len);
  return (s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_itoa>:
  endbr64
  movslq %edi,%rcx
  movabs $0xcccccccccccccccd,%rdi
  mov    %rcx,%r8
  shr    $0x3f,%rcx
  mov    %r8d,%esi
  neg    %esi
  cmovs  %r8d,%esi
  mov    %esi,%esi
  mov    %rsi,%rax
  mul    %rdi
  shr    $0x3,%rdx
  je     L19
  nop
L8:
  mov    %rdx,%rax
  add    $0x1,%rcx
  mul    %rdi
  shr    $0x3,%rdx
  jne    L8
L19:
  lea    0x1(%rcx),%rdx
  mov    %r8d,%edi
  jmp    1050 <itoa_cpy@plt>
```

assistant: ```c
char *ft_itoa(int n) {
  char *s;
  long nb;
  long tmp;
  size_t len;
  nb = n;
  tmp = n;
  len = 0;
  if (n < 0)
    ++len;
  if (nb < 0) {
    nb = nb * -1;
    tmp = nb;
  }
  while (tmp /= 10)
    ++len;
  ++len;
  s = itoa_cpy(n, nb, len);
  return (s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_itoa>:
  endbr64
  movslq %edi,%rcx
  movabs $0xcccccccccccccccd,%rdi
  mov    %rcx,%r8
  shr    $0x3f,%rcx
  mov    %r8d,%esi
  neg    %esi
  cmovs  %r8d,%esi
  mov    %esi,%esi
  mov    %rsi,%rax
  mul    %rdi
  shr    $0x3,%rdx
  je     L16
  nop
L3:
  mov    %rdx,%rax
  add    $0x1,%rcx
  mul    %rdi
  shr    $0x3,%rdx
  jne    L3
L16:
  lea    0x1(%rcx),%rdx
  mov    %r8d,%edi
  jmp    1050 <itoa_cpy@plt>
```

assistant: ```c
char *ft_itoa(int n) {
  char *s;
  long nb;
  long tmp;
  size_t len;
  nb = n;
  tmp = n;
  len = 0;
  if (n < 0)
    ++len;
  if (nb < 0) {
    nb = nb * -1;
    tmp = nb;
  }
  while (tmp /= 10)
    ++len;
  ++len;
  s = itoa_cpy(n, nb, len);
  return (s);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L2
  mov    -0x30(%rbp),%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    $0x1,%eax
  jmp    L29
L2:
  lea    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x10(%rbp),%xmm0
  jp     L21
  pxor   %xmm0,%xmm0
  ucomisd -0x10(%rbp),%xmm0
  je     L41
L21:
  mov    -0x18(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L41
  mov    -0x30(%rbp),%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    $0x1,%eax
  jmp    L29
L41:
  mov    $0x0,%eax
L29:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L63
  call   <__stack_chk_fail@plt>
L63:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D7
data_type: string
name: retrieve_data
arguments: label: D5
data_type: string

tool: D7: 0

tool: 

assistant: ```c
int isNumber(char *s, double *num) {
  char *end;
  double returnNum;
  if (strcmp(s, "0") == 0) {
    *num = 0;
    return 1;
  } else {
    returnNum = strtod(s, &end);
    if ((returnNum != 0.0) && (strcmp(end, "") == 0)) {
      *num = returnNum;
      return 1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isNumber>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L16
  movq   $0x0,0x0(%rbp)
  mov    $0x1,%eax
L14:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L43
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <strtod@plt>
  movq   %xmm0,%rbx
  ucomisd D2(%rip),%xmm0
  jp     L0
  mov    $0x0,%eax
  je     L14
L0:
  lea    D23(%rip),%rsi
  mov    (%rsp),%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L14
  mov    %rbx,0x0(%rbp)
  mov    $0x1,%eax
  jmp    L14
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
type: string
name: extract_symbolic_data
arguments: tag: D2
type: f64
name: extract_symbolic_data
arguments: tag: D23
type: f64

tool: D10: 0

tool: D2: 0

tool: D23: 0

assistant: ```c
int isNumber(char *s, double *num) {
  char *end;
  double returnNum;
  if (strcmp(s, "0") == 0) {
    *num = 0;
    return 1;
  } else {
    returnNum = strtod(s, &end);
    if ((returnNum != 0.0) && (strcmp(end, "") == 0)) {
      *num = returnNum;
      return 1;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isNumber>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  lea    D3(%rip),%rsi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L9
  movq   $0x0,(%rbx)
  mov    $0x1,%eax
L54:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L58
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <strtod@plt>
  ucomisd D24(%rip),%xmm0
  jp     L5
  jne    L5
L22:
  xor    %eax,%eax
  jmp    L54
  nopl   (%rax)
L5:
  mov    0x10(%rsp),%rdi
  lea    D16(%rip),%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <strcmp@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  jne    L22
  movsd  %xmm0,(%rbx)
  mov    $0x1,%eax
  jmp    L54
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string
name: extract_symbolic_data
arguments: label: D24
type: f64
name: extract_symbolic_data
arguments: label: D16
type: f64

tool: D3: 0

tool: D24: 0

tool: D16: 0

assistant: ```c
int isNumber(char *s, double *num) {
  char *end;
  double returnNum;
  if (strcmp(s, "0") == 0) {
    *num = 0;
    return 1;
  } else {
    returnNum = strtod(s, &end);
    if ((returnNum != 0.0) && (strcmp(end, "") == 0)) {
      *num = returnNum;
      return 1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isNumber>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  lea    D24(%rip),%rsi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L27
  movq   $0x0,(%rbx)
  mov    $0x1,%eax
L30:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L53
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L27:
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <strtod@plt>
  ucomisd D0(%rip),%xmm0
  jp     L67
  jne    L67
L68:
  xor    %eax,%eax
  jmp    L30
  nopl   (%rax)
L67:
  mov    0x10(%rsp),%rdi
  lea    D19(%rip),%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <strcmp@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  jne    L68
  movsd  %xmm0,(%rbx)
  mov    $0x1,%eax
  jmp    L30
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D0
datatype: f64
name: disassemble_data
arguments: data_tag: D19
datatype: f64

tool: D24: 0

tool: D0: 0

tool: D19: 0

assistant: ```c
int isNumber(char *s, double *num) {
  char *end;
  double returnNum;
  if (strcmp(s, "0") == 0) {
    *num = 0;
    return 1;
  } else {
    returnNum = strtod(s, &end);
    if ((returnNum != 0.0) && (strcmp(end, "") == 0)) {
      *num = returnNum;
      return 1;
    }
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<geteuid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <getenv@plt>
  test   %rax,%rax
  je     L15
  mov    $0x0,%eax
  jmp    L7
L15:
  mov    $0x64,%eax
L7:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D7: LIBFAKE_EUID_ROOT

assistant: ```c
uid_t geteuid(void) {
  if (getenv("LIBFAKE_EUID_ROOT") != NULL)
    return 0;
  else
    return 100;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<geteuid>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <getenv@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x64,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: string

tool: D7: LIBFAKE_EUID_ROOT

assistant: ```c
uid_t geteuid(void) {
  if (getenv("LIBFAKE_EUID_ROOT") != NULL)
    return 0;
  else
    return 100;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<geteuid>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <getenv@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0x64,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type: string

tool: D7: LIBFAKE_EUID_ROOT

assistant: ```c
uid_t geteuid(void) {
  if (getenv("LIBFAKE_EUID_ROOT") != NULL)
    return 0;
  else
    return 100;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<geteuid>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <getenv@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0x64,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
type_infered: string

tool: D6: LIBFAKE_EUID_ROOT

assistant: ```c
uid_t geteuid(void) {
  if (getenv("LIBFAKE_EUID_ROOT") != NULL)
    return 0;
  else
    return 100;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_upper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x40,-0x4(%rbp)
  jle    L26
  cmpl   $0x5a,-0x4(%rbp)
  jg     L26
  mov    $0x1,%eax
  jmp    L24
L26:
  mov    $0x0,%eax
L24:
  pop    %rbp
  ret
```

assistant: ```c
int is_upper(int c) {
  if (c >= 'A' && c <= 'Z')
    return 1;
  else
    return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<is_upper>:
  endbr64
  sub    $0x41,%edi
  cmp    $0x19,%edi
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_upper(int c) {
  if (c >= 'A' && c <= 'Z')
    return 1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_upper>:
  endbr64
  sub    $0x41,%edi
  xor    %eax,%eax
  cmp    $0x19,%edi
  setbe  %al
  ret
```

assistant: ```c
int is_upper(int c) {
  if (c >= 'A' && c <= 'Z')
    return 1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_upper>:
  endbr64
  sub    $0x41,%edi
  xor    %eax,%eax
  cmp    $0x19,%edi
  setbe  %al
  ret
```

assistant: ```c
int is_upper(int c) {
  if (c >= 'A' && c <= 'Z')
    return 1;
  else
    return 0;
}
```
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
  lea    -0x61000(%rsp),%r11
L62:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L62
  sub    $0xaa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x61a98(%rbp)
L5:
  lea    -0x61aa0(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x61aa0(%rbp),%eax
  test   %eax,%eax
  je     L34
  movl   $0x0,-0x61a9c(%rbp)
  jmp    L53
L20:
  lea    -0x61a90(%rbp),%rdx
  mov    -0x61a9c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x61a9c(%rbp)
L53:
  mov    -0x61aa0(%rbp),%eax
  cmp    %eax,-0x61a9c(%rbp)
  jl     L20
  movl   $0x1,-0x61a9c(%rbp)
  jmp    L25
L33:
  mov    -0x61a9c(%rbp),%eax
  cltq
  mov    -0x61a90(%rbp,%rax,4),%edx
  mov    -0x61a9c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x61a90(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x61a9c(%rbp),%eax
  cltq
  mov    %edx,-0x61a90(%rbp,%rax,4)
  mov    -0x61a9c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x61a90(%rbp,%rax,4),%eax
  mov    %eax,%edi
  call   <abs@plt>
  add    %rax,-0x61a98(%rbp)
  addl   $0x1,-0x61a9c(%rbp)
L25:
  mov    -0x61aa0(%rbp),%eax
  cmp    %eax,-0x61a9c(%rbp)
  jl     L33
  mov    -0x61a98(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movq   $0x0,-0x61a98(%rbp)
  jmp    L5
L34:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L74
  call   <__stack_chk_fail@plt>
L74:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D10
type: string
name: parse_data
arguments: data_label: D26
type: string

tool: D10: %d

tool: D26: %lld


assistant: ```c
int main() {
  int i, t;
  int a[100000];
  long long int sum = 0;
  while (1) {
    scanf("%d", &t);
    if (t == 0)
      break;
    for (i = 0; i < t; i++) {
      scanf("%d", &a[i]);
    }
    for (i = 1; i < t; i++) {
      a[i] += a[i - 1];
      sum += abs(a[i - 1]);
    }
    printf("%lld\n", sum);
    sum = 0;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x61000(%rsp),%r11
L82:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L82
  sub    $0xaa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x61a98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  lea    D11(%rip),%r12
  lea    D12(%rip),%r14
  jmp    L74
L33:
  mov    $0x0,%ebp
L35:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
L74:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L32
  jle    L12
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
L72:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L72
L12:
  cmpl   $0x1,0xc(%rsp)
  jle    L33
  mov    $0x1,%ebx
  mov    $0x0,%ebp
L20:
  mov    0xc(%rsp,%rbx,4),%edi
  add    %edi,0x10(%rsp,%rbx,4)
  call   <abs@plt>
  add    %rax,%rbp
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L20
  jmp    L35
L32:
  mov    0x61a98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x61aa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D11
datatype: string
name: unpack_data
arguments: label: D12
datatype: string

tool: D11: %d

tool: D12: %lld


assistant: ```c
int main() {
  int i, t;
  int a[100000];
  long long int sum = 0;
  while (1) {
    scanf("%d", &t);
    if (t == 0)
      break;
    for (i = 0; i < t; i++) {
      scanf("%d", &a[i]);
    }
    for (i = 1; i < t; i++) {
      a[i] += a[i - 1];
      sum += abs(a[i - 1]);
    }
    printf("%lld\n", sum);
    sum = 0;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x61000(%rsp),%r11
L8:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L8
  sub    $0xaa8,%rsp
  lea    D9(%rip),%r12
  lea    D5(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x61a98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  nopl   0x0(%rax)
L41:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L70
  jle    L59
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L68:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L68
L59:
  cmp    $0x1,%eax
  jle    L47
  mov    $0x1,%ebx
  xor    %r15d,%r15d
  lea    0x10(%rsp),%rbp
  nopl   0x0(%rax)
L75:
  mov    -0x4(%rbp,%rbx,4),%edi
  add    %edi,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  call   <abs@plt>
  add    %rax,%r15
  cmp    %ebx,0xc(%rsp)
  jg     L75
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L41
  nop
L70:
  mov    0x61a98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x61aa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  xor    %r15d,%r15d
  mov    %r14,%rdi
  mov    %r15,%rsi
  call   <printf@plt>
  jmp    L41
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D9
type_guessed: string
name: disassemble_data
arguments: tag: D5
type_guessed: string

tool: D9: %d

tool: D5: %lld


assistant: ```c
int main() {
  int i, t;
  int a[100000];
  long long int sum = 0;
  while (1) {
    scanf("%d", &t);
    if (t == 0)
      break;
    for (i = 0; i < t; i++) {
      scanf("%d", &a[i]);
    }
    for (i = 1; i < t; i++) {
      a[i] += a[i - 1];
      sum += abs(a[i - 1]);
    }
    printf("%lld\n", sum);
    sum = 0;
  }
  return 0;
}
```