ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L12
L6:
  mov    $0xa,%edi
  call   <putchar@plt>
L12:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D5(%rip),%rdx
  cmp    %al,(%rdx)
  je     L1
  cmp    $0xa,%al
  je     L6
  sub    $0x7,%eax
  movsbl %al,%edi
  call   <putchar@plt>
  jmp    L12
L1:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  char c;
  while ((c = getchar()) != EOF) {
    (c == 10) ? putchar(c) : putchar(c - 7);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    D9(%rip),%rbx
L25:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  je     L28
L34:
  cmp    $0xa,%al
  je     L9
  sub    $0x7,%eax
  movsbl %al,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  jne    L34
L28:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L25
```

assistant: ```c
int main() {
  char c;
  while ((c = getchar()) != EOF) {
    (c == 10) ? putchar(c) : putchar(c - 7);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D2(%rip),%rbx
L38:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  je     L1
L16:
  cmp    $0xa,%al
  je     L28
  sub    $0x7,%eax
  movsbl %al,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  jne    L16
L1:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L28:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L38
```

assistant: ```c
int main() {
  char c;
  while ((c = getchar()) != EOF) {
    (c == 10) ? putchar(c) : putchar(c - 7);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<call_checker>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <check_access@plt>
  cmp    $0x1,%eax
  jne    L12
  mov    $0xffffffff,%eax
  jmp    L56
L12:
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <check_access@plt>
  cmp    $0x1,%eax
  jne    L23
  mov    $0xffffffff,%eax
  jmp    L56
L23:
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pipe_fork@plt>
  cmp    $0xffffffff,%eax
  jne    L0
  mov    $0xffffffff,%eax
  jmp    L56
L0:
  mov    $0x0,%eax
L56:
  leave
  ret
```

assistant: ```c
int call_checker(char **s1, char **path, char **env, char **s2) {
  if ((check_access(s1, path)) == 1)
    return (-1);
  if ((check_access(s2, path)) == 1)
    return (-1);
  if ((pipe_fork(s1, s2, env, path)) == -1)
    return (-1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<call_checker>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L25
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L27
  mov    %rbx,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <pipe_fork@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  mov    $0xffffffff,%eax
  jmp    L30
L27:
  mov    $0xffffffff,%eax
  jmp    L30
```

assistant: ```c
int call_checker(char **s1, char **path, char **env, char **s2) {
  if ((check_access(s1, path)) == 1)
    return (-1);
  if ((check_access(s2, path)) == 1)
    return (-1);
  if ((pipe_fork(s1, s2, env, path)) == -1)
    return (-1);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<call_checker>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L18
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L18
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <pipe_fork@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L0:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L18:
  mov    $0xffffffff,%eax
  jmp    L0
```

assistant: ```c
int call_checker(char **s1, char **path, char **env, char **s2) {
  if ((check_access(s1, path)) == 1)
    return (-1);
  if ((check_access(s2, path)) == 1)
    return (-1);
  if ((pipe_fork(s1, s2, env, path)) == -1)
    return (-1);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<call_checker>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L15
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <check_access@plt>
  cmp    $0x1,%eax
  je     L15
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <pipe_fork@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L21:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L15:
  mov    $0xffffffff,%eax
  jmp    L21
```

assistant: ```c
int call_checker(char **s1, char **path, char **env, char **s2) {
  if ((check_access(s1, path)) == 1)
    return (-1);
  if ((check_access(s2, path)) == 1)
    return (-1);
  if ((pipe_fork(s1, s2, env, path)) == -1)
    return (-1);
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<print_addba_reject_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D35
datatype: string
name: parse_data
arguments: data_label: D1
datatype: string
name: parse_data
arguments: data_label: D10
datatype: string
name: parse_data
arguments: data_label: D2
datatype: string

tool: D35: 
Usage : addbareject <m0> <m1> ... <m7>

tool: D1: 
Options: <mx> : 1 enables rejection of ADDBA request for TidX.

tool: D10: 
                0 would accept any ADDBAs for TidX.

tool: D2: 
         empty - get the addbareject table


assistant: ```c
void print_addba_reject_usage(void) {
  printf("\nUsage : addbareject <m0> <m1> ... <m7>");
  printf("\nOptions: <mx> : 1 enables rejection of ADDBA request for TidX.");
  printf("\n                0 would accept any ADDBAs for TidX.");
  printf("\n         empty - get the addbareject table\n");
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_addba_reject_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
type: string
name: retrieve_data
arguments: label: D28
type: string
name: retrieve_data
arguments: label: D26
type: string
name: retrieve_data
arguments: label: D4
type: string

tool: D0: 
Usage : addbareject <m0> <m1> ... <m7>

tool: D28: 
Options: <mx> : 1 enables rejection of ADDBA request for TidX.

tool: D26: 
                0 would accept any ADDBAs for TidX.

tool: D4: 
         empty - get the addbareject table


assistant: ```c
void print_addba_reject_usage(void) {
  printf("\nUsage : addbareject <m0> <m1> ... <m7>");
  printf("\nOptions: <mx> : 1 enables rejection of ADDBA request for TidX.");
  printf("\n                0 would accept any ADDBAs for TidX.");
  printf("\n         empty - get the addbareject table\n");
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_addba_reject_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D5
type_guessed: string
name: read_data
arguments: label: D28
type_guessed: string
name: read_data
arguments: label: D30
type_guessed: string
name: read_data
arguments: label: D12
type_guessed: string

tool: D5: 
Usage : addbareject <m0> <m1> ... <m7>

tool: D28: 
Options: <mx> : 1 enables rejection of ADDBA request for TidX.

tool: D30: 
                0 would accept any ADDBAs for TidX.

tool: D12: 
         empty - get the addbareject table


assistant: ```c
void print_addba_reject_usage(void) {
  printf("\nUsage : addbareject <m0> <m1> ... <m7>");
  printf("\nOptions: <mx> : 1 enables rejection of ADDBA request for TidX.");
  printf("\n                0 would accept any ADDBAs for TidX.");
  printf("\n         empty - get the addbareject table\n");
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<print_addba_reject_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D18(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D2
datatype: string
name: retrieve_data
arguments: label: D10
datatype: string
name: retrieve_data
arguments: label: D26
datatype: string
name: retrieve_data
arguments: label: D18
datatype: string

tool: D2: 
Usage : addbareject <m0> <m1> ... <m7>

tool: D10: 
Options: <mx> : 1 enables rejection of ADDBA request for TidX.

tool: D26: 
                0 would accept any ADDBAs for TidX.

tool: D18: 
         empty - get the addbareject table


assistant: ```c
void print_addba_reject_usage(void) {
  printf("\nUsage : addbareject <m0> <m1> ... <m7>");
  printf("\nOptions: <mx> : 1 enables rejection of ADDBA request for TidX.");
  printf("\n                0 would accept any ADDBAs for TidX.");
  printf("\n         empty - get the addbareject table\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<base_mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <int_digits@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D6(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: D6: %d

assistant: ```c
char *base_mul(char *x, char *y) {
  int xd, yd, zd;
  int d;
  char *res;
  xd = strtol(x, NULL, 10);
  yd = strtol(y, NULL, 10);
  zd = xd * yd;
  d = int_digits(zd);
  res = calloc(d + 1, sizeof(char));
  snprintf(res, d, "%d", zd);
  res[d] = 0;
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<base_mul>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %eax,%ebp
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <strtol@plt>
  imul   %eax,%ebp
  mov    %ebp,%edi
  call   <int_digits@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    %ebp,%ecx
  lea    D1(%rip),%rdx
  mov    %ebx,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  movslq %ebx,%rbx
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string

tool: D1: %d

assistant: ```c
char *base_mul(char *x, char *y) {
  int xd, yd, zd;
  int d;
  char *res;
  xd = strtol(x, NULL, 10);
  yd = strtol(y, NULL, 10);
  zd = xd * yd;
  d = int_digits(zd);
  res = calloc(d + 1, sizeof(char));
  snprintf(res, d, "%d", zd);
  res[d] = 0;
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<base_mul>:
  endbr64
  push   %r13
  mov    $0xa,%edx
  mov    %rsi,%r13
  xor    %esi,%esi
  push   %r12
  push   %rbx
  call   <strtol@plt>
  mov    $0xa,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  mov    %eax,%r12d
  call   <strtol@plt>
  imul   %eax,%r12d
  mov    %r12d,%edi
  call   <int_digits@plt>
  mov    $0x1,%esi
  movslq %eax,%rbx
  lea    0x1(%rbx),%edi
  call   <calloc@plt>
  mov    %r12d,%ecx
  mov    %ebx,%esi
  lea    D2(%rip),%rdx
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <snprintf@plt>
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    %r13,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type: string

tool: D2: %d

assistant: ```c
char *base_mul(char *x, char *y) {
  int xd, yd, zd;
  int d;
  char *res;
  xd = strtol(x, NULL, 10);
  yd = strtol(y, NULL, 10);
  zd = xd * yd;
  d = int_digits(zd);
  res = calloc(d + 1, sizeof(char));
  snprintf(res, d, "%d", zd);
  res[d] = 0;
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<base_mul>:
  endbr64
  push   %r13
  mov    $0xa,%edx
  mov    %rsi,%r13
  xor    %esi,%esi
  push   %r12
  push   %rbx
  call   <strtol@plt>
  mov    $0xa,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  mov    %eax,%r12d
  call   <strtol@plt>
  imul   %eax,%r12d
  mov    %r12d,%edi
  call   <int_digits@plt>
  mov    $0x1,%esi
  movslq %eax,%rbx
  lea    0x1(%rbx),%edi
  call   <calloc@plt>
  mov    %r12d,%ecx
  mov    %ebx,%esi
  lea    D4(%rip),%rdx
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <snprintf@plt>
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    %r13,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
type: string

tool: D4: %d

assistant: ```c
char *base_mul(char *x, char *y) {
  int xd, yd, zd;
  int d;
  char *res;
  xd = strtol(x, NULL, 10);
  yd = strtol(y, NULL, 10);
  zd = xd * yd;
  d = int_digits(zd);
  res = calloc(d + 1, sizeof(char));
  snprintf(res, d, "%d", zd);
  res[d] = 0;
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L4
  mov    -0x18(%rbp),%eax
  add    $0x2,%eax
  jmp    L19
L4:
  mov    -0x14(%rbp),%eax
  add    $0x3,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int foo(int x, int y) {
  int unknown;
  if (unknown)
    return y + 2;
  return x + 3;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<foo>:
  endbr64
  lea    0x3(%rdi),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int unknown;
  if (unknown)
    return y + 2;
  return x + 3;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo>:
  endbr64
  lea    0x3(%rdi),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int unknown;
  if (unknown)
    return y + 2;
  return x + 3;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  lea    0x3(%rdi),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int unknown;
  if (unknown)
    return y + 2;
  return x + 3;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L55
L9:
  movl   $0x0,-0x78(%rbp)
  movl   $0x1,-0x74(%rbp)
  jmp    L98
L39:
  cmpl   $0x0,-0x74(%rbp)
  je     L96
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isblank@plt>
  test   %rax,%rax
  jne    L96
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  sub    $0x20,%eax
  mov    %eax,%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  movl   $0x0,-0x74(%rbp)
  jmp    L24
L96:
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isblank@plt>
  test   %rax,%rax
  je     L24
  movl   $0x1,-0x74(%rbp)
L24:
  addl   $0x1,-0x78(%rbp)
L98:
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  test   %al,%al
  je     L71
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L39
L71:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L55:
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  lea    -0x70(%rbp),%rax
  mov    $0x65,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L52
  call   <__stack_chk_fail@plt>
L52:
  leave
  ret
```

assistant: ```c
int main() {
  int i, space;
  char in[101];
  while (fgets(in, sizeof(in), stdin)) {
    i = 0;
    space = 1;
    while (in[i] && in[i] != '\n') {
      if (space && !isblank(in[i])) {
        in[i] = in[i] - 32;
        space = 0;
      } else if (isblank(in[i])) {
        space = 1;
      }
      i++;
    }
    printf(in);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    D7(%rip),%r14
  mov    %rsp,%r13
  mov    $0x1,%r12d
  jmp    L11
L67:
  movsbl (%rbx),%edi
  call   <isblank@plt>
  test   %rax,%rax
  cmovne %r12d,%ebp
L36:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L73
  cmp    $0xa,%dil
  je     L73
L46:
  test   %ebp,%ebp
  je     L67
  movsbl %dil,%edi
  call   <isblank@plt>
  test   %rax,%rax
  jne    L67
  subb   $0x20,(%rbx)
  mov    $0x0,%ebp
  jmp    L36
L73:
  mov    %r13,%rdi
  call   <printf@plt>
L11:
  mov    (%r14),%edx
  mov    $0x65,%esi
  mov    %r13,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L30
  movzbl (%rsp),%edi
  test   %dil,%dil
  je     L73
  cmp    $0xa,%dil
  setne  %al
  mov    %rsp,%rbx
  mov    $0x1,%ebp
  test   %al,%al
  jne    L46
  jmp    L73
L30:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, space;
  char in[101];
  while (fgets(in, sizeof(in), stdin)) {
    i = 0;
    space = 1;
    while (in[i] && in[i] != '\n') {
      if (space && !isblank(in[i])) {
        in[i] = in[i] - 32;
        space = 0;
      } else if (isblank(in[i])) {
        space = 1;
      }
      i++;
    }
    printf(in);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x70,%rsp
  mov    D4(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nop
L65:
  mov    0x0(%r13),%edx
  mov    $0x65,%esi
  mov    %r12,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L90
  movsbl (%rsp),%edi
  test   %dil,%dil
  je     L99
  mov    %r12,%rbx
  mov    $0x1,%r14d
  jmp    L87
  nopw   0x0(%rax,%rax,1)
L69:
  call   <isblank@plt>
  test   %rax,%rax
  jne    L33
  movsbl 0x1(%rbx),%edi
  subb   $0x20,(%rbx)
  xor    %r14d,%r14d
  add    $0x1,%rbx
  test   %dil,%dil
  je     L99
L87:
  cmp    $0xa,%dil
  je     L99
  test   %r14d,%r14d
  jne    L69
L33:
  movsbl (%rbx),%edi
  call   <isblank@plt>
  movsbl 0x1(%rbx),%edi
  test   %rax,%rax
  cmovne %ebp,%r14d
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L87
L99:
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L65
  xchg   %ax,%ax
L90:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, space;
  char in[101];
  while (fgets(in, sizeof(in), stdin)) {
    i = 0;
    space = 1;
    while (in[i] && in[i] != '\n') {
      if (space && !isblank(in[i])) {
        in[i] = in[i] - 32;
        space = 0;
      } else if (isblank(in[i])) {
        space = 1;
      }
      i++;
    }
    printf(in);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x70,%rsp
  mov    D7(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nop
L95:
  mov    (%r12),%edx
  mov    $0x65,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L77
  movsbl (%rsp),%edi
  test   %dil,%dil
  je     L64
  mov    %rbp,%r13
  mov    $0x1,%r14d
  jmp    L71
  nopw   0x0(%rax,%rax,1)
L52:
  movsbl 0x0(%r13),%edi
L72:
  call   <isblank@plt>
  movsbl 0x1(%r13),%edi
  test   %rax,%rax
  cmovne %ebx,%r14d
  add    $0x1,%r13
  test   %dil,%dil
  je     L64
L71:
  cmp    $0xa,%dil
  je     L64
  test   %r14d,%r14d
  je     L72
  call   <isblank@plt>
  test   %rax,%rax
  jne    L52
  movsbl 0x1(%r13),%edi
  subb   $0x20,0x0(%r13)
  xor    %r14d,%r14d
  add    $0x1,%r13
  test   %dil,%dil
  jne    L71
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L95
  nopl   (%rax)
L77:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, space;
  char in[101];
  while (fgets(in, sizeof(in), stdin)) {
    i = 0;
    space = 1;
    while (in[i] && in[i] != '\n') {
      if (space && !isblank(in[i])) {
        in[i] = in[i] - 32;
        space = 0;
      } else if (isblank(in[i])) {
        space = 1;
      }
      i++;
    }
    printf(in);
  }
  return 0;
}
```