ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<eatspace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
L7:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0x4(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L28
  cmpl   $0xa,-0x4(%rbp)
  jne    L57
  movl   $0x1,-0x8(%rbp)
L57:
  cmpl   $0x23,-0x4(%rbp)
  jne    L43
  cmpl   $0x0,-0x1c(%rbp)
  je     L43
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <eatline@plt>
  movl   $0xa,-0x4(%rbp)
  jmp    L37
L43:
  movl   $0x0,-0x1c(%rbp)
L37:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L7
  jmp    L13
L28:
  nop
L13:
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L53
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
L53:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int eatspace(FILE *in, int sol) {
  int nl = 0;
  int c;
  do {
    c = getc(in);
    if (c == EOF)
      break;
    if (c == '\n')
      nl = 1;
    if (c == '#' && sol) {
      eatline(in);
      c = '\n';
      continue;
    }
    sol = 0;
  } while (isspace(c));
  if (c != EOF)
    ungetc(c, in);
  return nl;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<eatspace>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    $0x0,%r13d
  mov    $0x1,%r14d
  jmp    L26
L47:
  mov    %r14d,%r13d
  mov    $0x0,%r12d
  jmp    L24
L89:
  mov    $0x0,%r12d
L24:
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L19
L26:
  mov    %rbp,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  mov    D1(%rip),%rax
  cmp    %ebx,(%rax)
  je     L65
  cmp    $0xa,%ebx
  je     L47
  cmp    $0x23,%ebx
  jne    L89
  test   %r12d,%r12d
  je     L89
  mov    %rbp,%rdi
  call   <eatline@plt>
  mov    $0xa,%ebx
  jmp    L24
L19:
  mov    D1(%rip),%rax
  cmp    %ebx,(%rax)
  je     L65
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <ungetc@plt>
L65:
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int eatspace(FILE *in, int sol) {
  int nl = 0;
  int c;
  do {
    c = getc(in);
    if (c == EOF)
      break;
    if (c == '\n')
      nl = 1;
    if (c == '#' && sol) {
      eatline(in);
      c = '\n';
      continue;
    }
    sol = 0;
  } while (isspace(c));
  if (c != EOF)
    ungetc(c, in);
  return nl;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eatspace>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D3(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax,%rax,1)
L73:
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebp
  cmp    %eax,0x0(%r13)
  je     L44
  cmp    $0xa,%eax
  je     L5
  cmp    $0x23,%eax
  jne    L49
  test   %ebx,%ebx
  jne    L75
L49:
  xor    %ebx,%ebx
L22:
  mov    %ebp,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L73
  cmp    %ebp,0x0(%r13)
  je     L44
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <ungetc@plt>
L44:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L75:
  mov    %r12,%rdi
  mov    $0xa,%ebp
  call   <eatline@plt>
  jmp    L22
  nop
L5:
  mov    $0x1,%r14d
  xor    %ebx,%ebx
  jmp    L22
```

assistant: ```c
int eatspace(FILE *in, int sol) {
  int nl = 0;
  int c;
  do {
    c = getc(in);
    if (c == EOF)
      break;
    if (c == '\n')
      nl = 1;
    if (c == '#' && sol) {
      eatline(in);
      c = '\n';
      continue;
    }
    sol = 0;
  } while (isspace(c));
  if (c != EOF)
    ungetc(c, in);
  return nl;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<eatspace>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D6(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax,%rax,1)
L59:
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebp
  cmp    %eax,0x0(%r13)
  je     L47
  cmp    $0xa,%eax
  je     L4
  cmp    $0x23,%eax
  jne    L49
  test   %ebx,%ebx
  jne    L53
L49:
  xor    %ebx,%ebx
L67:
  mov    %ebp,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L59
  cmp    %ebp,0x0(%r13)
  je     L47
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <ungetc@plt>
L47:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %r12,%rdi
  mov    $0xa,%ebp
  call   <eatline@plt>
  jmp    L67
  nop
L4:
  mov    $0x1,%r14d
  xor    %ebx,%ebx
  jmp    L67
```

assistant: ```c
int eatspace(FILE *in, int sol) {
  int nl = 0;
  int c;
  do {
    c = getc(in);
    if (c == EOF)
      break;
    if (c == '\n')
      nl = 1;
    if (c == '#' && sol) {
      eatline(in);
      c = '\n';
      continue;
    }
    sol = 0;
  } while (isspace(c));
  if (c != EOF)
    ungetc(c, in);
  return nl;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_n>:
  endbr64
  cmp    $0x4,%rdi
  ja     L26
  sub    $0x8,%rsp
  lea    D8(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <_n_0@plt>
L12:
  add    $0x8,%rsp
  ret
  mov    %rsi,%rdi
  call   <_n_1@plt>
  jmp    L12
  mov    %rsi,%rdi
  call   <_n_2@plt>
  jmp    L12
  mov    %rsi,%rdi
  call   <_n_3@plt>
  jmp    L12
  mov    $0x0,%eax
  call   <_n_4@plt>
  jmp    L12
L26:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double _n(const long i, const double omega, const double vertices_omegas[4]) {
  switch (i) {
  case 0:
    return _n_0();
  case 1:
    return _n_1(omega, vertices_omegas);
  case 2:
    return _n_2(omega, vertices_omegas);
  case 3:
    return _n_3(omega, vertices_omegas);
  case 4:
    return _n_4();
  };
  ;
  ;
  ;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_n>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  cmp    $0x4,%rax
  ja     L6
  lea    D1(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  jmp    10b0 <_n_3@plt>
  nopl   (%rax)
  xor    %eax,%eax
  jmp    10c0 <_n_4@plt>
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  jmp    10d0 <_n_0@plt>
  nopw   0x0(%rax,%rax,1)
  jmp    10a0 <_n_1@plt>
  nopl   (%rax)
  jmp    1090 <_n_2@plt>
L6:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double _n(const long i, const double omega, const double vertices_omegas[4]) {
  switch (i) {
  case 0:
    return _n_0();
  case 1:
    return _n_1(omega, vertices_omegas);
  case 2:
    return _n_2(omega, vertices_omegas);
  case 3:
    return _n_3(omega, vertices_omegas);
  case 4:
    return _n_4();
  };
  ;
  ;
  ;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_n>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  cmp    $0x4,%rax
  ja     L9
  lea    D3(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  jmp    10b0 <_n_3@plt>
  nopl   (%rax)
  xor    %eax,%eax
  jmp    10c0 <_n_4@plt>
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  jmp    10d0 <_n_0@plt>
  nopw   0x0(%rax,%rax,1)
  jmp    10a0 <_n_1@plt>
  nopl   (%rax)
  jmp    1090 <_n_2@plt>
L9:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double _n(const long i, const double omega, const double vertices_omegas[4]) {
  switch (i) {
  case 0:
    return _n_0();
  case 1:
    return _n_1(omega, vertices_omegas);
  case 2:
    return _n_2(omega, vertices_omegas);
  case 3:
    return _n_3(omega, vertices_omegas);
  case 4:
    return _n_4();
  };
  ;
  ;
  ;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<timeout_ms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%eax
  imul   $0x3e8,%eax,%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    %ecx,%eax
  cltq
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
unsigned long timeout_ms(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (tv.tv_sec * 1000) + (tv.tv_usec / 1000);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timeout_ms>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0x3e8,(%rsp),%eax
  mov    0x4(%rsp),%ecx
  movslq %ecx,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  cltq
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long timeout_ms(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (tv.tv_sec * 1000) + (tv.tv_usec / 1000);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<timeout_ms>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq 0x4(%rsp),%rdx
  imul   $0x3e8,(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  cltq
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long timeout_ms(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (tv.tv_sec * 1000) + (tv.tv_usec / 1000);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<timeout_ms>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq 0x4(%rsp),%rdx
  imul   $0x3e8,(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  cltq
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long timeout_ms(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (tv.tv_sec * 1000) + (tv.tv_usec / 1000);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<replaceBlank>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L37:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L43
  addl   $0x1,-0x8(%rbp)
L43:
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L45
L54:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L35
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  subl   $0x1,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x32,(%rax)
  subl   $0x1,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x25,(%rax)
  subl   $0x1,-0xc(%rbp)
L35:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  lea    -0x1(%rdx),%ecx
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0xc(%rbp)
L45:
  cmpl   $0x0,-0x4(%rbp)
  jg     L54
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *replaceBlank(char string[], int length) {
  int count = 0;
  int spaceNumbers = 0;
  while (string[count] != '\0') {
    if (string[count] == ' ') {
      spaceNumbers += 1;
    }
    count++;
  }
  int movePos = 2 * spaceNumbers;
  string[count + movePos] = '\0';
  while (movePos > 0) {
    if (string[count - 1] == ' ') {
      string[count - 1 + movePos] = '0';
      movePos--;
      string[count - 1 + movePos] = '2';
      movePos--;
      string[count - 1 + movePos] = '%';
      count--;
    }
    string[count - 1 + movePos] = string[count - 1];
    count--;
  }
  return &string[0];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<replaceBlank>:
  endbr64
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L48
  mov    $0x1,%ecx
  mov    $0x0,%edx
L34:
  cmp    $0x20,%sil
  sete   %al
  movzbl %al,%eax
  add    %eax,%edx
  mov    %ecx,%eax
  add    $0x1,%rcx
  movzbl -0x1(%rdi,%rcx,1),%esi
  test   %sil,%sil
  jne    L34
  add    %edx,%edx
  lea    (%rdx,%rax,1),%ecx
  movslq %ecx,%rcx
  movb   $0x0,(%rdi,%rcx,1)
  test   %edx,%edx
  jg     L35
L48:
  mov    %rdi,%rax
  ret
L24:
  movslq %eax,%rcx
  sub    $0x1,%eax
  movzbl -0x1(%rdi,%rcx,1),%esi
  lea    (%rax,%rdx,1),%ecx
  movslq %ecx,%rcx
  mov    %sil,(%rdi,%rcx,1)
  test   %edx,%edx
  jle    L48
L35:
  movslq %eax,%rcx
  cmpb   $0x20,-0x1(%rdi,%rcx,1)
  jne    L24
  sub    $0x1,%eax
  lea    (%rax,%rdx,1),%ecx
  movslq %ecx,%rcx
  movb   $0x30,(%rdi,%rcx,1)
  lea    -0x1(%rax,%rdx,1),%ecx
  movslq %ecx,%rcx
  movb   $0x32,(%rdi,%rcx,1)
  sub    $0x2,%edx
  lea    (%rax,%rdx,1),%ecx
  movslq %ecx,%rcx
  movb   $0x25,(%rdi,%rcx,1)
  jmp    L24
```

assistant: ```c
char *replaceBlank(char string[], int length) {
  int count = 0;
  int spaceNumbers = 0;
  while (string[count] != '\0') {
    if (string[count] == ' ') {
      spaceNumbers += 1;
    }
    count++;
  }
  int movePos = 2 * spaceNumbers;
  string[count + movePos] = '\0';
  while (movePos > 0) {
    if (string[count - 1] == ' ') {
      string[count - 1 + movePos] = '0';
      movePos--;
      string[count - 1 + movePos] = '2';
      movePos--;
      string[count - 1 + movePos] = '%';
      count--;
    }
    string[count - 1 + movePos] = string[count - 1];
    count--;
  }
  return &string[0];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<replaceBlank>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%r8
  test   %cl,%cl
  je     L24
  mov    $0x1,%edx
  xor    %esi,%esi
  nopl   0x0(%rax)
L22:
  xor    %eax,%eax
  cmp    $0x20,%cl
  sete   %al
  add    %eax,%esi
  mov    %edx,%eax
  add    $0x1,%rdx
  movzbl -0x1(%r8,%rdx,1),%ecx
  test   %cl,%cl
  jne    L22
  lea    (%rsi,%rsi,1),%edx
  lea    (%rdx,%rax,1),%ecx
  movslq %ecx,%rcx
  movb   $0x0,(%r8,%rcx,1)
  test   %esi,%esi
  je     L24
L4:
  movslq %eax,%rsi
  movzbl -0x1(%r8,%rsi,1),%ecx
  cmp    $0x20,%cl
  je     L37
L52:
  sub    $0x1,%eax
  lea    (%rdx,%rax,1),%esi
  movslq %esi,%rsi
  mov    %cl,(%r8,%rsi,1)
  movslq %eax,%rsi
  movzbl -0x1(%r8,%rsi,1),%ecx
  cmp    $0x20,%cl
  jne    L52
L37:
  lea    -0x1(%rax),%ecx
  sub    $0x2,%eax
  lea    (%rdx,%rcx,1),%edi
  movslq %edi,%rdi
  movb   $0x30,(%r8,%rdi,1)
  lea    -0x1(%rdx,%rcx,1),%edi
  sub    $0x2,%edx
  add    %edx,%ecx
  movslq %edi,%rdi
  movslq %ecx,%rcx
  movb   $0x32,(%r8,%rdi,1)
  movb   $0x25,(%r8,%rcx,1)
  lea    (%rax,%rdx,1),%ecx
  movzbl -0x2(%r8,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  test   %edx,%edx
  jg     L4
L24:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *replaceBlank(char string[], int length) {
  int count = 0;
  int spaceNumbers = 0;
  while (string[count] != '\0') {
    if (string[count] == ' ') {
      spaceNumbers += 1;
    }
    count++;
  }
  int movePos = 2 * spaceNumbers;
  string[count + movePos] = '\0';
  while (movePos > 0) {
    if (string[count - 1] == ' ') {
      string[count - 1 + movePos] = '0';
      movePos--;
      string[count - 1 + movePos] = '2';
      movePos--;
      string[count - 1 + movePos] = '%';
      count--;
    }
    string[count - 1 + movePos] = string[count - 1];
    count--;
  }
  return &string[0];
}
```
user: What is the c source code of the assembly code below?

```asm
<replaceBlank>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%r8
  test   %cl,%cl
  je     L3
  mov    $0x1,%edx
  xor    %esi,%esi
  nopl   0x0(%rax)
L9:
  xor    %eax,%eax
  cmp    $0x20,%cl
  sete   %al
  add    %eax,%esi
  mov    %edx,%eax
  add    $0x1,%rdx
  movzbl -0x1(%r8,%rdx,1),%ecx
  test   %cl,%cl
  jne    L9
  lea    (%rsi,%rsi,1),%edx
  lea    (%rdx,%rax,1),%ecx
  movslq %ecx,%rcx
  movb   $0x0,(%r8,%rcx,1)
  test   %esi,%esi
  je     L3
L48:
  movslq %eax,%rsi
  movzbl -0x1(%r8,%rsi,1),%ecx
  cmp    $0x20,%cl
  je     L56
L19:
  sub    $0x1,%eax
  lea    (%rax,%rdx,1),%esi
  movslq %esi,%rsi
  mov    %cl,(%r8,%rsi,1)
  movslq %eax,%rsi
  movzbl -0x1(%r8,%rsi,1),%ecx
  cmp    $0x20,%cl
  jne    L19
L56:
  lea    -0x1(%rax),%ecx
  sub    $0x2,%eax
  lea    (%rdx,%rcx,1),%edi
  movslq %edi,%rdi
  movb   $0x30,(%r8,%rdi,1)
  lea    -0x1(%rdx,%rcx,1),%edi
  sub    $0x2,%edx
  add    %edx,%ecx
  movslq %edi,%rdi
  movslq %ecx,%rcx
  movb   $0x32,(%r8,%rdi,1)
  movb   $0x25,(%r8,%rcx,1)
  lea    (%rdx,%rax,1),%ecx
  movzbl -0x2(%r8,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r8,%rcx,1)
  test   %edx,%edx
  jg     L48
L3:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *replaceBlank(char string[], int length) {
  int count = 0;
  int spaceNumbers = 0;
  while (string[count] != '\0') {
    if (string[count] == ' ') {
      spaceNumbers += 1;
    }
    count++;
  }
  int movePos = 2 * spaceNumbers;
  string[count + movePos] = '\0';
  while (movePos > 0) {
    if (string[count - 1] == ' ') {
      string[count - 1 + movePos] = '0';
      movePos--;
      string[count - 1 + movePos] = '2';
      movePos--;
      string[count - 1 + movePos] = '%';
      count--;
    }
    string[count - 1 + movePos] = string[count - 1];
    count--;
  }
  return &string[0];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f_empty_while_loop_prefix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <prefix@plt>
L4:
  jmp    L4
```

assistant: ```c
void f_empty_while_loop_prefix() {
  prefix();
  while (1) {
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f_empty_while_loop_prefix>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <prefix@plt>
L4:
  jmp    L4
```

assistant: ```c
void f_empty_while_loop_prefix() {
  prefix();
  while (1) {
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f_empty_while_loop_prefix>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <prefix@plt>
L7:
  jmp    L7
```

assistant: ```c
void f_empty_while_loop_prefix() {
  prefix();
  while (1) {
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f_empty_while_loop_prefix>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <prefix@plt>
L2:
  jmp    L2
```

assistant: ```c
void f_empty_while_loop_prefix() {
  prefix();
  while (1) {
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<CountCharacters>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L1:
  mov    D4(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x5(%rbp)
  je     L32
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %al,-0x5(%rbp)
  cmpb   $0xa,-0x5(%rbp)
  jne    L1
  jmp    L31
L32:
  nop
L31:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <rewind@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int CountCharacters(FILE *InputFile) {
  int counter = 0;
  char c;
  while ((c = fgetc(InputFile)) != ('\n')) {
    if (c == EOF)
      break;
    counter++;
  }
  rewind(InputFile);
  return counter;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CountCharacters>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  jmp    L12
L26:
  add    $0x1,%ebx
L12:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%al
  je     L18
  mov    D6(%rip),%rdx
  cmp    %al,(%rdx)
  jne    L26
L18:
  mov    %rbp,%rdi
  call   <rewind@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int CountCharacters(FILE *InputFile) {
  int counter = 0;
  char c;
  while ((c = fgetc(InputFile)) != ('\n')) {
    if (c == EOF)
      break;
    counter++;
  }
  rewind(InputFile);
  return counter;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CountCharacters>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L1:
  mov    D0(%rip),%rdx
  cmp    %al,(%rdx)
  je     L13
  add    $0x1,%r12d
L26:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%al
  jne    L1
L13:
  mov    %rbp,%rdi
  call   <rewind@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int CountCharacters(FILE *InputFile) {
  int counter = 0;
  char c;
  while ((c = fgetc(InputFile)) != ('\n')) {
    if (c == EOF)
      break;
    counter++;
  }
  rewind(InputFile);
  return counter;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<CountCharacters>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L0:
  mov    D3(%rip),%rdx
  cmp    %al,(%rdx)
  je     L9
  add    $0x1,%r12d
L23:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%al
  jne    L0
L9:
  mov    %rbp,%rdi
  call   <rewind@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int CountCharacters(FILE *InputFile) {
  int counter = 0;
  char c;
  while ((c = fgetc(InputFile)) != ('\n')) {
    if (c == EOF)
      break;
    counter++;
  }
  rewind(InputFile);
  return counter;
}
```
user: What is the c source code of the assembly code below?

```asm
<fun1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L8
L19:
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    (%rax),%eax
  imul   %ecx,%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  ja     L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fun1(int *x, int *y, unsigned int n) {
  int i, j;
  int dot = 0;
  for (i = 0; i < n; i++)
    dot += *(x++) * *(y++);
  return dot;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fun1>:
  endbr64
  test   %edx,%edx
  je     L7
  mov    %edx,%edx
  mov    $0x0,%eax
  mov    $0x0,%r8d
L28:
  mov    (%rdi,%rax,4),%ecx
  imul   (%rsi,%rax,4),%ecx
  add    %ecx,%r8d
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L28
L6:
  mov    %r8d,%eax
  ret
L7:
  mov    $0x0,%r8d
  jmp    L6
```

assistant: ```c
int fun1(int *x, int *y, unsigned int n) {
  int i, j;
  int dot = 0;
  for (i = 0; i < n; i++)
    dot += *(x++) * *(y++);
  return dot;
}
```