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

```asm
<skip>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    (%rdi),%r12
  mov    %r12,%rdi
  call   <strcspn@plt>
  movslq %eax,%rbx
  add    %r12,%rbx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  cltq
  add    %rbx,%rax
  cmp    %rax,%rbx
  jae    L18
L15:
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
  cmp    %rbx,%rax
  jne    L15
L18:
  mov    %rax,0x0(%rbp)
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *skip(char **buf, const char *delimiters) {
  char *p, *begin_word, *end_word, *end_delimiters;
  begin_word = *buf;
  end_word = begin_word + strcspn(begin_word, delimiters);
  end_delimiters = end_word + strspn(end_word, delimiters);
  for (p = end_word; p < end_delimiters; p++) {
    *p = '\0';
  }
  *buf = end_delimiters;
  return begin_word;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<skip>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strcspn@plt>
  mov    %r12,%rsi
  movslq %eax,%rbp
  add    %r13,%rbp
  mov    %rbp,%rdi
  call   <strspn@plt>
  movslq %eax,%rdx
  lea    0x0(%rbp,%rdx,1),%r12
  cmp    %r12,%rbp
  jae    L3
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L3:
  mov    %r12,(%rbx)
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *skip(char **buf, const char *delimiters) {
  char *p, *begin_word, *end_word, *end_delimiters;
  begin_word = *buf;
  end_word = begin_word + strcspn(begin_word, delimiters);
  end_delimiters = end_word + strspn(end_word, delimiters);
  for (p = end_word; p < end_delimiters; p++) {
    *p = '\0';
  }
  *buf = end_delimiters;
  return begin_word;
}
```
user: Translate the below assembly code into C programming language:

```asm
<skip>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strcspn@plt>
  mov    %r12,%rsi
  movslq %eax,%rbp
  add    %r13,%rbp
  mov    %rbp,%rdi
  call   <strspn@plt>
  movslq %eax,%rdx
  lea    0x0(%rbp,%rdx,1),%r12
  cmp    %r12,%rbp
  jae    L4
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L4:
  mov    %r12,(%rbx)
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *skip(char **buf, const char *delimiters) {
  char *p, *begin_word, *end_word, *end_delimiters;
  begin_word = *buf;
  end_word = begin_word + strcspn(begin_word, delimiters);
  end_delimiters = end_word + strspn(end_word, delimiters);
  for (p = end_word; p < end_delimiters; p++) {
    *p = '\0';
  }
  *buf = end_delimiters;
  return begin_word;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<wavnfreq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  movsd  %xmm0,-0x28(%rbp)
  mov    %edi,-0x2c(%rbp)
  mov    %esi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %r9,-0x50(%rbp)
  mov    -0x40(%rbp),%rbx
  mov    -0x48(%rbp),%r12
  mov    -0x50(%rbp),%r14
  mov    $0x0,%r13d
  jmp    L0
L4:
  movsd  (%rbx),%xmm1
  movsd  D5(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r12)
  mov    %r14,%rax
  lea    0x4(%rax),%r14
  movl   $0x0,(%rax)
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rbx
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%r12
  add    $0x1,%r13d
L0:
  cmp    -0x2c(%rbp),%r13d
  jl     L4
  mov    $0x0,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type_infered: f64

tool: D5: 299792458

assistant: ```c
int wavnfreq(double dummy, int nwavn, int swavn, int sfreq, const double wavn[], double freq[], int stat[]) {
  register int iwavn, *statp;
  register const double *wavnp;
  register double *freqp;
  (void)dummy;
  wavnp = wavn;
  freqp = freq;
  statp = stat;
  for (iwavn = 0; iwavn < nwavn; iwavn++) {
    *freqp = (*wavnp) * 2.99792458e8;
    *(statp++) = 0;
    wavnp += swavn;
    freqp += sfreq;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wavnfreq>:
  endbr64
  mov    %rcx,%rax
  test   %edi,%edi
  jle    L9
  movslq %esi,%rsi
  lea    0x0(,%rsi,8),%rcx
  movslq %edx,%rdx
  shl    $0x3,%rdx
  mov    %edi,%edi
  lea    (%r9,%rdi,4),%rsi
  movsd  D8(%rip),%xmm1
L12:
  movapd %xmm1,%xmm0
  mulsd  (%rax),%xmm0
  movsd  %xmm0,(%r8)
  add    $0x4,%r9
  movl   $0x0,-0x4(%r9)
  add    %rcx,%rax
  add    %rdx,%r8
  cmp    %rsi,%r9
  jne    L12
L9:
  mov    $0x0,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type: f64

tool: D8: 299792458

assistant: ```c
int wavnfreq(double dummy, int nwavn, int swavn, int sfreq, const double wavn[], double freq[], int stat[]) {
  register int iwavn, *statp;
  register const double *wavnp;
  register double *freqp;
  (void)dummy;
  wavnp = wavn;
  freqp = freq;
  statp = stat;
  for (iwavn = 0; iwavn < nwavn; iwavn++) {
    *freqp = (*wavnp) * 2.99792458e8;
    *(statp++) = 0;
    wavnp += swavn;
    freqp += sfreq;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<wavnfreq>:
  endbr64
  mov    %rcx,%rax
  test   %edi,%edi
  jle    L7
  movslq %esi,%rsi
  movslq %edx,%rdx
  movslq %edi,%rdi
  movsd  D5(%rip),%xmm1
  lea    0x0(,%rsi,8),%rcx
  shl    $0x3,%rdx
  lea    (%r9,%rdi,4),%rsi
  nopl   0x0(%rax)
L0:
  movsd  (%rax),%xmm0
  add    $0x4,%r9
  add    %rcx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8)
  add    %rdx,%r8
  movl   $0x0,-0x4(%r9)
  cmp    %rsi,%r9
  jne    L0
L7:
  xor    %eax,%eax
  ret
```

tool_calls: name: read_data
arguments: label: D5
data_type: f64

tool: D5: 299792458

assistant: ```c
int wavnfreq(double dummy, int nwavn, int swavn, int sfreq, const double wavn[], double freq[], int stat[]) {
  register int iwavn, *statp;
  register const double *wavnp;
  register double *freqp;
  (void)dummy;
  wavnp = wavn;
  freqp = freq;
  statp = stat;
  for (iwavn = 0; iwavn < nwavn; iwavn++) {
    *freqp = (*wavnp) * 2.99792458e8;
    *(statp++) = 0;
    wavnp += swavn;
    freqp += sfreq;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wavnfreq>:
  endbr64
  mov    %esi,%r11d
  mov    %edx,%r10d
  mov    %rcx,%rsi
  test   %edi,%edi
  jle    L25
  movslq %r11d,%rax
  movslq %edx,%rdx
  movslq %edi,%rcx
  movsd  D4(%rip),%xmm1
  shl    $0x3,%rax
  shl    $0x3,%rdx
  lea    (%r9,%rcx,4),%rcx
  cmp    $0x1,%r10d
  jne    L27
  cmp    $0x1,%r11d
  jne    L27
  nopw   0x0(%rax,%rax,1)
L32:
  movsd  (%rsi),%xmm0
  add    $0x4,%r9
  add    %rax,%rsi
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8)
  add    %rdx,%r8
  movl   $0x0,-0x4(%r9)
  cmp    %rcx,%r9
  jne    L32
L25:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  movsd  (%rsi),%xmm0
  add    $0x4,%r9
  add    %rax,%rsi
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8)
  add    %rdx,%r8
  movl   $0x0,-0x4(%r9)
  cmp    %rcx,%r9
  jne    L27
  jmp    L25
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type: f64

tool: D4: 299792458

assistant: ```c
int wavnfreq(double dummy, int nwavn, int swavn, int sfreq, const double wavn[], double freq[], int stat[]) {
  register int iwavn, *statp;
  register const double *wavnp;
  register double *freqp;
  (void)dummy;
  wavnp = wavn;
  freqp = freq;
  statp = stat;
  for (iwavn = 0; iwavn < nwavn; iwavn++) {
    *freqp = (*wavnp) * 2.99792458e8;
    *(statp++) = 0;
    wavnp += swavn;
    freqp += sfreq;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<calc_precedence>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movsbl -0x14(%rbp),%eax
  cmp    $0x5e,%eax
  je     L127
  cmp    $0x5e,%eax
  jg     L57
  cmp    $0x2f,%eax
  je     L75
  cmp    $0x2f,%eax
  jg     L57
  cmp    $0x2d,%eax
  je     L3
  cmp    $0x2d,%eax
  jg     L57
  cmp    $0x2a,%eax
  je     L73
  cmp    $0x2b,%eax
  je     L80
  jmp    L57
L127:
  movl   $0x1,-0x4(%rbp)
  jmp    L57
L75:
  movl   $0x2,-0x4(%rbp)
  jmp    L57
L73:
  movl   $0x3,-0x4(%rbp)
  jmp    L57
L3:
  movl   $0x4,-0x4(%rbp)
  jmp    L57
L80:
  movl   $0x5,-0x4(%rbp)
  nop
L57:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int calc_precedence(char operator) {
  int returnPrecedence = 0;
  switch (operator) {
  case '^':
    returnPrecedence = 1;
    break;
  case '/':
    returnPrecedence = 2;
    break;
  case '*':
    returnPrecedence = 3;
    break;
  case '-':
    returnPrecedence = 4;
    break;
  case '+':
    returnPrecedence = 5;
    break;
  }
  return returnPrecedence;
}
```
user: Translate the below assembly code into C programming language:

```asm
<calc_precedence>:
  endbr64
  cmp    $0x2d,%dil
  je     L13
  jle    L7
  mov    $0x2,%eax
  cmp    $0x2f,%dil
  je     L32
  cmp    $0x5e,%dil
  sete   %al
  movzbl %al,%eax
  ret
L7:
  mov    $0x3,%eax
  cmp    $0x2a,%dil
  je     L32
  cmp    $0x2b,%dil
  sete   %al
  movzbl %al,%eax
  lea    (%rax,%rax,4),%eax
  ret
L13:
  mov    $0x4,%eax
L32:
  ret
```

assistant: ```c
int calc_precedence(char operator) {
  int returnPrecedence = 0;
  switch (operator) {
  case '^':
    returnPrecedence = 1;
    break;
  case '/':
    returnPrecedence = 2;
    break;
  case '*':
    returnPrecedence = 3;
    break;
  case '-':
    returnPrecedence = 4;
    break;
  case '+':
    returnPrecedence = 5;
    break;
  }
  return returnPrecedence;
}
```
user: What is the c source code of the assembly code below:

```asm
<calc_precedence>:
  endbr64
  cmp    $0x2d,%dil
  je     L17
  jle    L39
  mov    $0x2,%eax
  cmp    $0x2f,%dil
  je     L21
  xor    %eax,%eax
  cmp    $0x5e,%dil
  sete   %al
  ret
  nopl   0x0(%rax)
L39:
  mov    $0x3,%eax
  cmp    $0x2a,%dil
  je     L21
  xor    %eax,%eax
  cmp    $0x2b,%dil
  sete   %al
  lea    (%rax,%rax,4),%eax
  ret
L17:
  mov    $0x4,%eax
L21:
  ret
```

assistant: ```c
int calc_precedence(char operator) {
  int returnPrecedence = 0;
  switch (operator) {
  case '^':
    returnPrecedence = 1;
    break;
  case '/':
    returnPrecedence = 2;
    break;
  case '*':
    returnPrecedence = 3;
    break;
  case '-':
    returnPrecedence = 4;
    break;
  case '+':
    returnPrecedence = 5;
    break;
  }
  return returnPrecedence;
}
```
user: What is the c source code of the assembly code below:

```asm
<calc_precedence>:
  endbr64
  cmp    $0x2d,%dil
  je     L13
  jle    L37
  mov    $0x2,%eax
  cmp    $0x2f,%dil
  je     L7
  xor    %eax,%eax
  cmp    $0x5e,%dil
  sete   %al
  ret
  nopl   0x0(%rax)
L37:
  mov    $0x3,%eax
  cmp    $0x2a,%dil
  je     L7
  xor    %eax,%eax
  cmp    $0x2b,%dil
  sete   %al
  lea    (%rax,%rax,4),%eax
  ret
L13:
  mov    $0x4,%eax
L7:
  ret
```

assistant: ```c
int calc_precedence(char operator) {
  int returnPrecedence = 0;
  switch (operator) {
  case '^':
    returnPrecedence = 1;
    break;
  case '/':
    returnPrecedence = 2;
    break;
  case '*':
    returnPrecedence = 3;
    break;
  case '-':
    returnPrecedence = 4;
    break;
  case '+':
    returnPrecedence = 5;
    break;
  }
  return returnPrecedence;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<destroy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void destroy(int **distance, int i, int j) {
  distance[i][j] = 0;
  distance[j][i] = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<destroy>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  mov    (%rdi,%rsi,8),%rax
  movl   $0x0,(%rax,%rdx,4)
  mov    (%rdi,%rdx,8),%rax
  movl   $0x0,(%rax,%rsi,4)
  ret
```

assistant: ```c
void destroy(int **distance, int i, int j) {
  distance[i][j] = 0;
  distance[j][i] = 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<destroy>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  mov    (%rdi,%rsi,8),%rax
  movl   $0x0,(%rax,%rdx,4)
  mov    (%rdi,%rdx,8),%rax
  movl   $0x0,(%rax,%rsi,4)
  ret
```

assistant: ```c
void destroy(int **distance, int i, int j) {
  distance[i][j] = 0;
  distance[j][i] = 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<cosh>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  movsd  %xmm0,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L1
L7:
  mov    -0x1c(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <dpow@plt>
  movq   %xmm0,%rbx
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edi
  call   <fact@plt>
  movq   %rbx,%xmm1
  divsd  %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L1:
  cmpl   $0x1e,-0x1c(%rbp)
  jle    L7
  movsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
double cosh(double n) {
  double sum = 0;
  int i;
  for (i = 0; i <= 30; i++) {
    sum += dpow(n, 2 * i) / fact(2 * i);
  }
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<cosh>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movq   %xmm0,%r12
  mov    $0x0,%ebp
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
L9:
  mov    %ebp,%edi
  movq   %r12,%xmm0
  call   <dpow@plt>
  movq   %xmm0,%rbx
  mov    %ebp,%edi
  call   <fact@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm0
  divsd  %xmm1,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  add    $0x2,%ebp
  cmp    $0x3e,%ebp
  jne    L9
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double cosh(double n) {
  double sum = 0;
  int i;
  for (i = 0; i <= 30; i++) {
    sum += dpow(n, 2 * i) / fact(2 * i);
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cosh>:
  endbr64
  push   %r14
  pxor   %xmm4,%xmm4
  movq   %xmm0,%r14
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  movsd  %xmm4,(%rsp)
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %ebx,%edi
  movq   %r14,%xmm0
  call   <dpow@plt>
  mov    %ebx,%edi
  add    $0x2,%ebx
  movsd  %xmm0,0x8(%rsp)
  call   <fact@plt>
  movsd  0x8(%rsp),%xmm1
  divsd  %xmm0,%xmm1
  movsd  (%rsp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsp)
  cmp    $0x3e,%ebx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %r14
  ret
```

assistant: ```c
double cosh(double n) {
  double sum = 0;
  int i;
  for (i = 0; i <= 30; i++) {
    sum += dpow(n, 2 * i) / fact(2 * i);
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cosh>:
  endbr64
  push   %r14
  pxor   %xmm4,%xmm4
  movq   %xmm0,%r14
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  movsd  %xmm4,(%rsp)
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%edi
  movq   %r14,%xmm0
  call   <dpow@plt>
  mov    %ebx,%edi
  add    $0x2,%ebx
  movsd  %xmm0,0x8(%rsp)
  call   <fact@plt>
  movsd  0x8(%rsp),%xmm1
  divsd  %xmm0,%xmm1
  movsd  (%rsp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsp)
  cmp    $0x3e,%ebx
  jne    L6
  add    $0x18,%rsp
  pop    %rbx
  pop    %r14
  ret
```

assistant: ```c
double cosh(double n) {
  double sum = 0;
  int i;
  for (i = 0; i <= 30; i++) {
    sum += dpow(n, 2 * i) / fact(2 * i);
  }
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<satMul3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  xor    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  xor    -0x14(%rbp),%eax
  or     %edx,%eax
  sar    $0x1f,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  or     -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x1f,%eax
  xor    $0x80000000,%eax
  and    -0x4(%rbp),%eax
  xor    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int satMul3(int x) {
  int x2, x3;
  int k;
  x2 = x + x;
  x3 = x2 + x;
  k = (((x3 ^ x) | (x2 ^ x)) >> 31);
  return (k | x3) ^ (((x >> 31) ^ (1 << 31)) & k);
}
```
user: What is the c source code of the assembly code below?

```asm
<satMul3>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  lea    (%rdi,%rax,1),%ecx
  mov    %edi,%edx
  xor    %ecx,%edx
  xor    %edi,%eax
  or     %eax,%edx
  sar    $0x1f,%edx
  sar    $0x1f,%edi
  lea    -0x80000000(%rdi),%eax
  and    %edx,%eax
  or     %edx,%ecx
  xor    %ecx,%eax
  ret
```

assistant: ```c
int satMul3(int x) {
  int x2, x3;
  int k;
  x2 = x + x;
  x3 = x2 + x;
  k = (((x3 ^ x) | (x2 ^ x)) >> 31);
  return (k | x3) ^ (((x >> 31) ^ (1 << 31)) & k);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<satMul3>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  mov    %edi,%edx
  lea    (%rdi,%rax,1),%ecx
  xor    %edi,%eax
  sar    $0x1f,%edi
  xor    %ecx,%edx
  or     %eax,%edx
  lea    -0x80000000(%rdi),%eax
  sar    $0x1f,%edx
  and    %edx,%eax
  or     %edx,%ecx
  xor    %ecx,%eax
  ret
```

assistant: ```c
int satMul3(int x) {
  int x2, x3;
  int k;
  x2 = x + x;
  x3 = x2 + x;
  k = (((x3 ^ x) | (x2 ^ x)) >> 31);
  return (k | x3) ^ (((x >> 31) ^ (1 << 31)) & k);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<satMul3>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  mov    %edi,%edx
  lea    (%rdi,%rax,1),%ecx
  xor    %edi,%eax
  sar    $0x1f,%edi
  xor    %ecx,%edx
  or     %eax,%edx
  lea    -0x80000000(%rdi),%eax
  sar    $0x1f,%edx
  and    %edx,%eax
  or     %edx,%ecx
  xor    %ecx,%eax
  ret
```

assistant: ```c
int satMul3(int x) {
  int x2, x3;
  int k;
  x2 = x + x;
  x3 = x2 + x;
  k = (((x3 ^ x) | (x2 ^ x)) >> 31);
  return (k | x3) ^ (((x >> 31) ^ (1 << 31)) & k);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<UART_InUDec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <UART_InChar@plt>
  mov    %al,-0x9(%rbp)
  jmp    L13
L15:
  cmpb   $0x2f,-0x9(%rbp)
  jle    L47
  cmpb   $0x39,-0x9(%rbp)
  jg     L47
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movsbl -0x9(%rbp),%eax
  sub    $0x30,%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <UART_OutChar@plt>
  jmp    L35
L47:
  cmpb   $0x8,-0x9(%rbp)
  jne    L35
  cmpl   $0x0,-0x4(%rbp)
  je     L35
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <UART_OutChar@plt>
L35:
  mov    $0x0,%eax
  call   <UART_InChar@plt>
  mov    %al,-0x9(%rbp)
L13:
  cmpb   $0xd,-0x9(%rbp)
  jne    L15
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint32_t UART_InUDec(void) {
  uint32_t number = 0, length = 0;
  char character;
  character = UART_InChar();
  while (character != 0x0D) {
    if ((character >= '0') && (character <= '9')) {
      number = 10 * number + (character - '0');
      length++;
      UART_OutChar(character);
    } else if ((character == 0x08) && length) {
      number /= 10;
      length--;
      UART_OutChar(character);
    }
    character = UART_InChar();
  }
  return number;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<UART_InUDec>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <UART_InChar@plt>
  cmp    $0xd,%al
  je     L25
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L10
L42:
  lea    (%rbx,%rbx,4),%edx
  movsbl %al,%edi
  lea    -0x30(%rdi,%rdx,2),%ebx
  add    $0x1,%ebp
  call   <UART_OutChar@plt>
L1:
  mov    $0x0,%eax
  call   <UART_InChar@plt>
  cmp    $0xd,%al
  je     L61
L10:
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L42
  cmp    $0x8,%al
  jne    L1
  test   %ebp,%ebp
  je     L1
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ebx
  sub    %ebx,%eax
  mov    %eax,%ebx
  sub    $0x1,%ebp
  mov    $0x8,%edi
  call   <UART_OutChar@plt>
  jmp    L1
L25:
  mov    $0x0,%ebx
L61:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint32_t UART_InUDec(void) {
  uint32_t number = 0, length = 0;
  char character;
  character = UART_InChar();
  while (character != 0x0D) {
    if ((character >= '0') && (character <= '9')) {
      number = 10 * number + (character - '0');
      length++;
      UART_OutChar(character);
    } else if ((character == 0x08) && length) {
      number /= 10;
      length--;
      UART_OutChar(character);
    }
    character = UART_InChar();
  }
  return number;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<UART_InUDec>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <UART_InChar@plt>
  cmp    $0xd,%al
  je     L10
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  mov    $0xcccccccd,%ebp
  jmp    L18
  nop
L33:
  cmp    $0x8,%al
  jne    L4
  test   %ebx,%ebx
  je     L4
  mov    %r12d,%r12d
  mov    $0x8,%edi
  sub    $0x1,%ebx
  imul   %rbp,%r12
  shr    $0x23,%r12
  call   <UART_OutChar@plt>
L4:
  xor    %eax,%eax
  call   <UART_InChar@plt>
  cmp    $0xd,%al
  je     L61
L18:
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L33
  lea    (%r12,%r12,4),%edx
  movsbl %al,%edi
  add    $0x1,%ebx
  lea    -0x30(%rdi,%rdx,2),%r12d
  call   <UART_OutChar@plt>
  xor    %eax,%eax
  call   <UART_InChar@plt>
  cmp    $0xd,%al
  jne    L18
L61:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint32_t UART_InUDec(void) {
  uint32_t number = 0, length = 0;
  char character;
  character = UART_InChar();
  while (character != 0x0D) {
    if ((character >= '0') && (character <= '9')) {
      number = 10 * number + (character - '0');
      length++;
      UART_OutChar(character);
    } else if ((character == 0x08) && length) {
      number /= 10;
      length--;
      UART_OutChar(character);
    }
    character = UART_InChar();
  }
  return number;
}
```