ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fc8180_set_filter>:
  endbr64
  mov    %edx,%r8d
  mov    $0x5c490,%eax
  xor    %edx,%edx
  mov    $0xcccccccd,%ecx
  div    %r8d
  mov    $0xd1b71759,%edx
  imul   %esi,%eax
  imul   %rdx,%rax
  shr    $0x2d,%rax
  mov    %eax,%edx
  mov    %eax,%esi
  imul   %rcx,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%esi
  cmp    $0x4,%esi
  jbe    L1
  lea    0x1(%rdx),%eax
  movzbl %al,%eax
  ret
  nopl   (%rax)
L1:
  mov    $0xffffffcd,%edx
  mul    %dl
  shr    $0xb,%ax
  movzbl %al,%eax
  ret
```

assistant: ```c
unsigned int fc8180_set_filter(void *handle, unsigned int ref_clock, unsigned int filter_bw) {
  unsigned int div_cal = 0;
  unsigned char div = 0;
  div_cal = (378000 / filter_bw) * ref_clock / 10000;
  if ((div_cal % 10) >= 5)
    div = (unsigned char)(div_cal / 10) + 1;
  else
    div = (unsigned char)div_cal / 10;
  return div;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add_leading_zeros>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L21
L24:
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  sub    $0x1,%edx
  sub    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
L21:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L24
  movl   $0x0,-0x8(%rbp)
  jmp    L7
L28:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x8(%rbp)
L7:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L28
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void add_leading_zeros(char *inout, int size) {
  int i, j;
  j = strlen(inout);
  for (i = 0; i < (j + 1); i++) {
    inout[size - 1 - i] = inout[j - i];
  }
  for (i = 0; i < (size - 1 - j); i++) {
    inout[i] = '0';
  }
  inout[size - 1] = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<add_leading_zeros>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <strlen@plt>
  test   %eax,%eax
  js     L0
  movslq %eax,%r8
  lea    (%rbx,%r8,1),%rdx
  lea    -0x1(%rbx,%r8,1),%rdi
  mov    %eax,%ecx
  sub    %rcx,%rdi
  movslq %ebp,%r9
L12:
  movzbl (%rdx),%esi
  mov    %rdx,%rcx
  sub    %r8,%rcx
  mov    %sil,-0x1(%rcx,%r9,1)
  sub    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L12
L0:
  lea    -0x1(%rbp),%edx
  sub    %eax,%edx
  test   %edx,%edx
  jle    L19
  mov    %rbx,%rax
  lea    -0x1(%rdx),%edx
  lea    0x1(%rbx,%rdx,1),%rdx
L9:
  movb   $0x30,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L9
L19:
  movslq %ebp,%rbp
  movb   $0x0,-0x1(%rbx,%rbp,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void add_leading_zeros(char *inout, int size) {
  int i, j;
  j = strlen(inout);
  for (i = 0; i < (j + 1); i++) {
    inout[size - 1 - i] = inout[j - i];
  }
  for (i = 0; i < (size - 1 - j); i++) {
    inout[i] = '0';
  }
  inout[size - 1] = '\0';
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<add_leading_zeros>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rbx),%edx
  test   %eax,%eax
  js     L15
  movslq %eax,%r8
  lea    -0x1(%rbp,%r8,1),%r9
  lea    0x0(%rbp,%r8,1),%rcx
  sub    %r8,%r9
  nopl   0x0(%rax)
L22:
  movzbl (%rcx),%edi
  mov    %rcx,%rsi
  sub    $0x1,%rcx
  sub    %r8,%rsi
  mov    %dil,-0x1(%rsi,%rbx,1)
  cmp    %r9,%rcx
  jne    L22
L15:
  sub    %eax,%edx
  test   %edx,%edx
  jle    L0
  movslq %edx,%rdx
  mov    $0x30,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L0:
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void add_leading_zeros(char *inout, int size) {
  int i, j;
  j = strlen(inout);
  for (i = 0; i < (j + 1); i++) {
    inout[size - 1 - i] = inout[j - i];
  }
  for (i = 0; i < (size - 1 - j); i++) {
    inout[i] = '0';
  }
  inout[size - 1] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add_leading_zeros>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rbx),%edx
  test   %eax,%eax
  js     L3
  movslq %eax,%r8
  lea    -0x1(%rbp,%r8,1),%r9
  lea    0x0(%rbp,%r8,1),%rcx
  sub    %r8,%r9
  nopl   0x0(%rax)
L14:
  movzbl (%rcx),%edi
  mov    %rcx,%rsi
  sub    $0x1,%rcx
  sub    %r8,%rsi
  mov    %dil,-0x1(%rsi,%rbx,1)
  cmp    %rcx,%r9
  jne    L14
L3:
  sub    %eax,%edx
  test   %edx,%edx
  jle    L20
  movslq %edx,%rdx
  mov    $0x30,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L20:
  movb   $0x0,-0x1(%rbp,%rbx,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void add_leading_zeros(char *inout, int size) {
  int i, j;
  j = strlen(inout);
  for (i = 0; i < (j + 1); i++) {
    inout[size - 1 - i] = inout[j - i];
  }
  for (i = 0; i < (size - 1 - j); i++) {
    inout[i] = '0';
  }
  inout[size - 1] = '\0';
}
```
user: What is the c source code of the assembly code below?

```asm
<fibonacciRek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  movl   $0x0,-0x14(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jne    L2
  mov    $0x1,%eax
  jmp    L19
L2:
  cmpl   $0x0,-0x24(%rbp)
  jne    L18
  mov    $0x0,%eax
  jmp    L19
L18:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibonacciRek>
  mov    %eax,%ebx
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibonacciRek>
  add    %ebx,%eax
L19:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fibonacciRek(int n) {
  int temp = 0;
  if (n == 1) {
    return 1;
  } else if (n == 0) {
    return 0;
  } else {
    return fibonacciRek(n - 1) + fibonacciRek(n - 2);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fibonacciRek>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  cmp    $0x1,%edi
  je     L21
  test   %edi,%edi
  jne    L15
L21:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L15:
  lea    -0x1(%rdi),%edi
  call   <fibonacciRek>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fibonacciRek>
  lea    0x0(%rbp,%rax,1),%ebx
  jmp    L21
```

assistant: ```c
int fibonacciRek(int n) {
  int temp = 0;
  if (n == 1) {
    return 1;
  } else if (n == 0) {
    return 0;
  } else {
    return fibonacciRek(n - 1) + fibonacciRek(n - 2);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<simple_hash_function>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L15
L17:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L15:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L17
  mov    -0xc(%rbp),%eax
  cltd
  shr    $0x16,%edx
  add    %edx,%eax
  and    $0x3ff,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int simple_hash_function(char *str) {
  int len = 0;
  int idx = 0;
  int sum = 0;
  int hash = 0;
  len = strlen(str);
  for (idx = 0; idx < len; idx++)
    sum += str[idx];
  hash = sum % 1024;
  return hash;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<simple_hash_function>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L21
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rsi
  mov    $0x0,%eax
L10:
  movsbl (%rdx),%ecx
  add    %ecx,%eax
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L10
L3:
  cltd
  shr    $0x16,%edx
  add    %edx,%eax
  and    $0x3ff,%eax
  sub    %edx,%eax
  pop    %rbx
  ret
L21:
  mov    $0x0,%eax
  jmp    L3
```

assistant: ```c
int simple_hash_function(char *str) {
  int len = 0;
  int idx = 0;
  int sum = 0;
  int hash = 0;
  len = strlen(str);
  for (idx = 0; idx < len; idx++)
    sum += str[idx];
  hash = sum % 1024;
  return hash;
}
```
user: What is the c source code of the assembly code below?

```asm
<simple_hash_function>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L17
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rcx
  xor    %eax,%eax
  xchg   %ax,%ax
L0:
  movsbl (%rdi),%edx
  add    $0x1,%rdi
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L0
  cltd
  pop    %rbx
  shr    $0x16,%edx
  add    %edx,%eax
  and    $0x3ff,%eax
  sub    %edx,%eax
  ret
  nopl   (%rax)
L17:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int simple_hash_function(char *str) {
  int len = 0;
  int idx = 0;
  int sum = 0;
  int hash = 0;
  len = strlen(str);
  for (idx = 0; idx < len; idx++)
    sum += str[idx];
  hash = sum % 1024;
  return hash;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ighmm_dvector_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9,-0x40(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fputs@plt>
  cmpl   $0x0,-0x24(%rbp)
  jle    L0
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  lea    D22(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
L0:
  movl   $0x1,-0x4(%rbp)
  jmp    L5
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x38(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L8
  mov    -0x40(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  lea    D4(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D22
type: string
name: disassemble_data
arguments: tag: D9
type: string
name: disassemble_data
arguments: tag: D4
type: string

tool: D22: %3d

tool: D9: %s %3d

tool: D4: %s


assistant: ```c
void ighmm_dvector_print(FILE *file, int *vector, int len, char *tab, char *separator, char *ending) {
  int j;
  fprintf(file, "%s", tab);
  if (len > 0)
    fprintf(file, "%3d", vector[0]);
  for (j = 1; j < len; j++)
    fprintf(file, "%s %3d", separator, vector[j]);
  fprintf(file, "%s\n", ending);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ighmm_dvector_print>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    %rcx,%rdi
  mov    %r8,%r12
  mov    %r9,%r15
  mov    %rbp,%rsi
  call   <fputs@plt>
  test   %r13d,%r13d
  jg     L1
L13:
  cmp    $0x1,%r13d
  jle    L25
  lea    0x4(%r14),%rbx
  lea    -0x2(%r13),%eax
  lea    0x8(%r14,%rax,4),%r14
  lea    D14(%rip),%r13
L31:
  mov    (%rbx),%ecx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  add    $0x4,%rbx
  cmp    %r14,%rbx
  jne    L31
L25:
  mov    %r15,%rdx
  lea    D25(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  mov    (%r14),%edx
  lea    D23(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  jmp    L13
```

tool_calls: name: unpack_data
arguments: tag: D14
datatype: string
name: unpack_data
arguments: tag: D25
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string

tool: D14: %s %3d

tool: D25: %s


tool: D23: %3d

assistant: ```c
void ighmm_dvector_print(FILE *file, int *vector, int len, char *tab, char *separator, char *ending) {
  int j;
  fprintf(file, "%s", tab);
  if (len > 0)
    fprintf(file, "%3d", vector[0]);
  for (j = 1; j < len; j++)
    fprintf(file, "%s %3d", separator, vector[j]);
  fprintf(file, "%s\n", ending);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ighmm_dvector_print>:
  endbr64
  push   %r15
  mov    %r9,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rcx,%rdi
  push   %rbx
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <fputs@plt>
  test   %r13d,%r13d
  jg     L27
L9:
  add    $0x8,%rsp
  mov    %r15,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbx
  lea    D18(%rip),%rsi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <fprintf@plt>
  nopl   0x0(%rax,%rax,1)
L27:
  mov    (%r14),%edx
  xor    %eax,%eax
  lea    D7(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  cmp    $0x1,%r13d
  jle    L9
  lea    -0x2(%r13),%eax
  lea    0x4(%r14),%rbx
  lea    D10(%rip),%r13
  lea    0x8(%r14,%rax,4),%r14
  cs nopw 0x0(%rax,%rax,1)
L25:
  mov    (%rbx),%ecx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <fprintf@plt>
  cmp    %r14,%rbx
  jne    L25
  jmp    L9
```

tool_calls: name: extract_symbolic_data
arguments: label: D18
datatype: string
name: extract_symbolic_data
arguments: label: D7
datatype: string
name: extract_symbolic_data
arguments: label: D10
datatype: string

tool: D18: %s


tool: D7: %3d

tool: D10: %s %3d

assistant: ```c
void ighmm_dvector_print(FILE *file, int *vector, int len, char *tab, char *separator, char *ending) {
  int j;
  fprintf(file, "%s", tab);
  if (len > 0)
    fprintf(file, "%3d", vector[0]);
  for (j = 1; j < len; j++)
    fprintf(file, "%s %3d", separator, vector[j]);
  fprintf(file, "%s\n", ending);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ighmm_dvector_print>:
  endbr64
  push   %r15
  mov    %r9,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rcx,%rdi
  push   %rbx
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <fputs@plt>
  test   %r13d,%r13d
  jg     L18
L33:
  add    $0x8,%rsp
  mov    %r15,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbx
  lea    D15(%rip),%rsi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <fprintf@plt>
  nopl   0x0(%rax,%rax,1)
L18:
  mov    (%r14),%edx
  xor    %eax,%eax
  lea    D24(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  cmp    $0x1,%r13d
  jle    L33
  lea    -0x2(%r13),%eax
  lea    0x4(%r14),%rbx
  lea    D5(%rip),%r13
  lea    0x8(%r14,%rax,4),%r14
  cs nopw 0x0(%rax,%rax,1)
L36:
  mov    (%rbx),%ecx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <fprintf@plt>
  cmp    %r14,%rbx
  jne    L36
  jmp    L33
```

tool_calls: name: retrieve_data
arguments: data_label: D15
data_type: string
name: retrieve_data
arguments: data_label: D24
data_type: string
name: retrieve_data
arguments: data_label: D5
data_type: string

tool: D15: %s


tool: D24: %3d

tool: D5: %s %3d

assistant: ```c
void ighmm_dvector_print(FILE *file, int *vector, int len, char *tab, char *separator, char *ending) {
  int j;
  fprintf(file, "%s", tab);
  if (len > 0)
    fprintf(file, "%3d", vector[0]);
  for (j = 1; j < len; j++)
    fprintf(file, "%s %3d", separator, vector[j]);
  fprintf(file, "%s\n", ending);
}
```
user: Translate the below assembly code into C programming language:

```asm
<truncate_to>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L12
  mov    -0x8(%rbp),%eax
  jmp    L2
L12:
  mov    -0x4(%rbp),%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int truncate_to(int number, int limit) {
  if (number > limit) {
    return limit;
  }
  return number;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<truncate_to>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int truncate_to(int number, int limit) {
  if (number > limit) {
    return limit;
  }
  return number;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<truncate_to>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int truncate_to(int number, int limit) {
  if (number > limit) {
    return limit;
  }
  return number;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<truncate_to>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int truncate_to(int number, int limit) {
  if (number > limit) {
    return limit;
  }
  return number;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <initGPIO@plt>
  mov    $0x0,%eax
  call   <initWiFIREadc@plt>
  mov    $0x0,%eax
  call   <initTimer2AndIC3@plt>
L2:
  mov    $0x0,%eax
  call   <setADCpos@plt>
  jmp    L2
```

assistant: ```c
int32_t main(void) {
  initGPIO();
  initWiFIREadc();
  initTimer2AndIC3();
  while (1)
    setADCpos();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <initGPIO@plt>
  mov    $0x0,%eax
  call   <initWiFIREadc@plt>
  mov    $0x0,%eax
  call   <initTimer2AndIC3@plt>
L8:
  mov    $0x0,%eax
  call   <setADCpos@plt>
  jmp    L8
```

assistant: ```c
int32_t main(void) {
  initGPIO();
  initWiFIREadc();
  initTimer2AndIC3();
  while (1)
    setADCpos();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initGPIO@plt>
  xor    %eax,%eax
  call   <initWiFIREadc@plt>
  xor    %eax,%eax
  call   <initTimer2AndIC3@plt>
  nopl   (%rax)
L6:
  xor    %eax,%eax
  call   <setADCpos@plt>
  jmp    L6
```

assistant: ```c
int32_t main(void) {
  initGPIO();
  initWiFIREadc();
  initTimer2AndIC3();
  while (1)
    setADCpos();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initGPIO@plt>
  xor    %eax,%eax
  call   <initWiFIREadc@plt>
  xor    %eax,%eax
  call   <initTimer2AndIC3@plt>
  nopl   (%rax)
L9:
  xor    %eax,%eax
  call   <setADCpos@plt>
  jmp    L9
```

assistant: ```c
int32_t main(void) {
  initGPIO();
  initWiFIREadc();
  initTimer2AndIC3();
  while (1)
    setADCpos();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movl   $0x0,-0x18(%rbp)
L0:
  mov    -0x34(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x30(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %rdx,%rdi
  call   <dig2char@plt>
  mov    %al,(%rbx)
  mov    -0x34(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,-0x28(%rbp)
  addl   $0x1,-0x18(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L0
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L13
L21:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x19(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x19(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x18(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x18(%rbp)
  jl     L21
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int print_base(long unsigned int uval, char buf[], unsigned int base) {
  int i = 0;
  do {
    buf[i] = dig2char(uval % base);
    uval = uval / base;
    i++;
  } while (0 != uval);
  buf[i] = 0;
  int len = i;
  for (i = 0; i < (len / 2); i++) {
    char c = buf[i];
    buf[i] = buf[len - i - 1];
    buf[len - i - 1] = c;
  }
  return len;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_base>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    $0x1,%ebp
  mov    %edx,%r12d
L4:
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %r12
  mov    %rdx,%rdi
  mov    %rax,%r14
  call   <dig2char@plt>
  mov    %al,-0x1(%r13,%rbp,1)
  mov    %rbx,%rax
  mov    %r14,%rbx
  mov    %ebp,%edi
  add    $0x1,%rbp
  cmp    %rax,%r12
  jbe    L4
  movslq %edi,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %edi,%r8d
  shr    $0x1f,%r8d
  add    %edi,%r8d
  sar    %r8d
  cmp    $0x1,%edi
  jle    L29
  lea    -0x1(%r13,%rax,1),%rdx
  mov    $0x0,%eax
L22:
  movzbl 0x0(%r13,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,0x0(%r13,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%r8d
  jg     L22
L29:
  mov    %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int print_base(long unsigned int uval, char buf[], unsigned int base) {
  int i = 0;
  do {
    buf[i] = dig2char(uval % base);
    uval = uval / base;
    i++;
  } while (0 != uval);
  buf[i] = 0;
  int len = i;
  for (i = 0; i < (len / 2); i++) {
    char c = buf[i];
    buf[i] = buf[len - i - 1];
    buf[len - i - 1] = c;
  }
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_base>:
  endbr64
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %rbp,%rax
  xor    %edx,%edx
  div    %r13
  mov    %rdx,%rdi
  mov    %rax,%r14
  call   <dig2char@plt>
  mov    %r12d,%r8d
  mov    %al,-0x1(%rbx,%r12,1)
  mov    %rbp,%rax
  add    $0x1,%r12
  mov    %r14,%rbp
  cmp    %r13,%rax
  jae    L15
  mov    %r8d,%esi
  movslq %r8d,%rcx
  sar    %esi
  movb   $0x0,(%rbx,%rcx,1)
  je     L6
  lea    -0x2(%rcx,%rbx,1),%rdi
  lea    -0x1(%rcx,%rbx,1),%rdx
  mov    %rbx,%rax
  lea    -0x1(%rsi),%ecx
  sub    %rcx,%rdi
  nopl   0x0(%rax,%rax,1)
L0:
  movzbl (%rax),%ecx
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  add    $0x1,%rax
  mov    %sil,-0x1(%rax)
  mov    %cl,0x1(%rdx)
  cmp    %rdi,%rdx
  jne    L0
L6:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int print_base(long unsigned int uval, char buf[], unsigned int base) {
  int i = 0;
  do {
    buf[i] = dig2char(uval % base);
    uval = uval / base;
    i++;
  } while (0 != uval);
  buf[i] = 0;
  int len = i;
  for (i = 0; i < (len / 2); i++) {
    char c = buf[i];
    buf[i] = buf[len - i - 1];
    buf[len - i - 1] = c;
  }
  return len;
}
```