ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rdx
  add    $0x3686696a,%rdx
  mov    %rdx,%rax
  xor    $0xffffffffce5d000b,%rax
  test   $0x1,%dl
  jne    L19
  mov    %eax,%ecx
  and    $0xe,%ecx
  or     $0x1,%ecx
  ror    %cl,%rax
L13:
  or     $0x279d3baa,%rax
  add    $0x160c5c90,%rax
  mov    %rax,(%rsi)
  ret
L19:
  mov    %rax,%rdx
  imul   %rax,%rdx
  shl    $0x4,%rdx
  and    $0x3f0,%edx
  or     %rax,%rdx
  lea    0x0(,%rdx,8),%rax
  movzbl %al,%eax
  or     %rdx,%rax
  jmp    L13
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    if (state[0UL] & 1UL) {
      state[0UL] = (state[0UL] >> ((state[0UL] & 15UL) | 1UL)) | (state[0UL] << (64 - ((state[0UL] & 15UL) | 1UL)));
    } else {
      state[0UL] |= (state[0UL] * state[0UL] & 63UL) << 4UL;
      state[0UL] |= (state[0UL] & 31UL) << 3UL;
    }
    output[0UL] = (state[0UL] | 664615850UL) + 369908880UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rdx
  add    $0x3686696a,%rdx
  mov    %rdx,%rax
  xor    $0xffffffffce5d000b,%rax
  and    $0x1,%edx
  jne    L3
  mov    %eax,%ecx
  and    $0xe,%ecx
  or     $0x1,%ecx
  ror    %cl,%rax
  or     $0x279d3baa,%rax
  add    $0x160c5c90,%rax
  mov    %rax,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %rax,%rdx
  imul   %rax,%rdx
  shl    $0x4,%rdx
  and    $0x3f0,%edx
  or     %rax,%rdx
  lea    0x0(,%rdx,8),%rax
  movzbl %al,%eax
  or     %rdx,%rax
  or     $0x279d3baa,%rax
  add    $0x160c5c90,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    if (state[0UL] & 1UL) {
      state[0UL] = (state[0UL] >> ((state[0UL] & 15UL) | 1UL)) | (state[0UL] << (64 - ((state[0UL] & 15UL) | 1UL)));
    } else {
      state[0UL] |= (state[0UL] * state[0UL] & 63UL) << 4UL;
      state[0UL] |= (state[0UL] & 31UL) << 3UL;
    }
    output[0UL] = (state[0UL] | 664615850UL) + 369908880UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rdx
  add    $0x3686696a,%rdx
  mov    %rdx,%rax
  xor    $0xffffffffce5d000b,%rax
  and    $0x1,%edx
  jne    L0
  mov    %eax,%ecx
  and    $0xe,%ecx
  or     $0x1,%ecx
  ror    %cl,%rax
  or     $0x279d3baa,%rax
  add    $0x160c5c90,%rax
  mov    %rax,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %rax,%rdx
  imul   %rax,%rdx
  shl    $0x4,%rdx
  and    $0x3f0,%edx
  or     %rax,%rdx
  lea    0x0(,%rdx,8),%rax
  movzbl %al,%eax
  or     %rdx,%rax
  or     $0x279d3baa,%rax
  add    $0x160c5c90,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    if (state[0UL] & 1UL) {
      state[0UL] = (state[0UL] >> ((state[0UL] & 15UL) | 1UL)) | (state[0UL] << (64 - ((state[0UL] & 15UL) | 1UL)));
    } else {
      state[0UL] |= (state[0UL] * state[0UL] & 63UL) << 4UL;
      state[0UL] |= (state[0UL] & 31UL) << 3UL;
    }
    output[0UL] = (state[0UL] | 664615850UL) + 369908880UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_sample_path>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,%r15
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%r12
  mov    %r8d,%ebx
  mov    %r9d,%r13d
  lea    0x0(,%r8,8),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %ebx,%ebx
  jle    L68
  mov    %r14,%rax
  mov    %ebx,%edx
  lea    (%r14,%rdx,8),%rcx
  lea    -0x1(%r13),%edx
  imul   %ebx,%edx
  movslq %edx,%rdx
  lea    (%r15,%rdx,8),%rdx
  sub    %r14,%rdx
L37:
  movsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L37
L68:
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <_normalize@plt>
  movslq %r13d,%rax
  mov    (%rsp),%rdi
  lea    -0x4(%rdi,%rax,4),%rbp
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <_random_choice@plt>
  mov    %eax,0x0(%rbp)
  mov    %r13d,%r15d
  sub    $0x2,%r15d
  js     L27
  movslq %r15d,%rax
  shl    $0x2,%rax
  mov    (%rsp),%rdi
  lea    (%rdi,%rax,1),%rbp
  imul   %ebx,%r15d
  lea    -0x4(%rdi,%rax,1),%rdx
  lea    -0x2(%r13),%eax
  shl    $0x2,%rax
  sub    %rax,%rdx
  mov    %rdx,(%rsp)
  lea    -0x1(%rbx),%eax
  lea    0x8(%r14,%rax,8),%r13
  jmp    L39
L8:
  movslq %r15d,%rax
  mov    0x8(%rsp),%rdi
  lea    (%rdi,%rax,8),%rsi
  sub    %r14,%rsi
  mov    %r14,%rax
  mov    $0x0,%ecx
L73:
  mov    %ecx,%edx
  add    0x4(%rbp),%edx
  movslq %edx,%rdx
  movsd  (%r12,%rdx,8),%xmm0
  mulsd  (%rsi,%rax,1),%xmm0
  movsd  %xmm0,(%rax)
  add    %ebx,%ecx
  add    $0x8,%rax
  cmp    %r13,%rax
  jne    L73
L54:
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <_normalize@plt>
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <_random_choice@plt>
  mov    %eax,0x0(%rbp)
  sub    $0x4,%rbp
  sub    %ebx,%r15d
  cmp    (%rsp),%rbp
  je     L27
L39:
  test   %ebx,%ebx
  jg     L8
  jmp    L54
L27:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void _sample_path(int *path, const double *alpha, const double *A, const double *pobs, const int N, const int T) {
  int i, t;
  double *psel = (double *)malloc(N * sizeof(double));
  srand(time(NULL));
  for (i = 0; i < N; i++) {
    psel[i] = alpha[(T - 1) * N + i];
  }
  _normalize(psel, N);
  path[T - 1] = _random_choice(psel, N);
  for (t = T - 2; t >= 0; t--) {
    for (i = 0; i < N; i++) {
      psel[i] = alpha[t * N + i] * A[i * N + path[t + 1]];
    }
    _normalize(psel, N);
    path[t] = _random_choice(psel, N);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_sample_path>:
  endbr64
  push   %r15
  mov    %r9d,%r15d
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  movslq %r8d,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rbp,8),%edi
  mov    %rbp,%r14
  sub    $0x28,%rsp
  mov    %rsi,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  call   <malloc@plt>
  xor    %edi,%edi
  mov    %rax,%r12
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %ebp,%ebp
  jle    L30
  mov    %ebp,%edx
  mov    %r12,%rax
  lea    (%r12,%rdx,8),%rsi
  lea    -0x1(%r15),%edx
  imul   %ebp,%edx
  movslq %edx,%rdx
  lea    0x0(%r13,%rdx,8),%rdx
  sub    %r12,%rdx
  nopw   0x0(%rax,%rax,1)
L7:
  movsd  (%rdx,%rax,1),%xmm0
  add    $0x8,%rax
  movsd  %xmm0,-0x8(%rax)
  cmp    %rsi,%rax
  jne    L7
L30:
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <_normalize@plt>
  movslq %r15d,%rax
  mov    %r14d,%esi
  mov    %r12,%rdi
  lea    -0x4(%rbx,%rax,4),%r13
  call   <_random_choice@plt>
  sub    $0x2,%r15d
  mov    %eax,0x0(%r13)
  mov    %r15d,%r8d
  js     L0
  imul   %r14d,%r8d
  lea    -0x1(%r14),%eax
  movslq %r15d,%r15
  shl    $0x3,%rbp
  lea    0x8(%r12,%rax,8),%r13
  cs nopw 0x0(%rax,%rax,1)
L58:
  test   %r14d,%r14d
  jle    L39
  lea    0x1(%r15),%eax
  mov    0x18(%rsp),%rcx
  cltq
  movslq (%rbx,%rax,4),%rax
  lea    (%rcx,%rax,8),%rdx
  mov    0x10(%rsp),%rcx
  movslq %r8d,%rax
  lea    (%rcx,%rax,8),%rsi
  mov    %r12,%rax
  sub    %r12,%rsi
  nopw   0x0(%rax,%rax,1)
L6:
  movsd  (%rsi,%rax,1),%xmm0
  mulsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    %rbp,%rdx
  movsd  %xmm0,-0x8(%rax)
  cmp    %rax,%r13
  jne    L6
L39:
  mov    %r14d,%esi
  mov    %r12,%rdi
  mov    %r8d,0xc(%rsp)
  call   <_normalize@plt>
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <_random_choice@plt>
  mov    0xc(%rsp),%r8d
  mov    %eax,(%rbx,%r15,4)
  sub    $0x1,%r15
  sub    %r14d,%r8d
  test   %r15d,%r15d
  jns    L58
L0:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void _sample_path(int *path, const double *alpha, const double *A, const double *pobs, const int N, const int T) {
  int i, t;
  double *psel = (double *)malloc(N * sizeof(double));
  srand(time(NULL));
  for (i = 0; i < N; i++) {
    psel[i] = alpha[(T - 1) * N + i];
  }
  _normalize(psel, N);
  path[T - 1] = _random_choice(psel, N);
  for (t = T - 2; t >= 0; t--) {
    for (i = 0; i < N; i++) {
      psel[i] = alpha[t * N + i] * A[i * N + path[t + 1]];
    }
    _normalize(psel, N);
    path[t] = _random_choice(psel, N);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sendString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x820,%rsp
  mov    %edi,-0x814(%rbp)
  mov    %rsi,-0x820(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x820(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x810(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x810(%rbp),%rsi
  mov    -0x814(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <send@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void sendString(int client, const char *strToSend) {
  char buf[2048];
  strcpy(buf, strToSend);
  send(client, buf, strlen(buf), 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
<sendString>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%ecx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <send@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sendString(int client, const char *strToSend) {
  char buf[2048];
  strcpy(buf, strToSend);
  send(client, buf, strlen(buf), 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sendString>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %eax,%edx
  call   <send@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x818,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sendString(int client, const char *strToSend) {
  char buf[2048];
  strcpy(buf, strToSend);
  send(client, buf, strlen(buf), 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<sendString>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %eax,%edx
  call   <send@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x818,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sendString(int client, const char *strToSend) {
  char buf[2048];
  strcpy(buf, strToSend);
  send(client, buf, strlen(buf), 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<secondsToHuman>:
  endbr64
  push   %rbx
  mov    %rsi,%r9
  movabs $0x1845c8a0ce512957,%rdx
  mov    %rsi,%rax
  imul   %rdx
  sar    $0xd,%rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  mov    %rdx,%rcx
  sub    %rax,%rcx
  imul   $0x15180,%rcx,%rax
  sub    %rax,%r9
  movabs $0x48d159e26af37c05,%rdx
  mov    %r9,%rax
  imul   %rdx
  sar    $0xa,%rdx
  mov    %r9,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  mov    %rdx,%r11
  imul   $0xe10,%rdx,%rax
  sub    %rax,%r9
  movabs $0x8888888888888889,%r8
  mov    %r9,%rax
  imul   %r8
  lea    (%rdx,%r9,1),%rax
  sar    $0x5,%rax
  mov    %r9,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,%rdx
  shl    $0x4,%rdx
  sub    %rax,%rdx
  shl    $0x2,%rdx
  sub    %rdx,%r9
  test   %ecx,%ecx
  jg     L29
  mov    %r9d,%edx
  test   %r11d,%r11d
  jg     L43
  test   %eax,%eax
  jle    L57
  mov    %r9d,%ecx
  mov    %eax,%edx
  lea    D26(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L33
L29:
  mov    %ecx,%esi
  mov    %eax,%r8d
  mov    %r11d,%ecx
  mov    %esi,%edx
  lea    D32(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
L33:
  pop    %rbx
  ret
L43:
  mov    %r9d,%r8d
  mov    %eax,%ecx
  mov    %r11d,%edx
  lea    D18(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L33
L57:
  lea    D5(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L33
```

tool_calls: name: retrieve_data
arguments: data_label: D26
type_guessed: string
name: retrieve_data
arguments: data_label: D32
type_guessed: string
name: retrieve_data
arguments: data_label: D18
type_guessed: string
name: retrieve_data
arguments: data_label: D5
type_guessed: string

tool: D26:       %02dm%02ds

tool: D32: %02dd%02dh%02dm%02ds

tool: D18:    %02dh%02dm%02ds

tool: D5:          %02ds

assistant: ```c
void secondsToHuman(char *buffer, long elapsed) {
  int days;
  int hours;
  int minutes;
  int seconds;
  days = elapsed / 86400;
  elapsed = elapsed % 86400;
  hours = elapsed / 3600;
  elapsed = elapsed % 3600;
  minutes = elapsed / 60;
  seconds = elapsed % 60;
  if (days > 0) {
    sprintf(buffer, "%02dd%02dh%02dm%02ds", days, hours, minutes, seconds);
  } else if (hours > 0) {
    sprintf(buffer, "   %02dh%02dm%02ds", hours, minutes, seconds);
  } else if (minutes > 0) {
    sprintf(buffer, "      %02dm%02ds", minutes, seconds);
  } else {
    sprintf(buffer, "         %02ds", seconds);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<secondsToHuman>:
  endbr64
  movabs $0x1845c8a0ce512957,%rdx
  mov    %rsi,%rax
  push   %r12
  imul   %rdx
  mov    %rsi,%rax
  push   %rbx
  sar    $0x3f,%rax
  sar    $0xd,%rdx
  mov    %rdx,%rcx
  movabs $0x48d159e26af37c05,%rdx
  sub    %rax,%rcx
  imul   $0x15180,%rcx,%rax
  sub    %rax,%rsi
  mov    %rsi,%rax
  mov    %rsi,%rbx
  imul   %rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sar    $0xa,%rdx
  sub    %rax,%rdx
  mov    %edx,%r11d
  imul   $0xe10,%rdx,%rdx
  sub    %rdx,%rbx
  movabs $0x8888888888888889,%rdx
  mov    %rbx,%rax
  mov    %rbx,%r9
  imul   %rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rbx,1),%r8
  sar    $0x5,%r8
  sub    %rax,%r8
  mov    %r8,%rax
  shl    $0x4,%rax
  sub    %r8,%rax
  shl    $0x2,%rax
  sub    %rax,%r9
  test   %ecx,%ecx
  jg     L20
  mov    %r8d,%edx
  cmp    $0xe0f,%rsi
  jg     L4
  cmp    $0x3b,%rbx
  jle    L9
  pop    %rbx
  mov    %r9d,%ecx
  lea    D39(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
L4:
  mov    %edx,%ecx
  pop    %rbx
  mov    %r9d,%r8d
  mov    %r11d,%edx
  lea    D9(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
  nopl   0x0(%rax)
L20:
  mov    %ecx,%r10d
  pop    %rbx
  mov    %r11d,%ecx
  xor    %eax,%eax
  mov    %r10d,%edx
  lea    D21(%rip),%rsi
  pop    %r12
  jmp    1050 <sprintf@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  pop    %rbx
  mov    %r9d,%edx
  lea    D12(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D39
type_guessed: string
name: parse_data
arguments: data_tag: D9
type_guessed: string
name: parse_data
arguments: data_tag: D21
type_guessed: string
name: parse_data
arguments: data_tag: D12
type_guessed: string

tool: D39:       %02dm%02ds

tool: D9:    %02dh%02dm%02ds

tool: D21: %02dd%02dh%02dm%02ds

tool: D12:          %02ds

assistant: ```c
void secondsToHuman(char *buffer, long elapsed) {
  int days;
  int hours;
  int minutes;
  int seconds;
  days = elapsed / 86400;
  elapsed = elapsed % 86400;
  hours = elapsed / 3600;
  elapsed = elapsed % 3600;
  minutes = elapsed / 60;
  seconds = elapsed % 60;
  if (days > 0) {
    sprintf(buffer, "%02dd%02dh%02dm%02ds", days, hours, minutes, seconds);
  } else if (hours > 0) {
    sprintf(buffer, "   %02dh%02dm%02ds", hours, minutes, seconds);
  } else if (minutes > 0) {
    sprintf(buffer, "      %02dm%02ds", minutes, seconds);
  } else {
    sprintf(buffer, "         %02ds", seconds);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<secondsToHuman>:
  endbr64
  movabs $0x1845c8a0ce512957,%rdx
  mov    %rsi,%rax
  push   %r12
  imul   %rdx
  mov    %rsi,%rax
  push   %rbx
  sar    $0x3f,%rax
  sar    $0xd,%rdx
  mov    %rdx,%rcx
  movabs $0x48d159e26af37c05,%rdx
  sub    %rax,%rcx
  imul   $0x15180,%rcx,%rax
  sub    %rax,%rsi
  mov    %rsi,%rax
  mov    %rsi,%rbx
  imul   %rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sar    $0xa,%rdx
  sub    %rax,%rdx
  mov    %edx,%r11d
  imul   $0xe10,%rdx,%rdx
  sub    %rdx,%rbx
  movabs $0x8888888888888889,%rdx
  mov    %rbx,%rax
  mov    %rbx,%r9
  imul   %rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rbx,1),%r8
  sar    $0x5,%r8
  sub    %rax,%r8
  mov    %r8,%rax
  shl    $0x4,%rax
  sub    %r8,%rax
  shl    $0x2,%rax
  sub    %rax,%r9
  test   %ecx,%ecx
  jg     L6
  mov    %r8d,%edx
  cmp    $0xe0f,%rsi
  jg     L29
  cmp    $0x3b,%rbx
  jle    L8
  pop    %rbx
  mov    %r9d,%ecx
  lea    D7(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
L29:
  mov    %edx,%ecx
  pop    %rbx
  mov    %r9d,%r8d
  mov    %r11d,%edx
  lea    D39(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
  nopl   0x0(%rax)
L6:
  mov    %ecx,%r10d
  pop    %rbx
  mov    %r11d,%ecx
  xor    %eax,%eax
  mov    %r10d,%edx
  lea    D10(%rip),%rsi
  pop    %r12
  jmp    1050 <sprintf@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  pop    %rbx
  mov    %r9d,%edx
  lea    D18(%rip),%rsi
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <sprintf@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
type_infered: string
name: disassemble_data
arguments: label: D39
type_infered: string
name: disassemble_data
arguments: label: D10
type_infered: string
name: disassemble_data
arguments: label: D18
type_infered: string

tool: D7:       %02dm%02ds

tool: D39:    %02dh%02dm%02ds

tool: D10: %02dd%02dh%02dm%02ds

tool: D18:          %02ds

assistant: ```c
void secondsToHuman(char *buffer, long elapsed) {
  int days;
  int hours;
  int minutes;
  int seconds;
  days = elapsed / 86400;
  elapsed = elapsed % 86400;
  hours = elapsed / 3600;
  elapsed = elapsed % 3600;
  minutes = elapsed / 60;
  seconds = elapsed % 60;
  if (days > 0) {
    sprintf(buffer, "%02dd%02dh%02dm%02ds", days, hours, minutes, seconds);
  } else if (hours > 0) {
    sprintf(buffer, "   %02dh%02dm%02ds", hours, minutes, seconds);
  } else if (minutes > 0) {
    sprintf(buffer, "      %02dm%02ds", minutes, seconds);
  } else {
    sprintf(buffer, "         %02ds", seconds);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lenn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L17:
  addl   $0x1,-0x4(%rbp)
L29:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jae    L4
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
L4:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int lenn(char *pos, unsigned int max) {
  unsigned int p = 0;
  while (p < max && *pos++)
    p++;
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<lenn>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  je     L37
  mov    $0x0,%edx
L34:
  cmpb   $0x0,(%rdi)
  je     L3
  add    $0x1,%edx
  add    $0x1,%rdi
  cmp    %edx,%eax
  jne    L34
  jmp    L37
L3:
  mov    %edx,%eax
L37:
  ret
```

assistant: ```c
unsigned int lenn(char *pos, unsigned int max) {
  unsigned int p = 0;
  while (p < max && *pos++)
    p++;
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<lenn>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  je     L28
  xor    %edx,%edx
  jmp    L25
  xchg   %ax,%ax
L15:
  add    $0x1,%edx
  add    $0x1,%rdi
  cmp    %edx,%eax
  je     L28
L25:
  cmpb   $0x0,(%rdi)
  jne    L15
  mov    %edx,%eax
L28:
  ret
```

assistant: ```c
unsigned int lenn(char *pos, unsigned int max) {
  unsigned int p = 0;
  while (p < max && *pos++)
    p++;
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lenn>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  je     L22
  xor    %edx,%edx
  jmp    L15
  xchg   %ax,%ax
L13:
  add    $0x1,%edx
  add    $0x1,%rdi
  cmp    %edx,%eax
  je     L22
L15:
  cmpb   $0x0,(%rdi)
  jne    L13
  mov    %edx,%eax
L22:
  ret
```

assistant: ```c
unsigned int lenn(char *pos, unsigned int max) {
  unsigned int p = 0;
  while (p < max && *pos++)
    p++;
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<addmod8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  add    %rax,-0x8(%rbp)
  jmp    L8
L14:
  mov    -0x18(%rbp),%rax
  sub    %rax,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jge    L14
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint8_t addmod8(uint8_t a, uint8_t b, uint8_t m) {
  a += b;
  while (a >= m)
    a -= m;
  return a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<addmod8>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  cmp    %rdx,%rax
  jl     L17
L5:
  sub    %rdx,%rax
  cmp    %rax,%rdx
  jle    L5
L17:
  ret
```

assistant: ```c
uint8_t addmod8(uint8_t a, uint8_t b, uint8_t m) {
  a += b;
  while (a >= m)
    a -= m;
  return a;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<addmod8>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  cmp    %rdx,%rax
  jl     L11
  nopl   (%rax)
L3:
  sub    %rdx,%rax
  cmp    %rax,%rdx
  jle    L3
L11:
  ret
```

assistant: ```c
uint8_t addmod8(uint8_t a, uint8_t b, uint8_t m) {
  a += b;
  while (a >= m)
    a -= m;
  return a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addmod8>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  cmp    %rdx,%rax
  jl     L7
  nopl   (%rax)
L1:
  sub    %rdx,%rax
  cmp    %rax,%rdx
  jle    L1
L7:
  ret
```

assistant: ```c
uint8_t addmod8(uint8_t a, uint8_t b, uint8_t m) {
  a += b;
  while (a >= m)
    a -= m;
  return a;
}
```