ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_tube_name>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  cmpb   $0x20,(%rsi)
  jne    L12
L18:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  je     L18
L12:
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  test   %rax,%rax
  je     L8
  test   %r12,%r12
  je     L32
  mov    %rbx,(%r12)
L32:
  mov    $0x0,%edx
  test   %rbp,%rbp
  je     L49
  add    %rax,%rbx
  mov    %rbx,0x0(%rbp)
L49:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  mov    $0xffffffff,%edx
  jmp    L49
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D1: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-+/;.$_()

assistant: ```c
int read_tube_name(char **tubename, char *buf, char **end) {
  size_t len;
  while (buf[0] == ' ')
    buf++;
  len = strspn(buf, "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                    "abcdefghijklmnopqrstuvwxyz"
                    "0123456789-+/;.$_()");
  if (len == 0)
    return -1;
  if (tubename)
    *tubename = buf;
  if (end)
    *end = buf + len;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_tube_name>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  cmpb   $0x20,(%rsi)
  mov    %rsi,%rbx
  jne    L8
  cs nopw 0x0(%rax,%rax,1)
L0:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  je     L0
L8:
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  test   %rax,%rax
  je     L31
  test   %r12,%r12
  je     L38
  mov    %rbx,(%r12)
L38:
  xor    %r8d,%r8d
  test   %rbp,%rbp
  je     L49
  add    %rax,%rbx
  mov    %rbx,0x0(%rbp)
L49:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
L31:
  mov    $0xffffffff,%r8d
  jmp    L49
```

tool_calls: name: retrieve_data
arguments: label: D1
type: string

tool: D1: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-+/;.$_()

assistant: ```c
int read_tube_name(char **tubename, char *buf, char **end) {
  size_t len;
  while (buf[0] == ' ')
    buf++;
  len = strspn(buf, "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                    "abcdefghijklmnopqrstuvwxyz"
                    "0123456789-+/;.$_()");
  if (len == 0)
    return -1;
  if (tubename)
    *tubename = buf;
  if (end)
    *end = buf + len;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_tube_name>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  cmpb   $0x20,(%rsi)
  mov    %rsi,%rbx
  jne    L43
  cs nopw 0x0(%rax,%rax,1)
L59:
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  je     L59
L43:
  lea    D3(%rip),%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  test   %rax,%rax
  je     L3
  test   %r12,%r12
  je     L33
  mov    %rbx,(%r12)
L33:
  xor    %r8d,%r8d
  test   %rbp,%rbp
  je     L29
  add    %rax,%rbx
  mov    %rbx,0x0(%rbp)
L29:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    $0xffffffff,%r8d
  jmp    L29
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D3: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-+/;.$_()

assistant: ```c
int read_tube_name(char **tubename, char *buf, char **end) {
  size_t len;
  while (buf[0] == ' ')
    buf++;
  len = strspn(buf, "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                    "abcdefghijklmnopqrstuvwxyz"
                    "0123456789-+/;.$_()");
  if (len == 0)
    return -1;
  if (tubename)
    *tubename = buf;
  if (end)
    *end = buf + len;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  add    $0x696a,%ax
  xor    $0xb,%eax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  jmp    L21
L55:
  movzwl -0xa(%rbp),%eax
  cmp    %ax,-0xc(%rbp)
  jae    L28
  movzwl -0xa(%rbp),%eax
  cmp    %ax,-0xc(%rbp)
  jne    L0
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%eax
  movzwl -0xa(%rbp),%edx
  sub    %edx,%eax
  mov    %ax,-0xa(%rbp)
  jmp    L17
L0:
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movzwl -0xc(%rbp),%edx
  lea    -0xa(%rbp),%rcx
  movslq %edx,%rdx
  add    %rdx,%rdx
  add    %rcx,%rdx
  add    $0x1,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  jmp    L17
L28:
  movzwl -0xa(%rbp),%eax
  cmp    %ax,-0xc(%rbp)
  jb     L85
  movzwl -0xa(%rbp),%eax
  mov    %eax,%ecx
  movzwl -0xa(%rbp),%edx
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%eax
  add    %edx,%eax
  movzwl %ax,%eax
  shl    $0x3,%eax
  movzbl %al,%eax
  or     %ecx,%eax
  mov    %ax,-0xa(%rbp)
  jmp    L17
L85:
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%eax
  mov    %eax,%edx
  movzwl -0xc(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%eax
  movzwl %ax,%eax
  shl    $0x4,%eax
  and    $0x1f0,%ax
  or     %eax,%edx
  movzwl -0xc(%rbp),%eax
  cltq
  mov    %dx,-0xa(%rbp,%rax,2)
L17:
  movzwl -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0xc(%rbp)
L21:
  cmpw   $0x0,-0xc(%rbp)
  je     L55
  movzwl -0xa(%rbp),%eax
  lea    0x13fb(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned short)11;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] - state[0UL];
        } else {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
        }
      } else if (state[0UL] <= local1) {
        state[0UL] |= ((state[0UL] + state[local1]) & (unsigned short)31) << 3UL;
      } else {
        state[local1] |= (state[local1] & (unsigned short)31) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] - 674032645UL;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%ecx
  lea    0x696a(%rcx),%edx
  xor    $0xb,%edx
  mov    %edx,%eax
  cmp    $0x96a1,%cx
  je     L9
  movsbl %dh,%edx
  mov    %al,%ah
  mov    %dl,%al
L9:
  add    $0x13fb,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned short)11;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] - state[0UL];
        } else {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
        }
      } else if (state[0UL] <= local1) {
        state[0UL] |= ((state[0UL] + state[local1]) & (unsigned short)31) << 3UL;
      } else {
        state[local1] |= (state[local1] & (unsigned short)31) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] - 674032645UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%edx
  lea    0x696a(%rdx),%eax
  xor    $0xb,%eax
  mov    %eax,%ecx
  rol    $0x8,%cx
  cmp    $0x96a1,%dx
  cmovne %ecx,%eax
  add    $0x13fb,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned short)11;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] - state[0UL];
        } else {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
        }
      } else if (state[0UL] <= local1) {
        state[0UL] |= ((state[0UL] + state[local1]) & (unsigned short)31) << 3UL;
      } else {
        state[local1] |= (state[local1] & (unsigned short)31) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] - 674032645UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%edx
  lea    0x696a(%rdx),%eax
  xor    $0xb,%eax
  mov    %eax,%ecx
  rol    $0x8,%cx
  cmp    $0x96a1,%dx
  cmovne %ecx,%eax
  add    $0x13fb,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned short)11;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] - state[0UL];
        } else {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
        }
      } else if (state[0UL] <= local1) {
        state[0UL] |= ((state[0UL] + state[local1]) & (unsigned short)31) << 3UL;
      } else {
        state[local1] |= (state[local1] & (unsigned short)31) << 4UL;
      }
      local1++;
    }
    output[0UL] = state[0UL] - 674032645UL;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<av_gettime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rax
  imul   $0xf4240,%eax,%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int64_t av_gettime(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<av_gettime>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%eax
  add    (%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t av_gettime(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<av_gettime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%eax
  add    (%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t av_gettime(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<av_gettime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%eax
  add    (%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t av_gettime(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<__scan_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %al,-0x1(%rbp)
  cmpb   $0x2d,-0x1(%rbp)
  jne    L0
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <__scan_uint@plt>
  neg    %eax
  jmp    L13
L0:
  movsbl -0x1(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <__scan_uint@plt>
L13:
  leave
  ret
```

assistant: ```c
int __scan_int(FILE *stream, int width, int flags) {
  char c;
  c = getc(stream);
  if (c == '-') {
    return -__scan_uint(stream, width - 1, flags);
  } else {
    ungetc(c, stream);
    return __scan_uint(stream, width - 1, flags);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__scan_int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %edx,%r12d
  call   <getc@plt>
  cmp    $0x2d,%al
  je     L1
  movsbl %al,%edi
  mov    %rbp,%rsi
  call   <ungetc@plt>
  lea    -0x1(%rbx),%esi
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <__scan_uint@plt>
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  lea    -0x1(%rbx),%esi
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <__scan_uint@plt>
  neg    %eax
  jmp    L12
```

assistant: ```c
int __scan_int(FILE *stream, int width, int flags) {
  char c;
  c = getc(stream);
  if (c == '-') {
    return -__scan_uint(stream, width - 1, flags);
  } else {
    ungetc(c, stream);
    return __scan_uint(stream, width - 1, flags);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<__scan_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  sub    $0x1,%ebp
  call   <getc@plt>
  cmp    $0x2d,%al
  je     L4
  mov    %r13,%rsi
  movsbl %al,%edi
  call   <ungetc@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <__scan_uint@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <__scan_uint@plt>
  pop    %rbp
  pop    %r12
  neg    %eax
  pop    %r13
  ret
```

assistant: ```c
int __scan_int(FILE *stream, int width, int flags) {
  char c;
  c = getc(stream);
  if (c == '-') {
    return -__scan_uint(stream, width - 1, flags);
  } else {
    ungetc(c, stream);
    return __scan_uint(stream, width - 1, flags);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<__scan_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  sub    $0x1,%ebp
  call   <getc@plt>
  cmp    $0x2d,%al
  je     L5
  mov    %r13,%rsi
  movsbl %al,%edi
  call   <ungetc@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <__scan_uint@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <__scan_uint@plt>
  pop    %rbp
  pop    %r12
  neg    %eax
  pop    %r13
  ret
```

assistant: ```c
int __scan_int(FILE *stream, int width, int flags) {
  char c;
  c = getc(stream);
  if (c == '-') {
    return -__scan_uint(stream, width - 1, flags);
  } else {
    ungetc(c, stream);
    return __scan_uint(stream, width - 1, flags);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L12
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <reverse>
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L3
L12:
  nop
L3:
  leave
  ret
```

assistant: ```c
void reverse(char s[]) {
  if (s[0] == '\0') {
    return;
  } else {
    reverse(&s[1]);
    putchar(s[0]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  cmpb   $0x0,(%rdi)
  jne    L3
  ret
L3:
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%rdi),%rdi
  call   <reverse>
  movsbl (%rbx),%edi
  call   <putchar@plt>
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char s[]) {
  if (s[0] == '\0') {
    return;
  } else {
    reverse(&s[1]);
    putchar(s[0]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L5
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  movzbl 0x1(%rbx),%eax
  test   %al,%al
  je     L33
  movsbl 0x2(%rbx),%edi
  test   %dil,%dil
  jne    L2
L3:
  movsbl %al,%edi
  call   <putchar@plt>
  movsbl (%rbx),%edi
L33:
  pop    %rbx
  jmp    1050 <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  cmpb   $0x0,0x3(%rbx)
  je     L38
  lea    0x4(%rbx),%rdi
  call   <reverse>
  movsbl 0x3(%rbx),%edi
  call   <putchar@plt>
  movsbl 0x2(%rbx),%edi
L38:
  call   <putchar@plt>
  movzbl 0x1(%rbx),%eax
  jmp    L3
```

assistant: ```c
void reverse(char s[]) {
  if (s[0] == '\0') {
    return;
  } else {
    reverse(&s[1]);
    putchar(s[0]);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L20
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  movzbl 0x1(%rbx),%eax
  test   %al,%al
  je     L29
  movsbl 0x2(%rbx),%edi
  test   %dil,%dil
  jne    L19
L46:
  movsbl %al,%edi
  call   <putchar@plt>
  movsbl (%rbx),%edi
L29:
  pop    %rbx
  jmp    1050 <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L19:
  cmpb   $0x0,0x3(%rbx)
  je     L6
  lea    0x4(%rbx),%rdi
  call   <reverse>
  movsbl 0x3(%rbx),%edi
  call   <putchar@plt>
  movsbl 0x2(%rbx),%edi
L6:
  call   <putchar@plt>
  movzbl 0x1(%rbx),%eax
  jmp    L46
```

assistant: ```c
void reverse(char s[]) {
  if (s[0] == '\0') {
    return;
  } else {
    reverse(&s[1]);
    putchar(s[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
<sigaction_many>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L3
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L3:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <sigaction_many_ap@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int sigaction_many(const struct sigaction *sa, ...) {
  va_list ap;
  int r;
  va_start(ap, sa);
  r = sigaction_many_ap(sa, 0, ap);
  va_end(ap);
  return r;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sigaction_many>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <sigaction_many_ap@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int sigaction_many(const struct sigaction *sa, ...) {
  va_list ap;
  int r;
  va_start(ap, sa);
  r = sigaction_many_ap(sa, 0, ap);
  va_end(ap);
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sigaction_many>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  call   <sigaction_many_ap@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sigaction_many(const struct sigaction *sa, ...) {
  va_list ap;
  int r;
  va_start(ap, sa);
  r = sigaction_many_ap(sa, 0, ap);
  va_end(ap);
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sigaction_many>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  call   <sigaction_many_ap@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sigaction_many(const struct sigaction *sa, ...) {
  va_list ap;
  int r;
  va_start(ap, sa);
  r = sigaction_many_ap(sa, 0, ap);
  va_end(ap);
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<elapsed_usecs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0xf4240,%eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%rax
  sub    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int elapsed_usecs(struct timeval *t0, struct timeval *t1) {
  int elapsed = (t1->tv_sec - t0->tv_sec) * 1000000;
  elapsed += (t1->tv_usec - t0->tv_usec);
  return elapsed;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<elapsed_usecs>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  imul   $0xf4240,%eax,%eax
  sub    0x8(%rdi),%eax
  add    0x8(%rsi),%eax
  ret
```

assistant: ```c
int elapsed_usecs(struct timeval *t0, struct timeval *t1) {
  int elapsed = (t1->tv_sec - t0->tv_sec) * 1000000;
  elapsed += (t1->tv_usec - t0->tv_usec);
  return elapsed;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<elapsed_usecs>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  imul   $0xf4240,%eax,%eax
  sub    0x8(%rdi),%eax
  add    0x8(%rsi),%eax
  ret
```

assistant: ```c
int elapsed_usecs(struct timeval *t0, struct timeval *t1) {
  int elapsed = (t1->tv_sec - t0->tv_sec) * 1000000;
  elapsed += (t1->tv_usec - t0->tv_usec);
  return elapsed;
}
```
user: Translate the below assembly code into C programming language:

```asm
<elapsed_usecs>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  imul   $0xf4240,%eax,%eax
  sub    0x8(%rdi),%eax
  add    0x8(%rsi),%eax
  ret
```

assistant: ```c
int elapsed_usecs(struct timeval *t0, struct timeval *t1) {
  int elapsed = (t1->tv_sec - t0->tv_sec) * 1000000;
  elapsed += (t1->tv_usec - t0->tv_usec);
  return elapsed;
}
```