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

```asm
<fts_palloc>:
  endbr64
  push   %rbx
  mov    (%rdi),%rax
  mov    %rdi,%rbx
  mov    0x8(%rdi),%rdi
  lea    0x100(%rax,%rsi,1),%rsi
  cmp    %rsi,%rax
  jbe    L16
  test   %rdi,%rdi
  je     L32
  call   <free@plt>
L32:
  mov    D4(%rip),%rax
  movq   $0x0,0x8(%rbx)
  mov    (%rax),%edx
  mov    D15(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %rsi,(%rbx)
  call   <realloc@plt>
  test   %rax,%rax
  je     L38
  mov    %rax,0x8(%rbx)
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L38:
  mov    0x8(%rbx),%rdi
  test   %rdi,%rdi
  je     L15
  call   <free@plt>
L15:
  movq   $0x0,0x8(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int fts_palloc(FTS *sp, size_t more) {
  char *p;
  more += 256;
  if (sp->fts_pathlen + more < sp->fts_pathlen) {
    if (sp->fts_path)
      free(sp->fts_path);
    sp->fts_path = NULL;
    errno = ENAMETOOLONG;
    return (1);
  }
  sp->fts_pathlen += more;
  p = realloc(sp->fts_path, sp->fts_pathlen);
  if (p == NULL) {
    if (sp->fts_path)
      free(sp->fts_path);
    sp->fts_path = NULL;
    return (1);
  }
  sp->fts_path = p;
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<B_to_A>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x98(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x98(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x64(%rbp)
  movl   $0x0,-0x80(%rbp)
  movl   $0x0,-0x7c(%rbp)
  movl   $0x0,-0x78(%rbp)
  jmp    L72
L15:
  movl   $0x0,-0x74(%rbp)
  jmp    L63
L87:
  mov    -0x7c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x98(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x30(%rax),%edx
  mov    -0x74(%rbp),%eax
  cltq
  mov    %edx,-0x50(%rbp,%rax,4)
  addl   $0x1,-0x7c(%rbp)
  addl   $0x1,-0x74(%rbp)
L63:
  cmpl   $0x7,-0x74(%rbp)
  jle    L87
  movl   $0x7,-0x70(%rbp)
  movl   $0x0,-0x6c(%rbp)
  jmp    L73
L5:
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x70(%rbp),%edx
  mov    %edx,-0x30(%rbp,%rax,4)
  subl   $0x1,-0x70(%rbp)
  addl   $0x1,-0x6c(%rbp)
L73:
  cmpl   $0x7,-0x6c(%rbp)
  jle    L5
  movl   $0x0,-0x68(%rbp)
  jmp    L70
L79:
  mov    -0x68(%rbp),%eax
  cltq
  mov    -0x50(%rbp,%rax,4),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x60(%rbp)
  mov    -0x68(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x58(%rbp)
  mov    -0x58(%rbp),%rax
  movq   %rax,%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm1
  mulsd  -0x60(%rbp),%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x80(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x80(%rbp)
  addl   $0x1,-0x68(%rbp)
L70:
  cmpl   $0x7,-0x68(%rbp)
  jle    L79
  mov    -0x80(%rbp),%eax
  mov    %al,-0x81(%rbp)
  movl   $0x0,-0x80(%rbp)
  addl   $0x1,-0x78(%rbp)
L72:
  mov    -0x64(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  cmp    %eax,-0x78(%rbp)
  jl     L15
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

assistant: ```c
void B_to_A(const char *input) {
  int length = strlen(input);
  int binary[8];
  int asciiNum = 0;
  char ascii;
  int z = 0;
  int x;
  for (x = 0; x < length / 8; x++) {
    int a;
    for (a = 0; a < 8; a++) {
      binary[a] = (int)input[z] - 48;
      z++;
    }
    int power[8];
    int counter = 7;
    int x;
    for (x = 0; x < 8; x++) {
      power[x] = counter;
      counter--;
    }
    int y;
    for (y = 0; y < 8; y++) {
      double a = binary[y];
      double b = power[y];
      asciiNum += a * pow(2, b);
    }
    ascii = asciiNum;
    asciiNum = 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<B_to_A>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  lea    0x7(%rax),%r13d
  test   %eax,%eax
  cmovns %eax,%r13d
  sar    $0x3,%r13d
  mov    $0x0,%r12d
  lea    0x20(%rsp),%rbp
  cmp    $0x7,%eax
  jg     L27
L11:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  add    $0x1,%r12d
  cmp    %r13d,%r12d
  jge    L11
L27:
  movl   $0x7,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0x5,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x3,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  mov    %rsp,%rbx
L12:
  pxor   %xmm0,%xmm0
  cvtsi2sdl (%rbx),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L12
  jmp    L25
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void B_to_A(const char *input) {
  int length = strlen(input);
  int binary[8];
  int asciiNum = 0;
  char ascii;
  int z = 0;
  int x;
  for (x = 0; x < length / 8; x++) {
    int a;
    for (a = 0; a < 8; a++) {
      binary[a] = (int)input[z] - 48;
      z++;
    }
    int power[8];
    int counter = 7;
    int x;
    for (x = 0; x < 8; x++) {
      power[x] = counter;
      counter--;
    }
    int y;
    for (y = 0; y < 8; y++) {
      double a = binary[y];
      double b = power[y];
      asciiNum += a * pow(2, b);
    }
    ascii = asciiNum;
    asciiNum = 0;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<B_to_A>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  lea    0x14(%rsp),%rcx
  test   %eax,%eax
  lea    0x7(%rax),%r12d
  mov    %rcx,0x8(%rsp)
  cmovns %eax,%r12d
  sar    $0x3,%r12d
  cmp    $0x7,%eax
  jle    L37
  xor    %ebp,%ebp
  lea    0x30(%rsp),%rbx
  movabs $0x400000005,%r14
  movabs $0x200000003,%r13
  nop
L0:
  mov    %r14,0x18(%rsp)
  mov    0x8(%rsp),%r15
  mov    $0x7,%edx
  movl   $0x6,0x14(%rsp)
  mov    %r13,0x20(%rsp)
  movq   $0x1,0x28(%rsp)
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L34:
  mov    (%r15),%edx
  add    $0x4,%r15
L17:
  pxor   %xmm0,%xmm0
  mov    $0x2,%edi
  cvtsi2sd %edx,%xmm0
  call   <pow@plt>
  cmp    %rbx,%r15
  jne    L34
  add    $0x1,%ebp
  cmp    %ebp,%r12d
  jg     L0
L37:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void B_to_A(const char *input) {
  int length = strlen(input);
  int binary[8];
  int asciiNum = 0;
  char ascii;
  int z = 0;
  int x;
  for (x = 0; x < length / 8; x++) {
    int a;
    for (a = 0; a < 8; a++) {
      binary[a] = (int)input[z] - 48;
      z++;
    }
    int power[8];
    int counter = 7;
    int x;
    for (x = 0; x < 8; x++) {
      power[x] = counter;
      counter--;
    }
    int y;
    for (y = 0; y < 8; y++) {
      double a = binary[y];
      double b = power[y];
      asciiNum += a * pow(2, b);
    }
    ascii = asciiNum;
    asciiNum = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<B_to_A>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  test   %eax,%eax
  lea    0x7(%rax),%r13d
  cmovns %eax,%r13d
  sar    $0x3,%r13d
  cmp    $0x7,%eax
  jle    L34
  xor    %r12d,%r12d
  mov    %rsp,%r14
  lea    0x20(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L21:
  movdqa D12(%rip),%xmm1
  movdqa D13(%rip),%xmm2
  mov    %r14,%rbx
  movaps %xmm1,(%rsp)
  movaps %xmm2,0x10(%rsp)
  nopl   0x0(%rax)
L4:
  pxor   %xmm0,%xmm0
  mov    $0x2,%edi
  add    $0x4,%rbx
  cvtsi2sdl -0x4(%rbx),%xmm0
  call   <pow@plt>
  cmp    %rbp,%rbx
  jne    L4
  add    $0x1,%r12d
  cmp    %r12d,%r13d
  jg     L21
L34:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D12
datatype: byte[16]
name: parse_data
arguments: data_label: D13
datatype: byte[16]

tool: {"D12": [0x07,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D13": [0x03,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void B_to_A(const char *input) {
  int length = strlen(input);
  int binary[8];
  int asciiNum = 0;
  char ascii;
  int z = 0;
  int x;
  for (x = 0; x < length / 8; x++) {
    int a;
    for (a = 0; a < 8; a++) {
      binary[a] = (int)input[z] - 48;
      z++;
    }
    int power[8];
    int counter = 7;
    int x;
    for (x = 0; x < 8; x++) {
      power[x] = counter;
      counter--;
    }
    int y;
    for (y = 0; y < 8; y++) {
      double a = binary[y];
      double b = power[y];
      asciiNum += a * pow(2, b);
    }
    ascii = asciiNum;
    asciiNum = 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L11:
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L30
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L30
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L11
L30:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L0
L7:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jle    L64
  mov    $0x1,%eax
  jmp    L67
L64:
  mov    $0xffffffff,%eax
L67:
  mov    %eax,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strcmp(const char *a, const char *b) {
  int ret = 0;
  int i;
  for (i = 0; a[i] && b[i] && (a[i] == b[i]); i++)
    ;
  if (a[i] || b[i])
    ret = (a[i] > b[i]) ? 1 : -1;
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  mov    $0x0,%eax
  test   %dl,%dl
  je     L0
L15:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %dl,%cl
  jne    L13
  test   %cl,%cl
  je     L13
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L15
L0:
  mov    $0x0,%edx
  cmpb   $0x0,(%rsi,%rax,1)
  je     L31
  jmp    L33
L13:
  test   %dl,%dl
  je     L0
L33:
  cmp    %dl,(%rsi,%rax,1)
  setl   %dl
  movzbl %dl,%edx
  lea    -0x1(%rdx,%rdx,1),%edx
L31:
  mov    %edx,%eax
  ret
```

assistant: ```c
int strcmp(const char *a, const char *b) {
  int ret = 0;
  int i;
  for (i = 0; a[i] && b[i] && (a[i] == b[i]); i++)
    ;
  if (a[i] || b[i])
    ret = (a[i] > b[i]) ? 1 : -1;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L53
  jmp    L26
  nop
L41:
  test   %cl,%cl
  je     L56
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L8
L53:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %dl,%cl
  je     L41
L56:
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
  nopl   0x0(%rax)
L8:
  add    %rax,%rsi
L26:
  movzbl (%rsi),%ecx
  test   %cl,%cl
  je     L52
  xor    %edx,%edx
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
L52:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcmp(const char *a, const char *b) {
  int ret = 0;
  int i;
  for (i = 0; a[i] && b[i] && (a[i] == b[i]); i++)
    ;
  if (a[i] || b[i])
    ret = (a[i] > b[i]) ? 1 : -1;
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L59
  jmp    L43
  nop
L27:
  test   %cl,%cl
  je     L47
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L20
L59:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %dl,%cl
  je     L27
L47:
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
  nopl   0x0(%rax)
L20:
  add    %rax,%rsi
L43:
  movzbl (%rsi),%ecx
  test   %cl,%cl
  je     L39
  xor    %edx,%edx
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  ret
L39:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcmp(const char *a, const char *b) {
  int ret = 0;
  int i;
  for (i = 0; a[i] && b[i] && (a[i] == b[i]); i++)
    ;
  if (a[i] || b[i])
    ret = (a[i] > b[i]) ? 1 : -1;
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Curl_md5it>:
  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
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <MD5_Init@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <MD5_Update@plt>
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <MD5_Final@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void Curl_md5it(unsigned char *outbuffer, const unsigned char *input) {
  MD5_CTX ctx;
  MD5_Init(&ctx);
  MD5_Update(&ctx, input, (unsigned int)strlen((char *)input));
  MD5_Final(outbuffer, &ctx);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Curl_md5it>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <MD5_Init@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <MD5_Update@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <MD5_Final@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Curl_md5it(unsigned char *outbuffer, const unsigned char *input) {
  MD5_CTX ctx;
  MD5_Init(&ctx);
  MD5_Update(&ctx, input, (unsigned int)strlen((char *)input));
  MD5_Final(outbuffer, &ctx);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Curl_md5it>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  mov    %r13,%rdi
  call   <MD5_Init@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  call   <MD5_Update@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <MD5_Final@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Curl_md5it(unsigned char *outbuffer, const unsigned char *input) {
  MD5_CTX ctx;
  MD5_Init(&ctx);
  MD5_Update(&ctx, input, (unsigned int)strlen((char *)input));
  MD5_Final(outbuffer, &ctx);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Curl_md5it>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  mov    %r13,%rdi
  call   <MD5_Init@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  call   <MD5_Update@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <MD5_Final@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Curl_md5it(unsigned char *outbuffer, const unsigned char *input) {
  MD5_CTX ctx;
  MD5_Init(&ctx);
  MD5_Update(&ctx, input, (unsigned int)strlen((char *)input));
  MD5_Final(outbuffer, &ctx);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ceph_stable_mod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  and    -0xc(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L3
  mov    -0x4(%rbp),%eax
  and    -0xc(%rbp),%eax
  jmp    L2
L3:
  mov    -0xc(%rbp),%eax
  sar    %eax
  and    -0x4(%rbp),%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int ceph_stable_mod(int x, int b, int bmask) {
  if ((x & bmask) < b)
    return x & bmask;
  else
    return x & (bmask >> 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ceph_stable_mod>:
  endbr64
  mov    %edi,%eax
  and    %edx,%eax
  sar    %edx
  and    %edi,%edx
  cmp    %esi,%eax
  cmovge %edx,%eax
  ret
```

assistant: ```c
int ceph_stable_mod(int x, int b, int bmask) {
  if ((x & bmask) < b)
    return x & bmask;
  else
    return x & (bmask >> 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ceph_stable_mod>:
  endbr64
  mov    %edi,%eax
  and    %edx,%eax
  sar    %edx
  and    %edi,%edx
  cmp    %esi,%eax
  cmovge %edx,%eax
  ret
```

assistant: ```c
int ceph_stable_mod(int x, int b, int bmask) {
  if ((x & bmask) < b)
    return x & bmask;
  else
    return x & (bmask >> 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x4,%edi
  call   <omp_set_num_threads@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,-0x14(%rbp)
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x14(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tarea_uno@plt>
  pxor   %xmm3,%xmm3
  cvtsi2sdl -0x14(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tarea_dos@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x10(%rbp),%xmm0
  pxor   %xmm4,%xmm4
  cvtsi2sdl -0x14(%rbp),%xmm4
  movq   %xmm4,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  subsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D32
type_guessed: string
name: retrieve_data
arguments: label: D13
type_guessed: string
name: retrieve_data
arguments: label: D6
type_guessed: string
name: retrieve_data
arguments: label: D21
type_guessed: string

tool: D32: Hilo %i ejecutando tarea 1 


tool: D13: Hilo %i ejecutando tarea 2



tool: D6: Tiempo de salida hilo %i = %.4lf segundos.



tool: D21: Tiempo tardado = %.4lf segundos



assistant: ```c
int main() {
  double timeIni, timeFin, time_s1, time_s2;
  int nthread;
  omp_set_num_threads(4);
  timeIni = omp_get_wtime();
#pragma omp parallel private(nthread)
  {
    nthread = omp_get_thread_num();
#pragma omp sections nowait
    {
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 1 \n", nthread);
        tarea_uno();
      }
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 2\n\n", nthread);
        tarea_dos();
      }
    }
    printf("Tiempo de salida hilo %i = %.4lf segundos.\n\n", nthread, (omp_get_wtime() - timeIni));
  }
  timeFin = omp_get_wtime();
  printf("Tiempo tardado = %.4lf segundos\n\n", (timeFin - timeIni));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    $0x4,%edi
  call   <omp_set_num_threads@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  movsd  %xmm0,0x8(%rsp)
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movq   %xmm2,%rbx
  movapd %xmm2,%xmm0
  lea    D18(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tarea_uno@plt>
  movq   %rbx,%xmm0
  lea    D37(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tarea_dos@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  movapd %xmm0,%xmm1
  subsd  0x8(%rsp),%xmm1
  movq   %rbx,%xmm0
  lea    D35(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <omp_get_wtime@plt>
  subsd  0x8(%rsp),%xmm0
  lea    D8(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D18
type_infered: string
name: unpack_data
arguments: tag: D37
type_infered: string
name: unpack_data
arguments: tag: D35
type_infered: string
name: unpack_data
arguments: tag: D8
type_infered: string

tool: D18: Hilo %i ejecutando tarea 1 


tool: D37: Hilo %i ejecutando tarea 2



tool: D35: Tiempo de salida hilo %i = %.4lf segundos.



tool: D8: Tiempo tardado = %.4lf segundos



assistant: ```c
int main() {
  double timeIni, timeFin, time_s1, time_s2;
  int nthread;
  omp_set_num_threads(4);
  timeIni = omp_get_wtime();
#pragma omp parallel private(nthread)
  {
    nthread = omp_get_thread_num();
#pragma omp sections nowait
    {
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 1 \n", nthread);
        tarea_uno();
      }
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 2\n\n", nthread);
        tarea_dos();
      }
    }
    printf("Tiempo de salida hilo %i = %.4lf segundos.\n\n", nthread, (omp_get_wtime() - timeIni));
  }
  timeFin = omp_get_wtime();
  printf("Tiempo tardado = %.4lf segundos\n\n", (timeFin - timeIni));
  return 0;
}
```