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

```asm
<getMIfile>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  lea    D23(%rip),%rdi
  push   %r14
  mov    %rsi,%r14
  lea    D16(%rip),%rsi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x24,%ebx
  sub    $0x38,%rsp
  mov    %ecx,0x8(%rsp)
  mov    %r8d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  lea    D42(%rip),%rsi
  lea    D17(%rip),%rdi
  mov    %rax,%r12
  call   <fopen@plt>
  mov    %rax,%rbp
  lea    0x24(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x20(%rsp),%rax
  mov    %rax,0x10(%rsp)
  nopl   0x0(%rax)
L7:
  mov    0x10(%rsp),%rdx
  mov    0x18(%rsp),%rcx
  lea    D19(%rip),%rsi
  mov    %r12,%rdi
  call   <fscanf@plt>
  sub    $0x8,%rsp
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    0x14(%rsp),%eax
  mov    %r15,%rdi
  push   %rax
  mov    0x18(%rsp),%r9d
  mov    0x34(%rsp),%r8d
  mov    0x30(%rsp),%ecx
  call   <getMIpair@plt>
  mov    0x34(%rsp),%ecx
  mov    0x30(%rsp),%edx
  mov    %rbp,%rdi
  lea    D27(%rip),%rsi
  call   <fprintf@plt>
  pop    %rax
  pop    %rdx
  sub    $0x1,%ebx
  jne    L7
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x38,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <fclose@plt>
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D23
data_type: string
name: parse_data
arguments: label: D16
data_type: string
name: parse_data
arguments: label: D42
data_type: string
name: parse_data
arguments: label: D17
data_type: string
name: parse_data
arguments: label: D19
data_type: string
name: parse_data
arguments: label: D27
data_type: string

tool: D23: genepairs.txt

tool: D16: r

tool: D42: w

tool: D17: genepairsmi.txt

tool: D19: %d %d

tool: D27: %d %d %f


assistant: ```c
void getMIfile(const double *X, size_t ngenes, int nexp, int numbins, int splineord) {
  int i;
  int nlines = 36;
  FILE *fi, *fo;
  fi = fopen("genepairs.txt", "r");
  fo = fopen("genepairsmi.txt", "w");
  for (i = 0; i < nlines; i++) {
    int ng1, ng2;
    fscanf(fi, "%d %d", &ng1, &ng2);
    float mi = getMIpair(X, ngenes, nexp, ng1, ng2, numbins, splineord);
    fprintf(fo, "%d %d %f\n", ng1, ng2, mi);
  }
  fclose(fi);
  fclose(fo);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mystrcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void mystrcpy(char *to, char *from) {
  int i = 0;
  do {
    to[i] = from[i];
  } while (from[i++] != '\0');
}
```
user: Translate the below assembly code into C programming language:

```asm
<mystrcpy>:
  endbr64
  mov    $0x0,%eax
L0:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L0
  ret
```

assistant: ```c
void mystrcpy(char *to, char *from) {
  int i = 0;
  do {
    to[i] = from[i];
  } while (from[i++] != '\0');
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mystrcpy>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L5:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L5
  ret
```

assistant: ```c
void mystrcpy(char *to, char *from) {
  int i = 0;
  do {
    to[i] = from[i];
  } while (from[i++] != '\0');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mystrcpy>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L8:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L8
  ret
```

assistant: ```c
void mystrcpy(char *to, char *from) {
  int i = 0;
  do {
    to[i] = from[i];
  } while (from[i++] != '\0');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main_cmdline_parser_required>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main_cmdline_parser_required(struct main_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main_cmdline_parser_required>:
  endbr64
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int main_cmdline_parser_required(struct main_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main_cmdline_parser_required>:
  endbr64
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int main_cmdline_parser_required(struct main_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main_cmdline_parser_required>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int main_cmdline_parser_required(struct main_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<softfloat_isNaNF128M>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x7fff0000,%eax
  cmp    $0x7fff0000,%eax
  je     L0
  mov    $0x0,%eax
  jmp    L23
L0:
  mov    -0x4(%rbp),%eax
  movzwl %ax,%eax
  test   %eax,%eax
  jne    L39
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  or     %edx,%eax
  test   %eax,%eax
  je     L35
L39:
  mov    $0x1,%eax
  jmp    L23
L35:
  mov    $0x0,%eax
L23:
  pop    %rbp
  ret
```

assistant: ```c
bool softfloat_isNaNF128M(const uint32_t *aWPtr) {
  uint32_t uiA96;
  uiA96 = aWPtr[((4) - 1)];
  if ((uiA96 & 0x7FFF0000) != 0x7FFF0000)
    return false;
  return ((uiA96 & 0x0000FFFF) != 0) || ((aWPtr[(2)] | aWPtr[(1)] | aWPtr[(0)]) != 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<softfloat_isNaNF128M>:
  endbr64
  mov    0xc(%rdi),%edx
  mov    %edx,%ecx
  and    $0x7fff0000,%ecx
  mov    $0x0,%eax
  cmp    $0x7fff0000,%ecx
  jne    L1
  mov    $0x1,%eax
  test   %dx,%dx
  jne    L1
  mov    0x8(%rdi),%eax
  or     0x4(%rdi),%eax
  or     (%rdi),%eax
  setne  %al
  movzbl %al,%eax
L1:
  ret
```

assistant: ```c
bool softfloat_isNaNF128M(const uint32_t *aWPtr) {
  uint32_t uiA96;
  uiA96 = aWPtr[((4) - 1)];
  if ((uiA96 & 0x7FFF0000) != 0x7FFF0000)
    return false;
  return ((uiA96 & 0x0000FFFF) != 0) || ((aWPtr[(2)] | aWPtr[(1)] | aWPtr[(0)]) != 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<softfloat_isNaNF128M>:
  endbr64
  mov    0xc(%rdi),%edx
  xor    %eax,%eax
  mov    %edx,%ecx
  and    $0x7fff0000,%ecx
  cmp    $0x7fff0000,%ecx
  jne    L0
  mov    $0x1,%eax
  test   %dx,%dx
  je     L9
L0:
  ret
  nopl   0x0(%rax)
L9:
  mov    0x8(%rdi),%eax
  or     0x4(%rdi),%eax
  or     (%rdi),%eax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool softfloat_isNaNF128M(const uint32_t *aWPtr) {
  uint32_t uiA96;
  uiA96 = aWPtr[((4) - 1)];
  if ((uiA96 & 0x7FFF0000) != 0x7FFF0000)
    return false;
  return ((uiA96 & 0x0000FFFF) != 0) || ((aWPtr[(2)] | aWPtr[(1)] | aWPtr[(0)]) != 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<softfloat_isNaNF128M>:
  endbr64
  mov    0xc(%rdi),%edx
  xor    %eax,%eax
  mov    %edx,%ecx
  and    $0x7fff0000,%ecx
  cmp    $0x7fff0000,%ecx
  jne    L13
  mov    $0x1,%eax
  test   %dx,%dx
  je     L7
L13:
  ret
  nopl   0x0(%rax)
L7:
  mov    0x8(%rdi),%eax
  or     0x4(%rdi),%eax
  or     (%rdi),%eax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool softfloat_isNaNF128M(const uint32_t *aWPtr) {
  uint32_t uiA96;
  uiA96 = aWPtr[((4) - 1)];
  if ((uiA96 & 0x7FFF0000) != 0x7FFF0000)
    return false;
  return ((uiA96 & 0x0000FFFF) != 0) || ((aWPtr[(2)] | aWPtr[(1)] | aWPtr[(0)]) != 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<btostr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movzbl -0x1c(%rbp),%eax
  mov    $0x29,%edx
  mul    %dl
  shr    $0x8,%ax
  shr    $0x4,%al
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L4
  mov    -0x8(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L4:
  movzbl -0x1c(%rbp),%edx
  mov    $0x29,%ecx
  mov    %ecx,%eax
  mul    %dl
  shr    $0x8,%ax
  shr    $0x4,%al
  mov    $0x64,%ecx
  imul   %ecx,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    $0xffffffcd,%edx
  mul    %dl
  shr    $0x8,%ax
  shr    $0x3,%al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L17
  cmpl   $0x0,-0x8(%rbp)
  je     L19
L17:
  mov    -0x4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L19:
  movzbl -0x1c(%rbp),%ecx
  mov    $0xffffffcd,%edx
  mov    %edx,%eax
  mul    %cl
  shr    $0x8,%ax
  mov    %eax,%edx
  shr    $0x3,%dl
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    0x30(%rdx),%ecx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int btostr(char *p, unsigned char val) {
  unsigned int a, b, i = 0;
  if ((a = val / 100) != 0)
    *(p + (i++)) = a + '0';
  if ((b = val % 100 / 10) != 0 || a)
    *(p + (i++)) = b + '0';
  *(p + (i++)) = '0' + val % 10;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<btostr>:
  endbr64
  mov    %esi,%edx
  mov    $0x29,%ecx
  mov    %ecx,%eax
  mul    %sil
  shr    $0xc,%ax
  cmp    $0x63,%sil
  jbe    L7
  add    $0x30,%eax
  mov    %al,(%rdi)
  mov    %ecx,%eax
  mul    %sil
  shr    $0xc,%ax
  mov    $0x64,%ecx
  imul   %ecx,%eax
  mov    %esi,%ecx
  sub    %eax,%ecx
  mov    $0xffffffcd,%esi
  mov    %ecx,%eax
  mul    %sil
  shr    $0xb,%ax
  mov    $0x1,%ecx
L14:
  lea    0x1(%rcx),%esi
  mov    %ecx,%ecx
  add    $0x30,%eax
  mov    %al,(%rdi,%rcx,1)
L24:
  mov    %esi,%r8d
  mov    $0xffffffcd,%ecx
  mov    %ecx,%eax
  mul    %dl
  shr    $0xb,%ax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  add    $0x30,%edx
  mov    %dl,(%rdi,%r8,1)
  lea    0x1(%rsi),%eax
  ret
L7:
  mov    $0x29,%ecx
  mov    %ecx,%eax
  mul    %sil
  mov    %eax,%ecx
  shr    $0xc,%cx
  mov    $0x64,%esi
  mov    %ecx,%eax
  imul   %esi,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    $0x0,%esi
  cmp    $0x9,%cl
  jbe    L24
  mov    $0xffffffcd,%esi
  mov    %ecx,%eax
  mul    %sil
  shr    $0xb,%ax
  mov    $0x0,%ecx
  jmp    L14
```

assistant: ```c
int btostr(char *p, unsigned char val) {
  unsigned int a, b, i = 0;
  if ((a = val / 100) != 0)
    *(p + (i++)) = a + '0';
  if ((b = val % 100 / 10) != 0 || a)
    *(p + (i++)) = b + '0';
  *(p + (i++)) = '0' + val % 10;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<btostr>:
  endbr64
  mov    $0x29,%eax
  mov    %esi,%edx
  mul    %sil
  mov    $0x64,%esi
  mov    %eax,%ecx
  mov    %eax,%r8d
  shr    $0xc,%cx
  shr    $0x8,%r8w
  mov    %ecx,%eax
  mov    %edx,%ecx
  imul   %esi,%eax
  sub    %eax,%ecx
  mov    $0xffffffcd,%eax
  mul    %cl
  shr    $0xb,%ax
  cmp    $0x63,%dl
  jbe    L26
  shr    $0x4,%r8b
  lea    0x1(%rdi),%rsi
  mov    $0x2,%ecx
  add    $0x30,%r8d
  mov    %r8b,(%rdi)
  mov    $0x3,%r8d
L32:
  add    $0x30,%eax
  add    %rcx,%rdi
  mov    %al,(%rsi)
L25:
  mov    $0xffffffcd,%eax
  mul    %dl
  shr    $0xb,%ax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %r8d,%eax
  add    $0x30,%edx
  mov    %dl,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  cmp    $0x9,%cl
  ja     L5
  mov    $0x1,%r8d
  jmp    L25
  nopl   (%rax)
L5:
  mov    %rdi,%rsi
  mov    $0x2,%r8d
  mov    $0x1,%ecx
  jmp    L32
```

assistant: ```c
int btostr(char *p, unsigned char val) {
  unsigned int a, b, i = 0;
  if ((a = val / 100) != 0)
    *(p + (i++)) = a + '0';
  if ((b = val % 100 / 10) != 0 || a)
    *(p + (i++)) = b + '0';
  *(p + (i++)) = '0' + val % 10;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<btostr>:
  endbr64
  mov    $0x29,%eax
  mov    %esi,%edx
  mul    %sil
  mov    $0x64,%esi
  mov    %eax,%ecx
  mov    %eax,%r8d
  shr    $0xc,%cx
  shr    $0x8,%r8w
  mov    %ecx,%eax
  mov    %edx,%ecx
  imul   %esi,%eax
  sub    %eax,%ecx
  mov    $0xffffffcd,%eax
  mul    %cl
  shr    $0xb,%ax
  cmp    $0x63,%dl
  jbe    L20
  shr    $0x4,%r8b
  lea    0x1(%rdi),%rsi
  mov    $0x2,%ecx
  add    $0x30,%r8d
  mov    %r8b,(%rdi)
  mov    $0x3,%r8d
L24:
  add    $0x30,%eax
  add    %rcx,%rdi
  mov    %al,(%rsi)
L27:
  mov    $0xffffffcd,%eax
  mul    %dl
  shr    $0xb,%ax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %r8d,%eax
  add    $0x30,%edx
  mov    %dl,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  cmp    $0x9,%cl
  ja     L9
  mov    $0x1,%r8d
  jmp    L27
  nopl   (%rax)
L9:
  mov    %rdi,%rsi
  mov    $0x2,%r8d
  mov    $0x1,%ecx
  jmp    L24
```

assistant: ```c
int btostr(char *p, unsigned char val) {
  unsigned int a, b, i = 0;
  if ((a = val / 100) != 0)
    *(p + (i++)) = a + '0';
  if ((b = val % 100 / 10) != 0 || a)
    *(p + (i++)) = b + '0';
  *(p + (i++)) = '0' + val % 10;
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<utilMacCmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L149
  cmpq   $0x0,-0x20(%rbp)
  jne    L149
  mov    $0x0,%eax
  jmp    L19
L149:
  cmpq   $0x0,-0x18(%rbp)
  jne    L31
  cmpq   $0x0,-0x20(%rbp)
  je     L31
  mov    $0xffffffffffffffff,%rax
  jmp    L19
L31:
  cmpq   $0x0,-0x18(%rbp)
  je     L55
  cmpq   $0x0,-0x20(%rbp)
  jne    L55
  mov    $0x1,%eax
  jmp    L19
L55:
  movb   $0x0,-0x1(%rbp)
  jmp    L56
L39:
  movzbl -0x1(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  movzbl -0x1(%rbp),%ecx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jbe    L10
  mov    $0x1,%eax
  jmp    L19
L10:
  movzbl -0x1(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  movzbl -0x1(%rbp),%ecx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jae    L73
  mov    $0xffffffffffffffff,%rax
  jmp    L19
L73:
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L56:
  cmpb   $0x5,-0x1(%rbp)
  jbe    L39
  mov    $0x0,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
long utilMacCmp(unsigned char *byMac1, unsigned char *byMac2) {
  unsigned char iCnt;
  if (byMac1 == (0) && byMac2 == (0))
    return 0;
  if (byMac1 == (0) && byMac2 != (0))
    return -1;
  if (byMac1 != (0) && byMac2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 6; iCnt++) {
    if (byMac1[iCnt] > byMac2[iCnt])
      return 1;
    if (byMac1[iCnt] < byMac2[iCnt])
      return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<utilMacCmp>:
  endbr64
  mov    $0x0,%eax
  mov    %rdi,%rdx
  or     %rsi,%rdx
  je     L31
  test   %rdi,%rdi
  jne    L15
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  je     L15
L31:
  ret
L15:
  test   %rsi,%rsi
  jne    L64
  mov    $0x1,%eax
  test   %rdi,%rdi
  jne    L31
L64:
  mov    $0x0,%eax
L42:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,%cl
  ja     L58
  jb     L86
  add    $0x1,%rax
  cmp    $0x6,%rax
  jne    L42
  mov    $0x0,%eax
  ret
L58:
  mov    $0x1,%eax
  ret
L86:
  mov    $0xffffffffffffffff,%rax
  jmp    L31
```

assistant: ```c
long utilMacCmp(unsigned char *byMac1, unsigned char *byMac2) {
  unsigned char iCnt;
  if (byMac1 == (0) && byMac2 == (0))
    return 0;
  if (byMac1 == (0) && byMac2 != (0))
    return -1;
  if (byMac1 != (0) && byMac2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 6; iCnt++) {
    if (byMac1[iCnt] > byMac2[iCnt])
      return 1;
    if (byMac1[iCnt] < byMac2[iCnt])
      return -1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<utilMacCmp>:
  endbr64
  mov    %rdi,%rcx
  xor    %eax,%eax
  or     %rsi,%rcx
  je     L44
  test   %rdi,%rdi
  jne    L39
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  jne    L44
L5:
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L18
L44:
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  test   %rsi,%rsi
  je     L5
L18:
  xor    %eax,%eax
L1:
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,(%rdi,%rax,1)
  ja     L69
  jb     L41
  add    $0x1,%rax
  cmp    $0x6,%rax
  jne    L1
  xor    %eax,%eax
  ret
  nop
L69:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L41:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long utilMacCmp(unsigned char *byMac1, unsigned char *byMac2) {
  unsigned char iCnt;
  if (byMac1 == (0) && byMac2 == (0))
    return 0;
  if (byMac1 == (0) && byMac2 != (0))
    return -1;
  if (byMac1 != (0) && byMac2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 6; iCnt++) {
    if (byMac1[iCnt] > byMac2[iCnt])
      return 1;
    if (byMac1[iCnt] < byMac2[iCnt])
      return -1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<utilMacCmp>:
  endbr64
  mov    %rdi,%rdx
  xor    %eax,%eax
  or     %rsi,%rdx
  je     L106
  test   %rdi,%rdi
  jne    L82
  mov    $0xffffffffffffffff,%rax
  test   %rsi,%rsi
  jne    L106
L0:
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L112
L106:
  ret
  nopw   0x0(%rax,%rax,1)
L82:
  test   %rsi,%rsi
  je     L0
L112:
  movzbl (%rsi),%eax
  cmp    %al,(%rdi)
  ja     L43
  jb     L101
  movzbl 0x1(%rsi),%eax
  cmp    %al,0x1(%rdi)
  ja     L43
  jb     L101
  movzbl 0x2(%rsi),%eax
  cmp    %al,0x2(%rdi)
  ja     L43
  jb     L101
  movzbl 0x3(%rsi),%eax
  cmp    %al,0x3(%rdi)
  ja     L43
  jb     L101
  movzbl 0x4(%rsi),%eax
  cmp    %al,0x4(%rdi)
  ja     L43
  jb     L101
  movzbl 0x5(%rsi),%eax
  cmp    %al,0x5(%rdi)
  ja     L43
  sbb    %rax,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L101:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long utilMacCmp(unsigned char *byMac1, unsigned char *byMac2) {
  unsigned char iCnt;
  if (byMac1 == (0) && byMac2 == (0))
    return 0;
  if (byMac1 == (0) && byMac2 != (0))
    return -1;
  if (byMac1 != (0) && byMac2 == (0))
    return 1;
  for (iCnt = 0; iCnt < 6; iCnt++) {
    if (byMac1[iCnt] > byMac2[iCnt])
      return 1;
    if (byMac1[iCnt] < byMac2[iCnt])
      return -1;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cumSum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L5
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  add    %esi,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L2
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cumSum(int start_value, int *diff, int length, unsigned char *start_adress) {
  int i;
  ;
  ;
  start_adress[0] = start_value;
  ;
  ;
  for (i = 1; i <= length; i++) {
    ;
    ;
    start_adress[i] = start_adress[i - 1] + diff[i - 1];
  };
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cumSum>:
  endbr64
  mov    %dil,(%rcx)
  test   %edx,%edx
  jle    L11
  mov    %edx,%edx
  mov    $0x0,%eax
L4:
  movzbl (%rcx,%rax,1),%edi
  add    (%rsi,%rax,4),%dil
  mov    %dil,0x1(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L4
L11:
  ret
```

assistant: ```c
void cumSum(int start_value, int *diff, int length, unsigned char *start_adress) {
  int i;
  ;
  ;
  start_adress[0] = start_value;
  ;
  ;
  for (i = 1; i <= length; i++) {
    ;
    ;
    start_adress[i] = start_adress[i - 1] + diff[i - 1];
  };
}
```
user: What is the c source code of the assembly code below:

```asm
<cumSum>:
  endbr64
  mov    %dil,(%rcx)
  mov    %edi,%r8d
  test   %edx,%edx
  jle    L23
  movslq %edx,%rdx
  xor    %eax,%eax
  jmp    L10
  nopl   (%rax)
L17:
  movzbl (%rcx,%rax,1),%r8d
L10:
  add    (%rsi,%rax,4),%r8b
  mov    %r8b,0x1(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L23:
  ret
```

assistant: ```c
void cumSum(int start_value, int *diff, int length, unsigned char *start_adress) {
  int i;
  ;
  ;
  start_adress[0] = start_value;
  ;
  ;
  for (i = 1; i <= length; i++) {
    ;
    ;
    start_adress[i] = start_adress[i - 1] + diff[i - 1];
  };
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cumSum>:
  endbr64
  mov    %dil,(%rcx)
  mov    %edi,%r8d
  test   %edx,%edx
  jle    L14
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L15:
  add    (%rsi,%rax,4),%r8b
  mov    %r8b,0x1(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L15
L14:
  ret
```

assistant: ```c
void cumSum(int start_value, int *diff, int length, unsigned char *start_adress) {
  int i;
  ;
  ;
  start_adress[0] = start_value;
  ;
  ;
  for (i = 1; i <= length; i++) {
    ;
    ;
    start_adress[i] = start_adress[i - 1] + diff[i - 1];
  };
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<btic2f_log2u>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L15:
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  sar    %eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0x1,-0x8(%rbp)
  jg     L15
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int btic2f_log2u(int val) {
  int v, e;
  v = val;
  e = 0;
  while (v > 1) {
    v = (v + 1) >> 1;
    e++;
  }
  return (e);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<btic2f_log2u>:
  endbr64
  cmp    $0x1,%edi
  jle    L23
  mov    $0x0,%edx
L29:
  lea    0x1(%rdi),%eax
  mov    %eax,%edi
  sar    %edi
  add    $0x1,%edx
  cmp    $0x3,%eax
  jg     L29
L15:
  mov    %edx,%eax
  ret
L23:
  mov    $0x0,%edx
  jmp    L15
```

assistant: ```c
int btic2f_log2u(int val) {
  int v, e;
  v = val;
  e = 0;
  while (v > 1) {
    v = (v + 1) >> 1;
    e++;
  }
  return (e);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<btic2f_log2u>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L19
  nopl   0x0(%rax,%rax,1)
L10:
  add    $0x1,%edi
  add    $0x1,%eax
  sar    %edi
  cmp    $0x1,%edi
  jne    L10
  ret
  xchg   %ax,%ax
L19:
  ret
```

assistant: ```c
int btic2f_log2u(int val) {
  int v, e;
  v = val;
  e = 0;
  while (v > 1) {
    v = (v + 1) >> 1;
    e++;
  }
  return (e);
}
```