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

```asm
<uStrCat>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <uStrLen@plt>
  mov    %r12,%rdi
  movslq %eax,%rbp
  call   <uStrLen@plt>
  test   %eax,%eax
  js     L2
  cltq
  xor    %edx,%edx
  lea    (%rbx,%rbp,1),%rdi
  nopw   0x0(%rax,%rax,1)
L10:
  movzbl (%r12,%rdx,1),%ecx
  mov    %cl,(%rdi,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rcx,%rax
  jne    L10
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void uStrCat(char *str1, const char *str2) {
  int len1 = uStrLen(str1);
  int len2 = uStrLen(str2);
  int i;
  for (i = 0; i <= len2; i++) {
    str1[i + len1] = str2[i];
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<uStrCat>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <uStrLen@plt>
  mov    %rbx,%rdi
  mov    %eax,%r12d
  call   <uStrLen@plt>
  movslq %eax,%rdx
  test   %edx,%edx
  js     L16
  movslq %r12d,%rcx
  lea    0x1(%rbx),%rsi
  add    %rbp,%rcx
  mov    %rcx,%rax
  sub    %rsi,%rax
  cmp    $0xe,%rax
  jbe    L32
  cmp    $0x6,%edx
  jbe    L32
  lea    0x1(%rdx),%edi
  cmp    $0xe,%edx
  jbe    L139
  mov    %edi,%esi
  xor    %eax,%eax
  shr    $0x4,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax,%rax,1)
L60:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L60
  mov    %edi,%esi
  and    $0xfffffff0,%esi
  mov    %esi,%eax
  cmp    %esi,%edi
  je     L16
  mov    %edx,%r8d
  sub    %esi,%r8d
  lea    0x1(%r8),%edi
  cmp    $0x6,%r8d
  jbe    L50
L19:
  mov    (%rbx,%rsi,1),%r8
  mov    %r8,(%rcx,%rsi,1)
  mov    %edi,%ecx
  and    $0xfffffff8,%ecx
  add    %ecx,%eax
  cmp    %ecx,%edi
  je     L16
L50:
  movslq %eax,%rcx
  movzbl (%rbx,%rcx,1),%esi
  lea    (%r12,%rax,1),%ecx
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%edx
  jl     L16
  movslq %ecx,%rsi
  add    %r12d,%ecx
  movzbl (%rbx,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  lea    0x2(%rax),%ecx
  cmp    %ecx,%edx
  jl     L16
  movslq %ecx,%rsi
  add    %r12d,%ecx
  movzbl (%rbx,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  lea    0x3(%rax),%ecx
  cmp    %ecx,%edx
  jl     L16
  movslq %ecx,%rsi
  add    %r12d,%ecx
  movzbl (%rbx,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  lea    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jl     L16
  movslq %ecx,%rsi
  add    %r12d,%ecx
  movzbl (%rbx,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  lea    0x5(%rax),%ecx
  cmp    %ecx,%edx
  jl     L16
  movslq %ecx,%rsi
  add    %r12d,%ecx
  add    $0x6,%eax
  movzbl (%rbx,%rsi,1),%esi
  movslq %ecx,%rcx
  mov    %sil,0x0(%rbp,%rcx,1)
  cmp    %eax,%edx
  jl     L16
  movslq %eax,%rdx
  add    %r12d,%eax
  movzbl (%rbx,%rdx,1),%edx
  cltq
  mov    %dl,0x0(%rbp,%rax,1)
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L145:
  movzbl (%rbx,%rax,1),%esi
  mov    %sil,(%rcx,%rax,1)
  mov    %rax,%rsi
  add    $0x1,%rax
  cmp    %rdx,%rsi
  jne    L145
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L139:
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L19
```

assistant: ```c
void uStrCat(char *str1, const char *str2) {
  int len1 = uStrLen(str1);
  int len2 = uStrLen(str2);
  int i;
  for (i = 0; i <= len2; i++) {
    str1[i + len1] = str2[i];
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<align>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  add    %rax,-0x18(%rbp)
L2:
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uintptr_t align(uintptr_t ptr, size_t align) {
  uintptr_t mod = ptr % align;
  if (mod) {
    ptr -= mod;
    ptr += align;
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<align>:
  endbr64
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rsi
  add    %rdi,%rsi
  sub    %rdx,%rsi
  test   %rdx,%rdx
  cmovne %rsi,%rdi
  mov    %rdi,%rax
  ret
```

assistant: ```c
uintptr_t align(uintptr_t ptr, size_t align) {
  uintptr_t mod = ptr % align;
  if (mod) {
    ptr -= mod;
    ptr += align;
  }
  return ptr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<align>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  mov    %rdi,%r8
  div    %rsi
  add    %rdi,%rsi
  sub    %rdx,%rsi
  test   %rdx,%rdx
  cmovne %rsi,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
uintptr_t align(uintptr_t ptr, size_t align) {
  uintptr_t mod = ptr % align;
  if (mod) {
    ptr -= mod;
    ptr += align;
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<align>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  mov    %rdi,%r8
  div    %rsi
  add    %rdi,%rsi
  sub    %rdx,%rsi
  test   %rdx,%rdx
  cmovne %rsi,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
uintptr_t align(uintptr_t ptr, size_t align) {
  uintptr_t mod = ptr % align;
  if (mod) {
    ptr -= mod;
    ptr += align;
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<skipchar_s>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <skips@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L7
  mov    -0x8(%rbp),%rax
  jmp    L13
L7:
  movsbl -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <skipchar@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <skips@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L13:
  leave
  ret
```

assistant: ```c
char *skipchar_s(char *in, char c) {
  in = skips(in);
  if (*in == '\n')
    return in;
  in = skipchar(in, c);
  in = skips(in);
  return in;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<skipchar_s>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <skips@plt>
  cmpb   $0xa,(%rax)
  je     L9
  movsbl %bl,%esi
  mov    %rax,%rdi
  call   <skipchar@plt>
  mov    %rax,%rdi
  call   <skips@plt>
L9:
  pop    %rbx
  ret
```

assistant: ```c
char *skipchar_s(char *in, char c) {
  in = skips(in);
  if (*in == '\n')
    return in;
  in = skipchar(in, c);
  in = skips(in);
  return in;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<skipchar_s>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <skips@plt>
  cmpb   $0xa,(%rax)
  mov    %rax,%rdi
  je     L2
  movsbl %bl,%esi
  call   <skipchar@plt>
  pop    %rbx
  mov    %rax,%rdi
  jmp    1060 <skips@plt>
  nopl   (%rax)
L2:
  pop    %rbx
  ret
```

assistant: ```c
char *skipchar_s(char *in, char c) {
  in = skips(in);
  if (*in == '\n')
    return in;
  in = skipchar(in, c);
  in = skips(in);
  return in;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<skipchar_s>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <skips@plt>
  cmpb   $0xa,(%rax)
  mov    %rax,%rdi
  je     L5
  movsbl %bl,%esi
  call   <skipchar@plt>
  pop    %rbx
  mov    %rax,%rdi
  jmp    1060 <skips@plt>
  nopl   (%rax)
L5:
  pop    %rbx
  ret
```

assistant: ```c
char *skipchar_s(char *in, char c) {
  in = skips(in);
  if (*in == '\n')
    return in;
  in = skipchar(in, c);
  in = skips(in);
  return in;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<padcon_clear_parameters>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x2e0,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
int padcon_clear_parameters(void) {
  int fail, *a, *b;
  double *c;
  fail = _ada_use_c2phc4c(736, a, b, c, 0);
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<padcon_clear_parameters>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x2e0,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int padcon_clear_parameters(void) {
  int fail, *a, *b;
  double *c;
  fail = _ada_use_c2phc4c(736, a, b, c, 0);
  return fail;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<padcon_clear_parameters>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x2e0,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int padcon_clear_parameters(void) {
  int fail, *a, *b;
  double *c;
  fail = _ada_use_c2phc4c(736, a, b, c, 0);
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<padcon_clear_parameters>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x2e0,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int padcon_clear_parameters(void) {
  int fail, *a, *b;
  double *c;
  fail = _ada_use_c2phc4c(736, a, b, c, 0);
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L16:
  mov    $0x0,%eax
  call   <foo@plt>
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0x270f,-0x4(%rbp)
  jle    L16
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10000; i++)
    foo();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
L2:
  mov    $0x0,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L2
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10000; i++)
    foo();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L4
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10000; i++)
    foo();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L3
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10000; i++)
    foo();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<radixSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %esi,-0x8c(%rbp)
  mov    %edx,-0x90(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x90(%rbp),%edx
  mov    -0x8c(%rbp),%ecx
  mov    -0x88(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <maxInArray@plt>
  mov    %eax,-0x64(%rbp)
  movl   $0xa,-0x6c(%rbp)
  movl   $0x1,-0x68(%rbp)
  jmp    L9
L23:
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x8c(%rbp),%eax
  mov    %eax,-0x74(%rbp)
  jmp    L28
L41:
  mov    -0x74(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x74(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x6c(%rbp)
  mov    %edx,%eax
  cltd
  idivl  -0x68(%rbp)
  lea    -0x60(%rbp),%rdx
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <insert@plt>
  mov    -0x74(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x74(%rbp)
L28:
  mov    -0x74(%rbp),%eax
  cmp    %eax,-0x90(%rbp)
  jg     L41
  movl   $0x0,-0x70(%rbp)
  mov    -0x8c(%rbp),%eax
  mov    %eax,-0x74(%rbp)
  jmp    L14
L47:
  mov    -0x70(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,8),%rax
  lea    -0x74(%rbp),%rdx
  mov    -0x88(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <displayList@plt>
  addl   $0x1,-0x70(%rbp)
L14:
  cmpl   $0x9,-0x70(%rbp)
  jle    L47
  mov    -0x68(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x68(%rbp)
  mov    -0x6c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x6c(%rbp)
L9:
  mov    -0x68(%rbp),%eax
  cmp    -0x64(%rbp),%eax
  jl     L23
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

assistant: ```c
void radixSort(int *a, int start, int end) {
  int max = maxInArray(a, start, end);
  int j, k, p, pp;
  for (p = 10, pp = 1; pp < max; pp *= 10, p *= 10) {
    struct node *b[10] = {NULL};
    for (j = start; j < end; j++) {
      insert(&b[(int)((a[j] % p) / pp)], a[j]);
    }
    for (k = 0, j = start; k < 10; k++) {
      displayList(b[k], a, &j);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<radixSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %rdi,%r13
  mov    %esi,0x18(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  call   <maxInArray@plt>
  mov    %eax,0x1c(%rsp)
  cmp    $0x1,%eax
  jle    L45
  mov    $0x1,%r12d
  mov    $0xa,%ebp
  lea    0x30(%rsp),%r15
L2:
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x0,0x78(%rsp)
  mov    0x18(%rsp),%eax
  mov    %eax,0x2c(%rsp)
  cmp    0x14(%rsp),%eax
  jge    L24
L20:
  cltq
  mov    0x0(%r13,%rax,4),%esi
  mov    %esi,%eax
  cltd
  idiv   %ebp
  mov    %edx,%eax
  cltd
  idiv   %r12d
  cltq
  lea    (%r15,%rax,8),%rdi
  call   <insert@plt>
  mov    0x2c(%rsp),%eax
  add    $0x1,%eax
  mov    %eax,0x2c(%rsp)
  cmp    0x14(%rsp),%eax
  jl     L20
L24:
  mov    0x18(%rsp),%eax
  mov    %eax,0x2c(%rsp)
  lea    0x30(%rsp),%rbx
  lea    0x80(%rsp),%r14
  lea    0x2c(%rsp),%rax
  mov    %rax,0x8(%rsp)
L16:
  mov    0x8(%rsp),%rdx
  mov    %r13,%rsi
  mov    (%rbx),%rdi
  call   <displayList@plt>
  add    $0x8,%rbx
  cmp    %r14,%rbx
  jne    L16
  lea    (%r12,%r12,4),%eax
  add    %eax,%eax
  mov    %eax,%r12d
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebp,%ebp
  cmp    %eax,0x1c(%rsp)
  jg     L2
L45:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void radixSort(int *a, int start, int end) {
  int max = maxInArray(a, start, end);
  int j, k, p, pp;
  for (p = 10, pp = 1; pp < max; pp *= 10, p *= 10) {
    struct node *b[10] = {NULL};
    for (j = start; j < end; j++) {
      insert(&b[(int)((a[j] % p) / pp)], a[j]);
    }
    for (k = 0, j = start; k < 10; k++) {
      displayList(b[k], a, &j);
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<radixSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %esi,0x10(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  call   <maxInArray@plt>
  mov    %eax,0x14(%rsp)
  cmp    $0x1,%eax
  jle    L2
  lea    0x38(%rsp),%rax
  mov    $0x1,%ebx
  mov    $0xa,%ebp
  mov    %rax,0x18(%rsp)
  lea    0x80(%rsp),%r14
  lea    0x2c(%rsp),%r13
  cs nopw 0x0(%rax,%rax,1)
L63:
  mov    0x10(%rsp),%eax
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x30(%rsp)
  mov    %eax,0x2c(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  cmp    0xc(%rsp),%eax
  jge    L1
  lea    0x30(%rsp),%r15
  nopl   0x0(%rax)
L41:
  cltq
  mov    (%r12,%rax,4),%esi
  mov    %esi,%eax
  cltd
  idiv   %ebp
  mov    %edx,%eax
  cltd
  idiv   %ebx
  cltq
  lea    (%r15,%rax,8),%rdi
  call   <insert@plt>
  mov    0x2c(%rsp),%eax
  add    $0x1,%eax
  mov    %eax,0x2c(%rsp)
  cmp    0xc(%rsp),%eax
  jl     L41
  mov    0x30(%rsp),%rdi
L59:
  mov    0x10(%rsp),%eax
  mov    0x18(%rsp),%r15
  mov    %eax,0x2c(%rsp)
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L4:
  mov    (%r15),%rdi
  add    $0x8,%r15
L7:
  mov    %r13,%rdx
  mov    %r12,%rsi
  call   <displayList@plt>
  cmp    %r15,%r14
  jne    L4
  lea    (%rbx,%rbx,4),%ebx
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebx,%ebx
  add    %ebp,%ebp
  cmp    %ebx,0x14(%rsp)
  jg     L63
L2:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  xor    %edi,%edi
  jmp    L59
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void radixSort(int *a, int start, int end) {
  int max = maxInArray(a, start, end);
  int j, k, p, pp;
  for (p = 10, pp = 1; pp < max; pp *= 10, p *= 10) {
    struct node *b[10] = {NULL};
    for (j = start; j < end; j++) {
      insert(&b[(int)((a[j] % p) / pp)], a[j]);
    }
    for (k = 0, j = start; k < 10; k++) {
      displayList(b[k], a, &j);
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<radixSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %esi,0x18(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  call   <maxInArray@plt>
  mov    %eax,0x1c(%rsp)
  cmp    $0x1,%eax
  jle    L50
  mov    $0x1,%ebx
  mov    $0xa,%ebp
  lea    0x30(%rsp),%r15
  lea    0x80(%rsp),%rax
  lea    0x2c(%rsp),%r13
  mov    %rax,0x8(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L51:
  mov    0x18(%rsp),%eax
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x30(%rsp)
  mov    %eax,0x2c(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  cmp    0x14(%rsp),%eax
  jge    L48
  nopl   0x0(%rax,%rax,1)
L56:
  cltq
  mov    (%r12,%rax,4),%esi
  mov    %esi,%eax
  cltd
  idiv   %ebp
  mov    %edx,%eax
  cltd
  idiv   %ebx
  cltq
  lea    (%r15,%rax,8),%rdi
  call   <insert@plt>
  mov    0x2c(%rsp),%eax
  add    $0x1,%eax
  mov    %eax,0x2c(%rsp)
  cmp    0x14(%rsp),%eax
  jl     L56
L48:
  mov    0x18(%rsp),%eax
  mov    %r15,%r14
  mov    %eax,0x2c(%rsp)
  nopw   0x0(%rax,%rax,1)
L43:
  mov    (%r14),%rdi
  mov    %r13,%rdx
  mov    %r12,%rsi
  add    $0x8,%r14
  call   <displayList@plt>
  cmp    %r14,0x8(%rsp)
  jne    L43
  lea    (%rbx,%rbx,4),%ebx
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebx,%ebx
  add    %ebp,%ebp
  cmp    %ebx,0x1c(%rsp)
  jg     L51
L50:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void radixSort(int *a, int start, int end) {
  int max = maxInArray(a, start, end);
  int j, k, p, pp;
  for (p = 10, pp = 1; pp < max; pp *= 10, p *= 10) {
    struct node *b[10] = {NULL};
    for (j = start; j < end; j++) {
      insert(&b[(int)((a[j] % p) / pp)], a[j]);
    }
    for (k = 0, j = start; k < 10; k++) {
      displayList(b[k], a, &j);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fc8300_es1_get_rssi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %rdx,-0x28(%rbp)
  mov    %ax,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0xd(%rbp)
  movl   $0x0,-0xc(%rbp)
  movzwl -0x1c(%rbp),%esi
  lea    -0xd(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0xd7,%edx
  mov    %rax,%rdi
  call   <fc8300_read@plt>
  movzbl -0xd(%rbp),%eax
  movzbl %al,%eax
  sub    $0x100,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int fc8300_es1_get_rssi(void *handle, unsigned short devid, int *rssi) {
  unsigned char reg_value = 0;
  int rssi_value = 0;
  fc8300_read(handle, devid, 0xd7, &reg_value);
  rssi_value = reg_value - 256;
  *rssi = rssi_value;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fc8300_es1_get_rssi>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x7(%rsp)
  lea    0x7(%rsp),%rcx
  movzwl %si,%esi
  mov    $0xd7,%edx
  call   <fc8300_read@plt>
  movzbl 0x7(%rsp),%eax
  sub    $0x100,%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fc8300_es1_get_rssi(void *handle, unsigned short devid, int *rssi) {
  unsigned char reg_value = 0;
  int rssi_value = 0;
  fc8300_read(handle, devid, 0xd7, &reg_value);
  rssi_value = reg_value - 256;
  *rssi = rssi_value;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fc8300_es1_get_rssi>:
  endbr64
  push   %rbx
  movzwl %si,%esi
  mov    %rdx,%rbx
  mov    $0xd7,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rcx
  movb   $0x0,0x7(%rsp)
  call   <fc8300_read@plt>
  movzbl 0x7(%rsp),%eax
  sub    $0x100,%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fc8300_es1_get_rssi(void *handle, unsigned short devid, int *rssi) {
  unsigned char reg_value = 0;
  int rssi_value = 0;
  fc8300_read(handle, devid, 0xd7, &reg_value);
  rssi_value = reg_value - 256;
  *rssi = rssi_value;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fc8300_es1_get_rssi>:
  endbr64
  push   %rbx
  movzwl %si,%esi
  mov    %rdx,%rbx
  mov    $0xd7,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rcx
  movb   $0x0,0x7(%rsp)
  call   <fc8300_read@plt>
  movzbl 0x7(%rsp),%eax
  sub    $0x100,%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fc8300_es1_get_rssi(void *handle, unsigned short devid, int *rssi) {
  unsigned char reg_value = 0;
  int rssi_value = 0;
  fc8300_read(handle, devid, 0xd7, &reg_value);
  rssi_value = reg_value - 256;
  *rssi = rssi_value;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  leave
  ret
```

assistant: ```c
int cmp(const void *pa, const void *pb) {
  const char **x = (const char **)pa;
  const char **y = (const char **)pb;
  return strcmp(*x, *y);
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmp>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cmp(const void *pa, const void *pb) {
  const char **x = (const char **)pa;
  const char **y = (const char **)pb;
  return strcmp(*x, *y);
}
```