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

```asm
<nextPermutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x1,-0x1c(%rbp)
  jle    L110
  movl   $0xffffffff,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L76
L36:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L6
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L100
L6:
  subl   $0x1,-0x4(%rbp)
L76:
  cmpl   $0x0,-0x4(%rbp)
  jns    L36
L100:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L43
L46:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L42
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L116
L42:
  subl   $0x1,-0x4(%rbp)
L43:
  cmpl   $0x0,-0x4(%rbp)
  jns    L46
L116:
  cmpl   $0x0,-0xc(%rbp)
  js     L13
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L13:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x1,-0x1c(%rbp)
  jmp    L73
L88:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x4(%rbp)
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L73:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L88
  jmp    L26
L110:
  nop
L26:
  leave
  ret
```

assistant: ```c
void nextPermutation(int *nums, int numsSize) {
  if (numsSize <= 1)
    return;
  int ind1, ind2;
  int i;
  ind1 = ind2 = -1;
  for (i = numsSize - 2; i >= 0; i--)
    if (nums[i] < nums[i + 1]) {
      ind1 = i;
      break;
    }
  for (i = numsSize - 1; i >= 0; i--)
    if (nums[i] > nums[ind1]) {
      ind2 = i;
      break;
    }
  if (ind1 >= 0)
    swap(&nums[ind1], &nums[ind2]);
  i = ind1 + 1;
  numsSize--;
  while (i < numsSize)
    swap(&nums[i++], &nums[numsSize--]);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<nextPermutation>:
  endbr64
  cmp    $0x1,%esi
  jle    L33
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r13d
  lea    -0x2(%rsi),%eax
  cltq
L77:
  mov    %eax,%r12d
  mov    0x4(%rbp,%rax,4),%ecx
  cmp    %ecx,0x0(%rbp,%rax,4)
  jl     L73
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L77
  mov    $0xffffffff,%r12d
L73:
  lea    -0x1(%r13),%ebx
  movslq %r12d,%r14
  lea    0x0(,%r14,4),%r15
  lea    0x0(%rbp,%r15,1),%rdi
  mov    (%rdi),%ecx
  movslq %ebx,%rax
L74:
  mov    %eax,%edx
  cmp    %ecx,0x0(%rbp,%rax,4)
  jg     L31
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L74
  mov    $0xffffffff,%edx
L31:
  test   %r12d,%r12d
  jns    L1
L53:
  lea    0x1(%r12),%eax
  cmp    %eax,%ebx
  jle    L13
  lea    0x4(%rbp,%r15,1),%r15
  movslq %r13d,%rax
  add    %r14,%rax
  lea    0x0(%rbp,%rax,2),%rbp
  add    %rbp,%rbp
L41:
  sub    $0x1,%ebx
  mov    %rbp,%rsi
  sub    %r15,%rsi
  mov    %r15,%rdi
  call   <swap@plt>
  add    $0x4,%r15
  mov    %r13d,%eax
  sub    %ebx,%eax
  add    %r12d,%eax
  cmp    %eax,%ebx
  jg     L41
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,4),%rsi
  call   <swap@plt>
  jmp    L53
L33:
  ret
```

assistant: ```c
void nextPermutation(int *nums, int numsSize) {
  if (numsSize <= 1)
    return;
  int ind1, ind2;
  int i;
  ind1 = ind2 = -1;
  for (i = numsSize - 2; i >= 0; i--)
    if (nums[i] < nums[i + 1]) {
      ind1 = i;
      break;
    }
  for (i = numsSize - 1; i >= 0; i--)
    if (nums[i] > nums[ind1]) {
      ind2 = i;
      break;
    }
  if (ind1 >= 0)
    swap(&nums[ind1], &nums[ind2]);
  i = ind1 + 1;
  numsSize--;
  while (i < numsSize)
    swap(&nums[i++], &nums[numsSize--]);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nextPermutation>:
  endbr64
  cmp    $0x1,%esi
  jle    L33
  lea    -0x2(%rsi),%edx
  push   %r14
  movslq %edx,%rax
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    (%rdi,%rax,4),%rdi
  nopl   (%rax)
L20:
  mov    (%rdi),%ecx
  cmp    0x4(%rdi),%ecx
  jl     L12
  sub    $0x1,%edx
  sub    $0x4,%rdi
  cmp    $0xffffffff,%edx
  jne    L20
  mov    -0x4(%rbx),%ecx
  lea    -0x4(%rbx),%rdi
  xor    %r13d,%r13d
L72:
  lea    -0x1(%r12),%ebp
  movslq %ebp,%rax
  nopl   0x0(%rax)
L60:
  movslq %eax,%rsi
  cmp    %ecx,(%rbx,%rax,4)
  jg     L78
  sub    $0x1,%rax
  movslq %eax,%rsi
  cmp    $0xffffffff,%eax
  jne    L60
L78:
  cmp    $0xffffffff,%edx
  je     L54
  lea    (%rbx,%rsi,4),%rsi
  call   <swap@plt>
  cmp    %r13d,%ebp
  jle    L44
L54:
  movslq %r13d,%rdx
  movslq %r12d,%rax
  add    %rdx,%rax
  lea    (%rbx,%rdx,4),%rbp
  lea    (%rbx,%rax,2),%rax
  mov    %r13d,%ebx
  lea    -0x4(%rax,%rax,1),%r14
  nopl   0x0(%rax,%rax,1)
L73:
  mov    %r14,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  sub    %rbp,%rsi
  add    $0x4,%rbp
  call   <swap@plt>
  mov    %r12d,%eax
  sub    %ebx,%eax
  lea    -0x1(%r13,%rax,1),%eax
  cmp    %ebx,%eax
  jg     L73
L44:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L12:
  lea    0x1(%rdx),%r13d
  jmp    L72
  nopl   0x0(%rax)
L33:
  ret
```

assistant: ```c
void nextPermutation(int *nums, int numsSize) {
  if (numsSize <= 1)
    return;
  int ind1, ind2;
  int i;
  ind1 = ind2 = -1;
  for (i = numsSize - 2; i >= 0; i--)
    if (nums[i] < nums[i + 1]) {
      ind1 = i;
      break;
    }
  for (i = numsSize - 1; i >= 0; i--)
    if (nums[i] > nums[ind1]) {
      ind2 = i;
      break;
    }
  if (ind1 >= 0)
    swap(&nums[ind1], &nums[ind2]);
  i = ind1 + 1;
  numsSize--;
  while (i < numsSize)
    swap(&nums[i++], &nums[numsSize--]);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nextPermutation>:
  endbr64
  cmp    $0x1,%esi
  jle    L34
  push   %r14
  movslq %esi,%r14
  lea    -0x2(%rsi),%edx
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  lea    0x0(,%r14,4),%rsi
  mov    -0x4(%rdi,%rsi,1),%eax
  lea    -0x8(%rdi,%rsi,1),%rdi
  nopl   0x0(%rax)
L43:
  mov    %eax,%ecx
  mov    (%rdi),%eax
  cmp    %ecx,%eax
  jl     L69
  sub    $0x1,%edx
  sub    $0x4,%rdi
  cmp    $0xffffffff,%edx
  jne    L43
  mov    -0x4(%r12),%eax
  lea    -0x4(%r12),%rdi
  xor    %ebp,%ebp
L89:
  lea    -0x1(%rbx),%r8d
  mov    %r14,%rcx
  sub    $0x4,%rsi
  sub    %r8,%rcx
  mov    %r8,%r13
  lea    -0x8(,%rcx,4),%rcx
  nopl   (%rax)
L23:
  cmp    %eax,(%r12,%rsi,1)
  jg     L24
  sub    $0x4,%rsi
  cmp    %rcx,%rsi
  jne    L23
  mov    $0xfffffffffffffffc,%rsi
L24:
  cmp    $0xffffffff,%edx
  je     L6
  add    %r12,%rsi
  call   <swap@plt>
  cmp    %ebp,%r13d
  jle    L77
L6:
  movslq %ebp,%rax
  lea    (%r12,%rax,4),%r13
  add    %r14,%rax
  lea    (%r12,%rax,2),%rax
  mov    %ebp,%r12d
  lea    -0x4(%rax,%rax,1),%r14
  cs nopw 0x0(%rax,%rax,1)
L83:
  mov    %r14,%rsi
  mov    %r13,%rdi
  add    $0x1,%r12d
  sub    %r13,%rsi
  add    $0x4,%r13
  call   <swap@plt>
  mov    %ebx,%eax
  sub    %r12d,%eax
  lea    -0x1(%rbp,%rax,1),%eax
  cmp    %r12d,%eax
  jg     L83
L77:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L69:
  lea    0x1(%rdx),%ebp
  jmp    L89
  nopl   0x0(%rax,%rax,1)
L34:
  ret
```

assistant: ```c
void nextPermutation(int *nums, int numsSize) {
  if (numsSize <= 1)
    return;
  int ind1, ind2;
  int i;
  ind1 = ind2 = -1;
  for (i = numsSize - 2; i >= 0; i--)
    if (nums[i] < nums[i + 1]) {
      ind1 = i;
      break;
    }
  for (i = numsSize - 1; i >= 0; i--)
    if (nums[i] > nums[ind1]) {
      ind2 = i;
      break;
    }
  if (ind1 >= 0)
    swap(&nums[ind1], &nums[ind2]);
  i = ind1 + 1;
  numsSize--;
  while (i < numsSize)
    swap(&nums[i++], &nums[numsSize--]);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_delim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  movzbl -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%al
  jne    L7
  mov    $0x1,%eax
  jmp    L16
L7:
  mov    $0x0,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int is_delim(char c, char delim) {
  if (c == delim)
    return (1);
  else
    return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<is_delim>:
  endbr64
  cmp    %sil,%dil
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_delim(char c, char delim) {
  if (c == delim)
    return (1);
  else
    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
<is_delim>:
  endbr64
  xor    %eax,%eax
  cmp    %sil,%dil
  sete   %al
  ret
```

assistant: ```c
int is_delim(char c, char delim) {
  if (c == delim)
    return (1);
  else
    return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<is_delim>:
  endbr64
  xor    %eax,%eax
  cmp    %sil,%dil
  sete   %al
  ret
```

assistant: ```c
int is_delim(char c, char delim) {
  if (c == delim)
    return (1);
  else
    return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<quantize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  mov    %edi,%eax
  mov    %al,-0x24(%rbp)
  movsd  -0x20(%rbp),%xmm0
  divsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movzbl -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %al,-0x9(%rbp)
  movsbl -0x9(%rbp),%eax
  neg    %eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  comisd -0x8(%rbp),%xmm0
  jb     L3
  movsbl -0x9(%rbp),%eax
  neg    %eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L29
L3:
  movsbl -0x9(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L29
  movsbl -0x9(%rbp),%eax
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x8(%rbp)
L29:
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  leave
  ret
```

assistant: ```c
char quantize(double D, double x, unsigned char B) {
  double qxd;
  char maxlev;
  qxd = floor(x / D);
  maxlev = 1 << (B - 1);
  if (qxd <= -maxlev)
    qxd = -maxlev;
  else if (qxd >= maxlev)
    qxd = maxlev - 1;
  return ((char)qxd);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<quantize>:
  endbr64
  divsd  %xmm0,%xmm1
  movapd %xmm1,%xmm2
  movsd  D28(%rip),%xmm3
  movapd %xmm1,%xmm0
  andpd  %xmm3,%xmm0
  movsd  D12(%rip),%xmm4
  ucomisd %xmm0,%xmm4
  jbe    L12
  cvttsd2si %xmm1,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movapd %xmm0,%xmm2
  cmpnlesd %xmm1,%xmm2
  movsd  D14(%rip),%xmm4
  andpd  %xmm4,%xmm2
  subsd  %xmm2,%xmm0
  andnpd %xmm1,%xmm3
  movapd %xmm0,%xmm2
  orpd   %xmm3,%xmm2
L12:
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  movsbl %al,%edx
  mov    %edx,%ecx
  neg    %ecx
  pxor   %xmm0,%xmm0
  cvtsi2sd %ecx,%xmm0
  comisd %xmm2,%xmm0
  jae    L11
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  comisd %xmm0,%xmm2
  jb     L9
  sub    $0x1,%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  jmp    L11
L9:
  movapd %xmm2,%xmm0
L11:
  cvttsd2si %xmm0,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
type: byte[8]
name: retrieve_data
arguments: data_tag: D12
type: byte[8]
name: retrieve_data
arguments: data_tag: D14
type: byte[8]

tool: {"D28": [0xff,0xff,0xff,0xff,0xff,0xff,0xff,0x7f]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0x30,0x43]}

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
char quantize(double D, double x, unsigned char B) {
  double qxd;
  char maxlev;
  qxd = floor(x / D);
  maxlev = 1 << (B - 1);
  if (qxd <= -maxlev)
    qxd = -maxlev;
  else if (qxd >= maxlev)
    qxd = maxlev - 1;
  return ((char)qxd);
}
```
user: Translate the below assembly code into C programming language:

```asm
<quantize>:
  endbr64
  divsd  %xmm0,%xmm1
  movsd  D17(%rip),%xmm2
  movsd  D10(%rip),%xmm3
  movapd %xmm1,%xmm0
  andpd  %xmm2,%xmm0
  ucomisd %xmm0,%xmm3
  jbe    L6
  cvttsd2si %xmm1,%rax
  pxor   %xmm0,%xmm0
  movsd  D22(%rip),%xmm4
  andnpd %xmm1,%xmm2
  cvtsi2sd %rax,%xmm0
  movapd %xmm0,%xmm3
  cmpnlesd %xmm1,%xmm3
  andpd  %xmm4,%xmm3
  subsd  %xmm3,%xmm0
  movapd %xmm0,%xmm1
  orpd   %xmm2,%xmm1
L6:
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%eax
  pxor   %xmm0,%xmm0
  shl    %cl,%eax
  movsbl %al,%eax
  mov    %eax,%edx
  neg    %edx
  cvtsi2sd %edx,%xmm0
  comisd %xmm1,%xmm0
  jae    L22
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  comisd %xmm0,%xmm1
  jb     L10
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
L22:
  cvttsd2si %xmm0,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  movapd %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
```

tool_calls: name: parse_data
arguments: label: D17
type: byte[8]
name: parse_data
arguments: label: D10
type: byte[8]
name: parse_data
arguments: label: D22
type: byte[8]

tool: {"D17": [0xff,0xff,0xff,0xff,0xff,0xff,0xff,0x7f]}

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0x30,0x43]}

tool: {"D22": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
char quantize(double D, double x, unsigned char B) {
  double qxd;
  char maxlev;
  qxd = floor(x / D);
  maxlev = 1 << (B - 1);
  if (qxd <= -maxlev)
    qxd = -maxlev;
  else if (qxd >= maxlev)
    qxd = maxlev - 1;
  return ((char)qxd);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<quantize>:
  endbr64
  divsd  %xmm0,%xmm1
  movsd  D15(%rip),%xmm2
  movsd  D17(%rip),%xmm3
  movapd %xmm1,%xmm0
  andpd  %xmm2,%xmm0
  ucomisd %xmm0,%xmm3
  jbe    L22
  cvttsd2si %xmm1,%rax
  pxor   %xmm0,%xmm0
  movsd  D16(%rip),%xmm4
  andnpd %xmm1,%xmm2
  cvtsi2sd %rax,%xmm0
  movapd %xmm0,%xmm3
  cmpnlesd %xmm1,%xmm3
  andpd  %xmm4,%xmm3
  subsd  %xmm3,%xmm0
  movapd %xmm0,%xmm1
  orpd   %xmm2,%xmm1
L22:
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%eax
  pxor   %xmm0,%xmm0
  shl    %cl,%eax
  movsbl %al,%eax
  mov    %eax,%edx
  neg    %edx
  cvtsi2sd %edx,%xmm0
  comisd %xmm1,%xmm0
  jae    L19
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  comisd %xmm0,%xmm1
  jb     L6
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
L19:
  cvttsd2si %xmm0,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  movapd %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D15
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D17
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D16
type_guessed: byte[8]

tool: {"D15": [0xff,0xff,0xff,0xff,0xff,0xff,0xff,0x7f]}

tool: {"D17": [0x00,0x00,0x00,0x00,0x00,0x00,0x30,0x43]}

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
char quantize(double D, double x, unsigned char B) {
  double qxd;
  char maxlev;
  qxd = floor(x / D);
  maxlev = 1 << (B - 1);
  if (qxd <= -maxlev)
    qxd = -maxlev;
  else if (qxd >= maxlev)
    qxd = maxlev - 1;
  return ((char)qxd);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<swap_xor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap_xor(uint32_t *x, uint32_t *y) {
  *x = *x ^ *y;
  *x = *x ^ *y;
  *y = *x ^ *y;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<swap_xor>:
  endbr64
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    %eax,(%rsi)
  ret
```

assistant: ```c
void swap_xor(uint32_t *x, uint32_t *y) {
  *x = *x ^ *y;
  *x = *x ^ *y;
  *y = *x ^ *y;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<swap_xor>:
  endbr64
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    %eax,(%rsi)
  ret
```

assistant: ```c
void swap_xor(uint32_t *x, uint32_t *y) {
  *x = *x ^ *y;
  *x = *x ^ *y;
  *y = *x ^ *y;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<swap_xor>:
  endbr64
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    %eax,(%rsi)
  ret
```

assistant: ```c
void swap_xor(uint32_t *x, uint32_t *y) {
  *x = *x ^ *y;
  *x = *x ^ *y;
  *y = *x ^ *y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmp23>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L67
  mov    -0x14(%rbp),%eax
  jmp    L21
L67:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x1,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L26
  mov    -0x14(%rbp),%eax
  jmp    L21
L26:
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x2,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L97
  mov    -0x14(%rbp),%eax
  jmp    L21
L97:
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x4,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L50
  mov    -0x14(%rbp),%eax
  jmp    L21
L50:
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x3,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L92
  mov    -0x14(%rbp),%eax
  jmp    L21
L92:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int cmp23(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cmp23>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L17
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L17
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L17
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
  jne    L17
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
L17:
  ret
```

assistant: ```c
int cmp23(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<cmp23>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L34
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L34
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L34
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
  jne    L34
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
L34:
  ret
```

assistant: ```c
int cmp23(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmp23>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L33
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L33
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L33
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
  jne    L33
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
L33:
  ret
```

assistant: ```c
int cmp23(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<uDMAIntUnregister>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <IntDisable@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <IntUnregister@plt>
  nop
  leave
  ret
```

assistant: ```c
void uDMAIntUnregister(uint32_t ui32IntChannel) {
  IntDisable(ui32IntChannel);
  IntUnregister(ui32IntChannel);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<uDMAIntUnregister>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <IntDisable@plt>
  mov    %ebx,%edi
  call   <IntUnregister@plt>
  pop    %rbx
  ret
```

assistant: ```c
void uDMAIntUnregister(uint32_t ui32IntChannel) {
  IntDisable(ui32IntChannel);
  IntUnregister(ui32IntChannel);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<uDMAIntUnregister>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  call   <IntDisable@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1060 <IntUnregister@plt>
```

assistant: ```c
void uDMAIntUnregister(uint32_t ui32IntChannel) {
  IntDisable(ui32IntChannel);
  IntUnregister(ui32IntChannel);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uDMAIntUnregister>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  call   <IntDisable@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1060 <IntUnregister@plt>
```

assistant: ```c
void uDMAIntUnregister(uint32_t ui32IntChannel) {
  IntDisable(ui32IntChannel);
  IntUnregister(ui32IntChannel);
}
```