ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<subtract>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subtract(int a, int b) {
  int c = 0;
  c = a - b;
  return c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<subtract>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subtract(int a, int b) {
  int c = 0;
  c = a - b;
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1d8(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rdi
  mov    $0x198,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  movl   $0x1,0x40(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L116
  lea    0x14(%rsp),%rbx
  mov    %rbx,%r12
  mov    $0x1,%ebp
  lea    D16(%rip),%r13
L97:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebp
  mov    0xc(%rsp),%eax
  add    $0x4,%r12
  cmp    %ebp,%eax
  jge    L97
  test   %eax,%eax
  jle    L116
  lea    -0x1(%rax),%eax
  lea    0x18(%rsp,%rax,4),%rsi
  jmp    L49
L66:
  add    $0x1,%rax
  cmp    $0x65,%rax
  je     L7
L114:
  mov    0x40(%rsp,%rax,4),%ecx
  test   %ecx,%ecx
  je     L66
  mov    %eax,%edx
  add    (%rbx),%edx
  movslq %edx,%rdx
  add    %ecx,0x40(%rsp,%rdx,4)
  jmp    L66
L7:
  add    $0x4,%rbx
  cmp    %rsi,%rbx
  je     L116
L49:
  mov    $0x0,%eax
  jmp    L114
L116:
  mov    $0x0,%ebx
  lea    D10(%rip),%rbp
L67:
  mov    0x40(%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x65,%rbx
  jne    L67
  lea    0x40(%rsp),%rax
  lea    0x1d4(%rsp),%rcx
  mov    $0x0,%esi
L22:
  mov    (%rax),%edx
  cmp    %edx,%esi
  cmovl  %edx,%esi
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L22
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x1d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L96
  mov    $0x0,%eax
  add    $0x1e8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L96:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D16
type_guessed: string
name: retrieve_data
arguments: tag: D10
type_guessed: string
name: retrieve_data
arguments: tag: D18
type_guessed: string

tool: D16: %d

tool: D10: coinChange[%d] = %d


tool: D18: Max Way to Give Money = %d


assistant: ```c
int main() {
  int i, j;
  int coinNumber;
  int coinNumberStore[10];
  int coinChange[102];
  memset(coinChange, 0, sizeof(coinChange));
  coinChange[0] = 1;
  scanf("%d", &coinNumber);
  for (i = 1; i <= coinNumber; i++) {
    scanf("%d", &coinNumberStore[i]);
  }
  for (i = 1; i <= coinNumber; i++) {
    for (j = 0; j <= 100; j++) {
      if (coinChange[j] != 0) {
        coinChange[j + coinNumberStore[i]] = coinChange[j] + coinChange[j + coinNumberStore[i]];
      }
    }
  }
  for (j = 0; j <= 100; j++) {
    printf("coinChange[%d] = %d\n", j, coinChange[j]);
  }
  int MaxWayToGiveMoney = 0;
  for (j = 0; j <= 100; j++) {
    if (coinChange[j] >= MaxWayToGiveMoney)
      MaxWayToGiveMoney = coinChange[j];
  }
  printf("Max Way to Give Money = %d\n", MaxWayToGiveMoney);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x198,%edx
  xor    %esi,%esi
  lea    D7(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1e0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1d8(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  lea    0xc(%rsp),%rsi
  mov    %r14,%rdi
  movl   $0x1,0x40(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L43
  lea    0x14(%rsp),%rbp
  mov    $0x1,%r12d
  mov    %rbp,%r13
L12:
  mov    %r13,%rsi
  mov    %r14,%rdi
  add    $0x1,%r12d
  add    $0x4,%r13
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %r12d,%eax
  jge    L12
  test   %eax,%eax
  jle    L43
  sub    $0x1,%eax
  lea    0x18(%rsp,%rax,4),%rsi
L79:
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L88:
  mov    (%rbx,%rax,4),%ecx
  test   %ecx,%ecx
  je     L16
  mov    0x0(%rbp),%edx
  add    %eax,%edx
  movslq %edx,%rdx
  add    %ecx,0x40(%rsp,%rdx,4)
L16:
  add    $0x1,%rax
  cmp    $0x65,%rax
  jne    L88
  add    $0x4,%rbp
  cmp    %rsi,%rbp
  jne    L79
L43:
  xor    %ebp,%ebp
  lea    D22(%rip),%r12
  nop
L70:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x65,%rbp
  jne    L70
  lea    0x1d4(%rsp),%rdx
  xor    %esi,%esi
  nopl   0x0(%rax,%rax,1)
L18:
  mov    (%rbx),%eax
  cmp    %eax,%esi
  cmovl  %eax,%esi
  add    $0x4,%rbx
  cmp    %rdx,%rbx
  jne    L18
  xor    %eax,%eax
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x1d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x1e0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
datatype: string
name: unpack_data
arguments: tag: D22
datatype: string
name: unpack_data
arguments: tag: D9
datatype: string

tool: D7: %d

tool: D22: coinChange[%d] = %d


tool: D9: Max Way to Give Money = %d


assistant: ```c
int main() {
  int i, j;
  int coinNumber;
  int coinNumberStore[10];
  int coinChange[102];
  memset(coinChange, 0, sizeof(coinChange));
  coinChange[0] = 1;
  scanf("%d", &coinNumber);
  for (i = 1; i <= coinNumber; i++) {
    scanf("%d", &coinNumberStore[i]);
  }
  for (i = 1; i <= coinNumber; i++) {
    for (j = 0; j <= 100; j++) {
      if (coinChange[j] != 0) {
        coinChange[j + coinNumberStore[i]] = coinChange[j] + coinChange[j + coinNumberStore[i]];
      }
    }
  }
  for (j = 0; j <= 100; j++) {
    printf("coinChange[%d] = %d\n", j, coinChange[j]);
  }
  int MaxWayToGiveMoney = 0;
  for (j = 0; j <= 100; j++) {
    if (coinChange[j] >= MaxWayToGiveMoney)
      MaxWayToGiveMoney = coinChange[j];
  }
  printf("Max Way to Give Money = %d\n", MaxWayToGiveMoney);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_sha512_be64dec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x38,%rax
  mov    (%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x30,%rax
  mov    (%rax),%rax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x28,%rax
  mov    (%rax),%rax
  shl    $0x10,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%rax
  shl    $0x18,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    %eax,%esi
  mov    $0x20,%eax
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %eax,%esi
  mov    $0x28,%eax
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %eax,%esi
  mov    $0x30,%eax
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %eax,%esi
  mov    $0x38,%eax
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t _sha512_be64dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint64_t)(p[7]) + ((uint64_t)(p[6]) << 8) + ((uint64_t)(p[5]) << 16) + ((uint64_t)(p[4]) << 24) + ((uint64_t)(p[3]) << 32) + ((uint64_t)(p[2]) << 40) + ((uint64_t)(p[1]) << 48) + ((uint64_t)(p[0]) << 56));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_sha512_be64dec>:
  endbr64
  mov    0x30(%rdi),%eax
  shl    $0x8,%eax
  add    0x38(%rdi),%eax
  mov    0x28(%rdi),%edx
  shl    $0x10,%edx
  add    %edx,%eax
  mov    0x20(%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint64_t _sha512_be64dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint64_t)(p[7]) + ((uint64_t)(p[6]) << 8) + ((uint64_t)(p[5]) << 16) + ((uint64_t)(p[4]) << 24) + ((uint64_t)(p[3]) << 32) + ((uint64_t)(p[2]) << 40) + ((uint64_t)(p[1]) << 48) + ((uint64_t)(p[0]) << 56));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_sha512_be64dec>:
  endbr64
  mov    0x30(%rdi),%eax
  mov    0x28(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  add    0x38(%rdi),%eax
  add    %edx,%eax
  mov    0x20(%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint64_t _sha512_be64dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint64_t)(p[7]) + ((uint64_t)(p[6]) << 8) + ((uint64_t)(p[5]) << 16) + ((uint64_t)(p[4]) << 24) + ((uint64_t)(p[3]) << 32) + ((uint64_t)(p[2]) << 40) + ((uint64_t)(p[1]) << 48) + ((uint64_t)(p[0]) << 56));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_sha512_be64dec>:
  endbr64
  mov    0x30(%rdi),%eax
  mov    0x28(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  add    0x38(%rdi),%eax
  add    %edx,%eax
  mov    0x20(%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint64_t _sha512_be64dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint64_t)(p[7]) + ((uint64_t)(p[6]) << 8) + ((uint64_t)(p[5]) << 16) + ((uint64_t)(p[4]) << 24) + ((uint64_t)(p[3]) << 32) + ((uint64_t)(p[2]) << 40) + ((uint64_t)(p[1]) << 48) + ((uint64_t)(p[0]) << 56));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<quickSort>:
  endbr64
  cmp    %rdx,%rsi
  jge    L57
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r10
  mov    %rdx,%rbp
  mov    %edx,%ebx
  cmp    %edx,%esi
  jge    L37
  movslq %esi,%rax
  lea    (%rdi,%rax,8),%r11
  mov    %esi,%r8d
  jmp    L41
L85:
  mov    %rdi,%rax
L80:
  mov    %eax,%r8d
  mov    %rdx,%r9
  mov    (%rdx),%rcx
  cmp    %rsi,%rcx
  jg     L65
  lea    0x1(%rax),%rdi
  add    $0x8,%rdx
  cmp    %rbp,%rax
  jne    L85
L65:
  movslq %ebx,%rax
  lea    (%r12,%rax,8),%rdi
  mov    (%rdi),%rdx
  cmp    %rsi,%rdx
  jle    L96
  lea    -0x8(%r12,%rax,8),%rax
L8:
  sub    $0x1,%ebx
  mov    %rax,%rdi
  sub    $0x8,%rax
  mov    0x8(%rax),%rdx
  cmp    %rsi,%rdx
  jg     L8
L96:
  cmp    %ebx,%r8d
  jge    L37
  mov    %rdx,(%r9)
  movslq %ecx,%rcx
  mov    %rcx,(%rdi)
L41:
  movslq %r8d,%rdx
  shl    $0x3,%rdx
  lea    (%r12,%rdx,1),%r9
  mov    (%r9),%rcx
  mov    (%r11),%rsi
  cmp    %rsi,%rcx
  jg     L65
  movslq %r8d,%rax
  cmp    %rax,%rbp
  jle    L65
  lea    0x1(%r8),%eax
  cltq
  lea    0x8(%r12,%rdx,1),%rdx
  jmp    L80
L37:
  movslq %r10d,%rax
  lea    (%r12,%rax,8),%rcx
  mov    (%rcx),%rax
  movslq %ebx,%rdx
  lea    (%r12,%rdx,8),%rdx
  mov    (%rdx),%rsi
  mov    %rsi,(%rcx)
  cltq
  mov    %rax,(%rdx)
  lea    -0x1(%rbx),%edx
  movslq %edx,%rdx
  mov    %r10,%rsi
  mov    %r12,%rdi
  call   <quickSort>
  lea    0x1(%rbx),%esi
  movslq %esi,%rsi
  mov    %rbp,%rdx
  mov    %r12,%rdi
  call   <quickSort>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L57:
  ret
```

assistant: ```c
void quickSort(long list[], long first, long last) {
  int i, j, v, temp;
  if (first < last) {
    v = first;
    i = first;
    j = last;
    while (i < j) {
      while (list[i] <= list[v] && i < last)
        i++;
      while (list[j] > list[v])
        j--;
      if (i < j) {
        temp = list[i];
        list[i] = list[j];
        list[j] = temp;
      }
    }
    temp = list[v];
    list[v] = list[j];
    list[j] = temp;
    quickSort(list, first, j - 1);
    quickSort(list, j + 1, last);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<quickSort>:
  endbr64
  cmp    %rdx,%rsi
  jge    L35
  movslq %edx,%rax
  push   %r14
  mov    %rsi,%r8
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  lea    (%rdi,%rax,8),%rbp
  push   %rbx
  mov    %edx,%ebx
L5:
  movslq %r8d,%rax
  mov    %r8d,%edi
  mov    %ebx,%r12d
  mov    %rbp,%r11
  lea    (%r14,%rax,8),%r9
  mov    (%r9),%rsi
  cmp    %ebx,%r8d
  jge    L18
  nopl   0x0(%rax,%rax,1)
L116:
  movslq %edi,%rax
  lea    0x0(,%rax,8),%rdx
  lea    (%r14,%rdx,1),%r10
  mov    (%r10),%rcx
  cmp    %rcx,%rsi
  jl     L34
  cmp    %rax,%r13
  jle    L34
  lea    0x1(%rdi),%eax
  lea    0x8(%r14,%rdx,1),%rdx
  cltq
  jmp    L94
L7:
  lea    0x1(%rax),%r11
  add    $0x8,%rdx
  cmp    %rax,%r13
  je     L34
  mov    %r11,%rax
L94:
  mov    (%rdx),%rcx
  mov    %eax,%edi
  mov    %rdx,%r10
  cmp    %rsi,%rcx
  jle    L7
L34:
  movslq %r12d,%rax
  lea    (%r14,%rax,8),%r11
  lea    -0x8(%r14,%rax,8),%rax
  mov    (%r11),%rdx
  cmp    %rdx,%rsi
  jge    L90
  nopl   0x0(%rax)
L19:
  mov    %rax,%r11
  mov    (%rax),%rdx
  sub    $0x8,%rax
  sub    $0x1,%r12d
  cmp    %rsi,%rdx
  jg     L19
L90:
  cmp    %r12d,%edi
  jl     L64
L18:
  mov    (%r11),%rax
  movslq %esi,%rsi
  lea    -0x1(%r12),%edx
  mov    %r14,%rdi
  movslq %edx,%rdx
  add    $0x1,%r12d
  mov    %rax,(%r9)
  mov    %rsi,(%r11)
  mov    %r8,%rsi
  call   <quickSort>
  movslq %r12d,%r8
  cmp    %r13,%r8
  jl     L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L64:
  movslq %ecx,%rcx
  mov    %rdx,(%r10)
  mov    %rcx,(%r11)
  mov    (%r9),%rsi
  jmp    L116
L35:
  ret
```

assistant: ```c
void quickSort(long list[], long first, long last) {
  int i, j, v, temp;
  if (first < last) {
    v = first;
    i = first;
    j = last;
    while (i < j) {
      while (list[i] <= list[v] && i < last)
        i++;
      while (list[j] > list[v])
        j--;
      if (i < j) {
        temp = list[i];
        list[i] = list[j];
        list[j] = temp;
      }
    }
    temp = list[v];
    list[v] = list[j];
    list[j] = temp;
    quickSort(list, first, j - 1);
    quickSort(list, j + 1, last);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<quickSort>:
  endbr64
  cmp    %rdx,%rsi
  jge    L21
  movslq %edx,%rax
  push   %r14
  mov    %rsi,%r8
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  lea    (%rdi,%rax,8),%rbp
  push   %rbx
  mov    %edx,%ebx
L112:
  movslq %r8d,%rax
  mov    0x0(%rbp),%rdx
  mov    %r8d,%edi
  mov    %ebx,%r12d
  lea    (%r14,%rax,8),%r9
  mov    %rbp,%r11
  mov    (%r9),%rsi
  cmp    %ebx,%r8d
  jge    L56
  nopl   0x0(%rax,%rax,1)
L38:
  movslq %edi,%rax
  lea    0x0(,%rax,8),%rdx
  lea    (%r14,%rdx,1),%r10
  mov    (%r10),%rcx
  cmp    %rcx,%rsi
  jl     L107
  cmp    %rax,%r13
  jle    L107
  lea    0x1(%rdi),%eax
  lea    0x8(%r14,%rdx,1),%rdx
  cltq
  jmp    L14
L62:
  lea    0x1(%rax),%r11
  add    $0x8,%rdx
  cmp    %rax,%r13
  je     L107
  mov    %r11,%rax
L14:
  mov    (%rdx),%rcx
  mov    %eax,%edi
  mov    %rdx,%r10
  cmp    %rsi,%rcx
  jle    L62
L107:
  movslq %r12d,%rax
  lea    (%r14,%rax,8),%r11
  lea    -0x8(%r14,%rax,8),%rax
  mov    (%r11),%rdx
  cmp    %rdx,%rsi
  jge    L116
  nopl   0x0(%rax)
L34:
  mov    %rax,%r11
  mov    (%rax),%rdx
  sub    $0x8,%rax
  sub    $0x1,%r12d
  cmp    %rsi,%rdx
  jg     L34
L116:
  cmp    %r12d,%edi
  jl     L48
L56:
  movslq %esi,%rsi
  mov    %rdx,(%r9)
  lea    -0x1(%r12),%edx
  mov    %r14,%rdi
  mov    %rsi,(%r11)
  movslq %edx,%rdx
  mov    %r8,%rsi
  add    $0x1,%r12d
  call   <quickSort>
  movslq %r12d,%r8
  cmp    %r13,%r8
  jl     L112
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L48:
  movslq %ecx,%rcx
  mov    %rdx,(%r10)
  mov    %rcx,(%r11)
  mov    (%r9),%rsi
  jmp    L38
L21:
  ret
```

assistant: ```c
void quickSort(long list[], long first, long last) {
  int i, j, v, temp;
  if (first < last) {
    v = first;
    i = first;
    j = last;
    while (i < j) {
      while (list[i] <= list[v] && i < last)
        i++;
      while (list[j] > list[v])
        j--;
      if (i < j) {
        temp = list[i];
        list[i] = list[j];
        list[j] = temp;
      }
    }
    temp = list[v];
    list[v] = list[j];
    list[j] = temp;
    quickSort(list, first, j - 1);
    quickSort(list, j + 1, last);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<comparator>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0xc(%rbp),%xmm0
  subss  -0x8(%rbp),%xmm0
  cvttss2si %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    -0x4(%rbp),%edx
  shr    $0x1f,%edx
  movzbl %dl,%edx
  sub    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int comparator(const void *first, const void *second) {
  float a = *((int *)first), b = *((int *)second);
  int result = a - b;
  return (0 < result) - (result < 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<comparator>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rdi),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl (%rsi),%xmm1
  subss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  test   %edx,%edx
  setg   %al
  movzbl %al,%eax
  shr    $0x1f,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int comparator(const void *first, const void *second) {
  float a = *((int *)first), b = *((int *)second);
  int result = a - b;
  return (0 < result) - (result < 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<comparator>:
  endbr64
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  xor    %eax,%eax
  cvtsi2ssl (%rdi),%xmm0
  cvtsi2ssl (%rsi),%xmm1
  subss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  test   %edx,%edx
  setg   %al
  shr    $0x1f,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int comparator(const void *first, const void *second) {
  float a = *((int *)first), b = *((int *)second);
  int result = a - b;
  return (0 < result) - (result < 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<comparator>:
  endbr64
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  xor    %eax,%eax
  cvtsi2ssl (%rdi),%xmm0
  cvtsi2ssl (%rsi),%xmm1
  subss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  test   %edx,%edx
  setg   %al
  shr    $0x1f,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int comparator(const void *first, const void *second) {
  float a = *((int *)first), b = *((int *)second);
  int result = a - b;
  return (0 < result) - (result < 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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x360,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jg     L96
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L88:
  addl   $0x1,-0x8(%rbp)
L3:
  cmpl   $0x63,-0x8(%rbp)
  jle    L88
  jmp    L115
L96:
  cmpl   $0x32,-0x4(%rbp)
  jg     L111
  movl   $0x0,-0x8(%rbp)
  jmp    L65
L102:
  addl   $0x1,-0x8(%rbp)
L65:
  cmpl   $0x64,-0x8(%rbp)
  jle    L102
L111:
  cmpl   $0x64,-0x4(%rbp)
  jg     L53
  movl   $0x0,-0x8(%rbp)
  jmp    L87
L105:
  addl   $0x1,-0x8(%rbp)
L87:
  cmpl   $0x65,-0x8(%rbp)
  jle    L105
  jmp    L115
L53:
  movl   $0x0,-0x8(%rbp)
L66:
  cmpl   $0x66,-0x8(%rbp)
  jle    L66
L115:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int c;
  int i;
  c = (24 * 6) * 6;
  if (c <= 10) {
    i = 0;
    while (i < 100) {
      i = i + 1;
    }
  } else {
    if (c <= 50) {
      i = 0;
      while (i < 101) {
        i = i + 1;
      }
    }
    if (c <= 100) {
      i = 0;
      while (i < 102) {
        i = i + 1;
      }
    } else {
      i = 0;
      while (i < 103) {
        i = i + 0;
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
L1:
  jmp    L1
```

assistant: ```c
int main() {
  int c;
  int i;
  c = (24 * 6) * 6;
  if (c <= 10) {
    i = 0;
    while (i < 100) {
      i = i + 1;
    }
  } else {
    if (c <= 50) {
      i = 0;
      while (i < 101) {
        i = i + 1;
      }
    }
    if (c <= 100) {
      i = 0;
      while (i < 102) {
        i = i + 1;
      }
    } else {
      i = 0;
      while (i < 103) {
        i = i + 0;
      }
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  nopl   0x0(%rax)
L6:
  jmp    L6
```

assistant: ```c
int main() {
  int c;
  int i;
  c = (24 * 6) * 6;
  if (c <= 10) {
    i = 0;
    while (i < 100) {
      i = i + 1;
    }
  } else {
    if (c <= 50) {
      i = 0;
      while (i < 101) {
        i = i + 1;
      }
    }
    if (c <= 100) {
      i = 0;
      while (i < 102) {
        i = i + 1;
      }
    } else {
      i = 0;
      while (i < 103) {
        i = i + 0;
      }
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  nopl   0x0(%rax)
L1:
  jmp    L1
```

assistant: ```c
int main() {
  int c;
  int i;
  c = (24 * 6) * 6;
  if (c <= 10) {
    i = 0;
    while (i < 100) {
      i = i + 1;
    }
  } else {
    if (c <= 50) {
      i = 0;
      while (i < 101) {
        i = i + 1;
      }
    }
    if (c <= 100) {
      i = 0;
      while (i < 102) {
        i = i + 1;
      }
    } else {
      i = 0;
      while (i < 103) {
        i = i + 0;
      }
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<free_dmatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L17
L13:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <free@plt>
  subl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L13
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void free_dmatrix(double **m, int nrl, int nrh, int ncl, int nch) {
  int i;
  for (i = nrh; i >= nrl; i--)
    free((void *)(m[i] + ncl));
  free((void *)(m + nrl));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<free_dmatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  cmp    %esi,%edx
  jl     L7
  movslq %ecx,%rcx
  lea    0x0(,%rcx,8),%r13
  movslq %edx,%rbx
L14:
  mov    %r13,%rdi
  add    (%r12,%rbx,8),%rdi
  call   <free@plt>
  sub    $0x1,%rbx
  cmp    %ebx,%ebp
  jle    L14
L7:
  movslq %ebp,%rbp
  lea    (%r12,%rbp,8),%rdi
  call   <free@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void free_dmatrix(double **m, int nrl, int nrh, int ncl, int nch) {
  int i;
  for (i = nrh; i >= nrl; i--)
    free((void *)(m[i] + ncl));
  free((void *)(m + nrl));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<free_dmatrix>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %ebp,%edx
  jl     L10
  movslq %ecx,%rcx
  movslq %edx,%rbx
  lea    0x0(,%rcx,8),%r13
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    (%r12,%rbx,8),%rdi
  sub    $0x1,%rbx
  add    %r13,%rdi
  call   <free@plt>
  cmp    %ebx,%ebp
  jle    L16
L10:
  add    $0x8,%rsp
  lea    (%r12,%rbp,8),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <free@plt>
```

assistant: ```c
void free_dmatrix(double **m, int nrl, int nrh, int ncl, int nch) {
  int i;
  for (i = nrh; i >= nrl; i--)
    free((void *)(m[i] + ncl));
  free((void *)(m + nrl));
}
```