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

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  xor    %eax,%eax
  call   <crmode@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %ebx,%edi
  call   <testReaction@plt>
  mov    %rax,-0x8(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    $0xb,%rbx
  jne    L5
  xor    %eax,%eax
  call   <endwin@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <reactionReport@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main() {
  int totalTests = 10;
  long resultTimes[totalTests];
  time_t t;
  srand((unsigned)time(&t));
  initscr();
  noecho();
  crmode();
  int i;
  for (i = 0; i < totalTests; i++) {
    resultTimes[i] = testReaction(i + 1);
  }
  endwin();
  reactionReport(resultTimes, totalTests);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x10(%rsp),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  xor    %eax,%eax
  call   <crmode@plt>
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %ebx,%edi
  call   <testReaction@plt>
  mov    %rax,-0x8(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    $0xb,%rbx
  jne    L18
  xor    %eax,%eax
  call   <endwin@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <reactionReport@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main() {
  int totalTests = 10;
  long resultTimes[totalTests];
  time_t t;
  srand((unsigned)time(&t));
  initscr();
  noecho();
  crmode();
  int i;
  for (i = 0; i < totalTests; i++) {
    resultTimes[i] = testReaction(i + 1);
  }
  endwin();
  reactionReport(resultTimes, totalTests);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    $0x2a,%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %eax,-0x4(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D0: Este codigo no se ejecuta nunca


assistant: ```c
int main() {
  int d = 0;
  int a = 0;
  a = 42 / d;
  printf("Este codigo no se ejecuta nunca\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
data_type: string

tool: D5: Este codigo no se ejecuta nunca


assistant: ```c
int main() {
  int d = 0;
  int a = 0;
  a = 42 / d;
  printf("Este codigo no se ejecuta nunca\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
data_type: string

tool: D1: Este codigo no se ejecuta nunca


assistant: ```c
int main() {
  int d = 0;
  int a = 0;
  a = 42 / d;
  printf("Este codigo no se ejecuta nunca\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type_guessed: string

tool: D8: Este codigo no se ejecuta nunca


assistant: ```c
int main() {
  int d = 0;
  int a = 0;
  a = 42 / d;
  printf("Este codigo no se ejecuta nunca\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1010(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x200,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x1,-0x102c(%rbp)
  jmp    L32
L95:
  movl   $0x0,-0x1020(%rbp)
  movl   $0x1,-0x1028(%rbp)
  jmp    L3
L5:
  mov    -0x102c(%rbp),%eax
  cltd
  idivl  -0x1028(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L23
  mov    -0x1020(%rbp),%eax
  cltq
  mov    -0x1028(%rbp),%edx
  mov    %edx,-0x1010(%rbp,%rax,4)
  mov    -0x102c(%rbp),%eax
  cmp    -0x1028(%rbp),%eax
  jne    L27
  mov    -0x1020(%rbp),%eax
  cltq
  movl   $0x0,-0x1010(%rbp,%rax,4)
  movl   $0x0,-0x1024(%rbp)
  movl   $0x0,-0x101c(%rbp)
  jmp    L78
L136:
  mov    -0x101c(%rbp),%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%eax
  add    %eax,-0x1024(%rbp)
  addl   $0x1,-0x101c(%rbp)
L78:
  mov    -0x101c(%rbp),%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L136
  mov    -0x1024(%rbp),%eax
  cmp    -0x102c(%rbp),%eax
  jne    L56
  movl   $0x0,-0x1018(%rbp)
  jmp    L58
L22:
  mov    -0x1018(%rbp),%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x1018(%rbp)
L58:
  mov    -0x1018(%rbp),%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L22
  mov    -0x1024(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L56:
  movl   $0x1,-0x1014(%rbp)
  jmp    L97
L81:
  mov    -0x1014(%rbp),%eax
  cltq
  movl   $0x0,-0x1010(%rbp,%rax,4)
  addl   $0x1,-0x1014(%rbp)
L97:
  cmpl   $0x3ff,-0x1014(%rbp)
  jle    L81
L27:
  addl   $0x1,-0x1020(%rbp)
L23:
  addl   $0x1,-0x1028(%rbp)
L3:
  mov    -0x1028(%rbp),%eax
  cmp    -0x102c(%rbp),%eax
  jle    L5
  addl   $0x1,-0x102c(%rbp)
L32:
  cmpl   $0x3e7,-0x102c(%rbp)
  jle    L95
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L51
  call   <__stack_chk_fail@plt>
L51:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D15
data_type: string

tool: D8: %d+

tool: D15: =%d


assistant: ```c
int main() {
  int i, j;
  int sum;
  int array[1024] = {};
  for (i = 1; i < 1000; i++) {
    int k = 0;
    for (j = 1; j <= i; j++) {
      if (0 == i % j) {
        array[k] = j;
        if (i == j) {
          array[k] = 0;
          sum = 0;
          for (int l = 0; 0 != array[l]; l++) {
            sum += array[l];
          }
          if (sum == i) {
            for (int n = 0; 0 != array[n]; n++) {
              printf("%d+", array[n]);
            }
            printf("=%d\n", sum);
          }
          for (int n = 1; n < 1024; n++) {
            array[n] = 0;
          }
        }
        k++;
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x200,%ecx
  rep stos %rax,%es:(%rdi)
  mov    $0x1,%r12d
  lea    0x1000(%rsp),%rbp
  lea    D12(%rip),%r15
  jmp    L78
L54:
  mov    %r15,%rdi
  call   <printf@plt>
  add    $0x4,%r14
  mov    -0x4(%r14),%esi
  test   %esi,%esi
  jne    L54
  mov    %r12d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
L123:
  lea    0x4(%rsp),%rax
L13:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L13
L44:
  add    $0x1,%r13d
L91:
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jg     L3
L6:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L91
  movslq %r13d,%rdx
  mov    %ebx,(%rsp,%rdx,4)
  cmp    %ebx,%r12d
  jne    L44
  movl   $0x0,(%rsp,%rdx,4)
  mov    (%rsp),%esi
  test   %esi,%esi
  je     L123
  lea    0x4(%rsp),%r14
  mov    %r14,%rcx
  mov    %esi,%edx
L22:
  add    %edx,%eax
  add    $0x4,%rcx
  mov    -0x4(%rcx),%edx
  test   %edx,%edx
  jne    L22
  cmp    %r12d,%eax
  jne    L123
  jmp    L54
L3:
  add    $0x1,%r12d
  cmp    $0x3e8,%r12d
  je     L119
L78:
  mov    $0x0,%r13d
  mov    $0x1,%ebx
  test   %r12d,%r12d
  jg     L6
  jmp    L3
L119:
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L138
  mov    $0x0,%eax
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L138:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type: string
name: retrieve_data
arguments: data_label: D3
type: string

tool: D12: %d+

tool: D3: =%d


assistant: ```c
int main() {
  int i, j;
  int sum;
  int array[1024] = {};
  for (i = 1; i < 1000; i++) {
    int k = 0;
    for (j = 1; j <= i; j++) {
      if (0 == i % j) {
        array[k] = j;
        if (i == j) {
          array[k] = 0;
          sum = 0;
          for (int l = 0; 0 != array[l]; l++) {
            sum += array[l];
          }
          if (sum == i) {
            for (int n = 0; 0 != array[n]; n++) {
              printf("%d+", array[n]);
            }
            printf("=%d\n", sum);
          }
          for (int n = 1; n < 1024; n++) {
            array[n] = 0;
          }
        }
        k++;
      }
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    $0x200,%ecx
  mov    $0x2,%r14d
  mov    $0x1,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  rep stos %rax,%es:(%rdi)
  nopl   0x0(%rax,%rax,1)
L57:
  xor    %r15d,%r15d
  mov    $0x1,%ebx
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L51:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L30
  movslq %r15d,%rax
  mov    %ebx,(%rsp,%rax,4)
  cmp    %ebx,%r12d
  je     L89
L109:
  add    $0x1,%r15d
L30:
  add    $0x1,%ebx
  cmp    %r14d,%ebx
  jne    L51
  add    $0x1,%r12d
  lea    0x1(%rbx),%r14d
  cmp    $0x3e8,%r12d
  jne    L57
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x1018,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L89:
  movl   $0x0,(%rsp,%rax,4)
  mov    (%rsp),%esi
  test   %esi,%esi
  je     L94
  mov    %r13,%rbp
  mov    %r13,%rcx
  mov    %esi,%eax
  nopl   0x0(%rax,%rax,1)
L33:
  add    $0x4,%rcx
  add    %eax,%edx
  mov    -0x4(%rcx),%eax
  test   %eax,%eax
  jne    L33
  cmp    %r12d,%edx
  je     L83
L94:
  lea    0x8(%rsp),%rdi
  mov    %r13,%rcx
  xor    %eax,%eax
  movq   $0x0,0x0(%r13)
  movq   $0x0,0xff8(%rsp)
  sub    %rdi,%rcx
  add    $0xffc,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  jmp    L109
  nopl   0x0(%rax)
L83:
  lea    D2(%rip),%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  mov    -0x4(%rbp),%esi
  test   %esi,%esi
  jne    L83
  mov    %r12d,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  jmp    L94
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
datatype: string
name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D2: %d+

tool: D0: =%d


assistant: ```c
int main() {
  int i, j;
  int sum;
  int array[1024] = {};
  for (i = 1; i < 1000; i++) {
    int k = 0;
    for (j = 1; j <= i; j++) {
      if (0 == i % j) {
        array[k] = j;
        if (i == j) {
          array[k] = 0;
          sum = 0;
          for (int l = 0; 0 != array[l]; l++) {
            sum += array[l];
          }
          if (sum == i) {
            for (int n = 0; 0 != array[n]; n++) {
              printf("%d+", array[n]);
            }
            printf("=%d\n", sum);
          }
          for (int n = 1; n < 1024; n++) {
            array[n] = 0;
          }
        }
        k++;
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    $0x200,%ecx
  mov    $0x2,%r14d
  mov    $0x1,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  rep stos %rax,%es:(%rdi)
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %r15d,%r15d
  mov    $0x1,%ebx
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L57:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L18
  movslq %r15d,%rax
  mov    %ebx,(%rsp,%rax,4)
  cmp    %ebx,%r12d
  je     L56
L96:
  add    $0x1,%r15d
L18:
  add    $0x1,%ebx
  cmp    %r14d,%ebx
  jne    L57
  add    $0x1,%r12d
  lea    0x1(%rbx),%r14d
  cmp    $0x3e8,%r12d
  jne    L16
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x1018,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L56:
  movl   $0x0,(%rsp,%rax,4)
  mov    (%rsp),%esi
  test   %esi,%esi
  je     L17
  mov    %r13,%rbp
  mov    %r13,%rcx
  mov    %esi,%eax
  nopl   0x0(%rax,%rax,1)
L1:
  add    $0x4,%rcx
  add    %eax,%edx
  mov    -0x4(%rcx),%eax
  test   %eax,%eax
  jne    L1
  cmp    %r12d,%edx
  je     L108
L17:
  lea    0x8(%rsp),%rdi
  mov    %r13,%rcx
  xor    %eax,%eax
  movq   $0x0,0x0(%r13)
  movq   $0x0,0xff8(%rsp)
  sub    %rdi,%rcx
  add    $0xffc,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  jmp    L96
  nopl   0x0(%rax)
L108:
  lea    D0(%rip),%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  mov    -0x4(%rbp),%esi
  test   %esi,%esi
  jne    L108
  mov    %r12d,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L17
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type_infered: string
name: extract_symbolic_data
arguments: label: D4
type_infered: string

tool: D0: %d+

tool: D4: =%d


assistant: ```c
int main() {
  int i, j;
  int sum;
  int array[1024] = {};
  for (i = 1; i < 1000; i++) {
    int k = 0;
    for (j = 1; j <= i; j++) {
      if (0 == i % j) {
        array[k] = j;
        if (i == j) {
          array[k] = 0;
          sum = 0;
          for (int l = 0; 0 != array[l]; l++) {
            sum += array[l];
          }
          if (sum == i) {
            for (int n = 0; 0 != array[n]; n++) {
              printf("%d+", array[n]);
            }
            printf("=%d\n", sum);
          }
          for (int n = 1; n < 1024; n++) {
            array[n] = 0;
          }
        }
        k++;
      }
    }
  }
  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
<bow_foilgain_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <bow_free@plt>
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <bow_free@plt>
  nop
  leave
  ret
```

assistant: ```c
void bow_foilgain_free(float **fig_per_wi_ci, int num_wi) {
  int wi;
  for (wi = 0; wi < num_wi; wi++)
    bow_free(fig_per_wi_ci[wi]);
  bow_free(fig_per_wi_ci);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bow_foilgain_free>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %esi,%esi
  jle    L8
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
L18:
  mov    (%rbx),%rdi
  mov    $0x0,%eax
  call   <bow_free@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L18
L8:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <bow_free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void bow_foilgain_free(float **fig_per_wi_ci, int num_wi) {
  int wi;
  for (wi = 0; wi < num_wi; wi++)
    bow_free(fig_per_wi_ci[wi]);
  bow_free(fig_per_wi_ci);
}
```
user: What is the c source code of the assembly code below:

```asm
<bow_foilgain_free>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L14
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nopw   0x0(%rax,%rax,1)
L1:
  mov    (%rbx),%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <bow_free@plt>
  cmp    %rbp,%rbx
  jne    L1
L14:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <bow_free@plt>
```

assistant: ```c
void bow_foilgain_free(float **fig_per_wi_ci, int num_wi) {
  int wi;
  for (wi = 0; wi < num_wi; wi++)
    bow_free(fig_per_wi_ci[wi]);
  bow_free(fig_per_wi_ci);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bow_foilgain_free>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L9
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nopw   0x0(%rax,%rax,1)
L4:
  mov    (%rbx),%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <bow_free@plt>
  cmp    %rbp,%rbx
  jne    L4
L9:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1050 <bow_free@plt>
```

assistant: ```c
void bow_foilgain_free(float **fig_per_wi_ci, int num_wi) {
  int wi;
  for (wi = 0; wi < num_wi; wi++)
    bow_free(fig_per_wi_ci[wi]);
  bow_free(fig_per_wi_ci);
}
```
user: What is the c source code of the assembly code below:

```asm
<phredToPhred64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x5d,-0x4(%rbp)
  jle    L20
  movl   $0x5d,-0x4(%rbp)
  jmp    L5
L20:
  cmpl   $0x0,-0x4(%rbp)
  jns    L5
  movl   $0x0,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  add    $0x40,%eax
  pop    %rbp
  ret
```

assistant: ```c
char phredToPhred64(int p) {
  if (p > 93)
    p = 93;
  else if (p < 0)
    p = 0;
  return ((char)(p + 64));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<phredToPhred64>:
  endbr64
  test   %edi,%edi
  mov    $0x0,%eax
  cmovns %edi,%eax
  mov    $0x5d,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0x40,%eax
  ret
```

assistant: ```c
char phredToPhred64(int p) {
  if (p > 93)
    p = 93;
  else if (p < 0)
    p = 0;
  return ((char)(p + 64));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<phredToPhred64>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  mov    $0x5d,%edx
  cmovns %edi,%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0x40,%eax
  ret
```

assistant: ```c
char phredToPhred64(int p) {
  if (p > 93)
    p = 93;
  else if (p < 0)
    p = 0;
  return ((char)(p + 64));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<phredToPhred64>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  mov    $0x5d,%edx
  cmovns %edi,%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0x40,%eax
  ret
```

assistant: ```c
char phredToPhred64(int p) {
  if (p > 93)
    p = 93;
  else if (p < 0)
    p = 0;
  return ((char)(p + 64));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rop3_189_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x1c(%rbp),%al
  mov    %al,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x20(%rbp),%al
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    %al,-0x2(%rbp)
  notb   -0x2(%rbp)
  movzbl -0x2(%rbp),%eax
  and    -0x1c(%rbp),%al
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x2(%rbp),%eax
  and    %edx,%eax
  or     %eax,%ecx
  mov    %ecx,%edx
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  and    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    %ecx,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_189_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk1 = S ^ *D;
  stk2 = T ^ *D;
  stk1 = stk1 & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: What is the c source code of the assembly code below:

```asm
<rop3_189_0_1>:
  endbr64
  mov    %esi,%ecx
  xor    (%rdi),%cl
  mov    %edx,%eax
  xor    (%rdi),%al
  and    %ecx,%eax
  mov    %eax,%ecx
  not    %ecx
  and    %esi,%ecx
  or     %edx,%eax
  not    %eax
  or     %ecx,%eax
  not    %esi
  and    %edx,%esi
  and    (%rdi),%sil
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_189_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk1 = S ^ *D;
  stk2 = T ^ *D;
  stk1 = stk1 & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rop3_189_0_1>:
  endbr64
  movzbl (%rdi),%ecx
  movzbl (%rdi),%eax
  xor    %esi,%ecx
  xor    %edx,%eax
  and    %ecx,%eax
  mov    %eax,%ecx
  or     %edx,%eax
  not    %ecx
  not    %eax
  and    %esi,%ecx
  not    %esi
  or     %ecx,%eax
  and    %edx,%esi
  and    (%rdi),%sil
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_189_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk1 = S ^ *D;
  stk2 = T ^ *D;
  stk1 = stk1 & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rop3_189_0_1>:
  endbr64
  movzbl (%rdi),%ecx
  movzbl (%rdi),%eax
  xor    %esi,%ecx
  xor    %edx,%eax
  and    %ecx,%eax
  mov    %eax,%ecx
  or     %edx,%eax
  not    %ecx
  not    %eax
  and    %esi,%ecx
  not    %esi
  or     %ecx,%eax
  and    %edx,%esi
  and    (%rdi),%sil
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_189_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk1 = S ^ *D;
  stk2 = T ^ *D;
  stk1 = stk1 & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0xf5,-0x4(%rbp)
  jne    L2
  cmpl   $0x24,-0x8(%rbp)
  jne    L2
  cmpl   $0x1bc,-0xc(%rbp)
  je     L29
L2:
  mov    $0x0,%eax
  call   <abort@plt>
L29:
  nop
  leave
  ret
```

assistant: ```c
void foo(int x, int y, int z) {
  if (x != 245 || y != 36 || z != 444)
    abort();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  cmp    $0x24,%esi
  setne  %al
  cmp    $0x1bc,%edx
  setne  %dl
  or     %dl,%al
  jne    L14
  cmp    $0xf5,%edi
  je     L16
L14:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <abort@plt>
  add    $0x8,%rsp
  ret
L16:
  ret
```

assistant: ```c
void foo(int x, int y, int z) {
  if (x != 245 || y != 36 || z != 444)
    abort();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<foo>:
  endbr64
  cmp    $0x24,%esi
  setne  %al
  cmp    $0x1bc,%edx
  setne  %dl
  or     %dl,%al
  jne    L0
  cmp    $0xf5,%edi
  je     L5
L0:
  xor    %eax,%eax
  jmp    1050 <abort@plt>
  cs nopw 0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void foo(int x, int y, int z) {
  if (x != 245 || y != 36 || z != 444)
    abort();
}
```