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

```asm
<crossxyjusevoter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    %r9,-0x48(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L13
L21:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L21
  movl   $0x0,-0x18(%rbp)
  jmp    L22
L62:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L23
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x3c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L35
L46:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
L35:
  mov    -0x14(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L46
L23:
  addl   $0x1,-0x18(%rbp)
L22:
  mov    -0x18(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L62
  nop
  pop    %rbp
  ret
```

assistant: ```c
void crossxyjusevoter(double **x, double **y, int n, int k, int p, double *xpy, int *usevoter) {
  int i, j;
  double *xrow, yip;
  for (j = 0; j < k; j++) {
    xpy[j] = 0.0;
  }
  for (i = 0; i < n; i++) {
    if (usevoter[i] > 0) {
      xrow = x[i];
      yip = y[i][p];
      for (j = 0; j < k; j++) {
        xpy[j] += xrow[j] * yip;
      }
    }
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<crossxyjusevoter>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%r10
  mov    %ecx,%r11d
  mov    0x18(%rsp),%rbx
  test   %ecx,%ecx
  jle    L34
  mov    %r9,%rax
  lea    -0x1(%rcx),%ecx
  lea    0x8(%r9,%rcx,8),%rcx
L82:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L82
L34:
  test   %edx,%edx
  jle    L9
  movslq %r8d,%r8
  lea    0x0(,%r8,8),%rbp
  mov    %edx,%r8d
  mov    $0x0,%edx
  mov    %r11d,%esi
  jmp    L2
L0:
  add    $0x1,%rdx
  cmp    %r8,%rdx
  je     L9
L2:
  cmpl   $0x0,(%rbx,%rdx,4)
  jle    L0
  mov    (%rdi,%rdx,8),%rcx
  mov    (%r10,%rdx,8),%rax
  movsd  (%rax,%rbp,1),%xmm1
  test   %r11d,%r11d
  jle    L0
  mov    $0x0,%eax
L87:
  movapd %xmm1,%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  addsd  (%r9,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L87
  jmp    L0
L9:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void crossxyjusevoter(double **x, double **y, int n, int k, int p, double *xpy, int *usevoter) {
  int i, j;
  double *xrow, yip;
  for (j = 0; j < k; j++) {
    xpy[j] = 0.0;
  }
  for (i = 0; i < n; i++) {
    if (usevoter[i] > 0) {
      xrow = x[i];
      yip = y[i][p];
      for (j = 0; j < k; j++) {
        xpy[j] += xrow[j] * yip;
      }
    }
  }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<crossxyjusevoter>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  movslq %r8d,%r14
  push   %r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    0x40(%rsp),%r13
  test   %ecx,%ecx
  jle    L29
  movslq %ecx,%rdx
  mov    %r9,%rdi
  xor    %esi,%esi
  shl    $0x3,%rdx
  call   <memset@plt>
  mov    %rax,%r9
L29:
  test   %r15d,%r15d
  jle    L40
  shl    $0x3,%r14
  movslq %r15d,%rdi
  xor    %edx,%edx
  movslq %r12d,%rsi
  jmp    L51
  nopl   0x0(%rax)
L6:
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  je     L40
L51:
  mov    0x0(%r13,%rdx,4),%eax
  test   %eax,%eax
  jle    L6
  mov    0x0(%rbp,%rdx,8),%rax
  mov    (%rbx,%rdx,8),%rcx
  movsd  (%rax,%r14,1),%xmm1
  test   %r12d,%r12d
  jle    L6
  xor    %eax,%eax
L28:
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  %xmm1,%xmm0
  addsd  (%r9,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L28
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  jne    L51
L40:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void crossxyjusevoter(double **x, double **y, int n, int k, int p, double *xpy, int *usevoter) {
  int i, j;
  double *xrow, yip;
  for (j = 0; j < k; j++) {
    xpy[j] = 0.0;
  }
  for (i = 0; i < n; i++) {
    if (usevoter[i] > 0) {
      xrow = x[i];
      yip = y[i][p];
      for (j = 0; j < k; j++) {
        xpy[j] += xrow[j] * yip;
      }
    }
  }
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<crossxyjusevoter>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%r13
  test   %ecx,%ecx
  jle    L5
  movslq %ecx,%rdx
  mov    %r9,%rdi
  xor    %esi,%esi
  shl    $0x3,%rdx
  call   <memset@plt>
  mov    %rax,%r9
L5:
  test   %r15d,%r15d
  jle    L7
  mov    %r12d,%eax
  movslq %r14d,%r8
  movslq %r15d,%r11
  shr    %eax
  shl    $0x3,%r8
  lea    -0x1(%r12),%r15d
  lea    -0x1(%rax),%edi
  mov    %r12d,%eax
  and    $0xfffffffe,%eax
  add    $0x1,%rdi
  movslq %eax,%r10
  mov    %eax,0xc(%rsp)
  shl    $0x4,%rdi
  xor    %eax,%eax
  shl    $0x3,%r10
  lea    (%r9,%r10,1),%rsi
  jmp    L23
L77:
  add    $0x1,%rax
  cmp    %r11,%rax
  je     L7
  nopl   0x0(%rax)
L23:
  mov    0x0(%r13,%rax,4),%edx
  test   %edx,%edx
  jle    L77
  mov    0x0(%rbp,%rax,8),%rcx
  mov    (%rbx,%rax,8),%rdx
  movsd  (%rcx,%r8,1),%xmm1
  test   %r12d,%r12d
  jle    L77
  lea    0x8(%rdx),%rcx
  cmp    %rcx,%r9
  je     L66
  cmp    $0x1,%r12d
  je     L66
  movapd %xmm1,%xmm2
  xor    %ecx,%ecx
  unpcklpd %xmm2,%xmm2
  cs nopw 0x0(%rax,%rax,1)
L15:
  movupd (%rdx,%rcx,1),%xmm0
  movupd (%r9,%rcx,1),%xmm3
  mulpd  %xmm2,%xmm0
  addpd  %xmm3,%xmm0
  movups %xmm0,(%r9,%rcx,1)
  add    $0x10,%rcx
  cmp    %rdi,%rcx
  jne    L15
  cmp    0xc(%rsp),%r12d
  je     L77
  mulsd  (%rdx,%r10,1),%xmm1
  movsd  (%rsi),%xmm0
  add    $0x1,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi)
  cmp    %r11,%rax
  jne    L23
L7:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L66:
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L40:
  movsd  (%rdx,%rcx,8),%xmm0
  mov    %rcx,%r14
  mulsd  %xmm1,%xmm0
  addsd  (%r9,%rcx,8),%xmm0
  movsd  %xmm0,(%r9,%rcx,8)
  add    $0x1,%rcx
  cmp    %r14,%r15
  jne    L40
  jmp    L77
```

assistant: ```c
void crossxyjusevoter(double **x, double **y, int n, int k, int p, double *xpy, int *usevoter) {
  int i, j;
  double *xrow, yip;
  for (j = 0; j < k; j++) {
    xpy[j] = 0.0;
  }
  for (i = 0; i < n; i++) {
    if (usevoter[i] > 0) {
      xrow = x[i];
      yip = y[i][p];
      for (j = 0; j < k; j++) {
        xpy[j] += xrow[j] * yip;
      }
    }
  }
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L8:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    $0x5f,%eax
  je     L35
  cmp    $0x80,%eax
  jne    L41
  mov    $0x0,%eax
  jmp    L91
L35:
  lea    -0xb(%rbp),%rax
  lea    0x1(%rax),%rdx
  lea    -0xb(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x2,-0x10(%rbp)
  je     L16
  mov    $0x1,%eax
  jmp    L91
L16:
  lea    -0x14(%rbp),%rdx
  lea    -0xb(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x80,-0x10(%rbp)
  je     L21
  cmpl   $0x0,-0x10(%rbp)
  jne    L41
L21:
  mov    $0x2,%eax
  jmp    L91
L41:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  nop
  jmp    L8
L91:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_infered: string
name: extract_symbolic_data
arguments: tag: D7
type_infered: string

tool: D18: %1[0-9A-Fa-f]%1[0-9A-Fa-f]

tool: D7: %2x

assistant: ```c
int main(int argc, char **argv) {
  while (1) {
    int c;
    switch (c = getchar()) {
    case EOF:
      return 0;
    case '_': {
      char hex[3];
      int i;
      i = scanf("%1[0-9A-Fa-f]%1[0-9A-Fa-f]", &hex[0], &hex[1]);
      if (i != 2) {
        return 1;
      }
      i = sscanf(hex, "%2x", &c);
      if (i == EOF || i == 0) {
        return 2;
      }
    }
    default:
      putchar(c);
      break;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rbx
  lea    0x6(%rsp),%r12
  lea    D17(%rip),%rbp
  lea    D14(%rip),%r13
  jmp    L24
L25:
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %eax,%r14d
  cmp    $0x2,%eax
  jne    L44
  mov    %rsp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   $0xffffff7f,%eax
  je     L1
L10:
  mov    (%rsp),%edi
  call   <putchar@plt>
L24:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,(%rsp)
  cmp    $0x5f,%eax
  je     L25
  cmp    $0x80,%eax
  jne    L10
  mov    $0x0,%r14d
L1:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    %r14d,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L44:
  mov    $0x1,%r14d
  jmp    L1
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D17
data_type: string
name: parse_data
arguments: data_tag: D14
data_type: string

tool: D17: %1[0-9A-Fa-f]%1[0-9A-Fa-f]

tool: D14: %2x

assistant: ```c
int main(int argc, char **argv) {
  while (1) {
    int c;
    switch (c = getchar()) {
    case EOF:
      return 0;
    case '_': {
      char hex[3];
      int i;
      i = scanf("%1[0-9A-Fa-f]%1[0-9A-Fa-f]", &hex[0], &hex[1]);
      if (i != 2) {
        return 1;
      }
      i = sscanf(hex, "%2x", &c);
      if (i == EOF || i == 0) {
        return 2;
      }
    }
    default:
      putchar(c);
      break;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  lea    D12(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rbx
  lea    0x6(%rsp),%r12
L17:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,(%rsp)
  mov    %eax,%edi
  cmp    $0x5f,%eax
  je     L19
  cmp    $0x80,%eax
  jne    L52
  xor    %r14d,%r14d
L49:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x10,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %eax,%r14d
  cmp    $0x2,%eax
  jne    L2
  mov    %rsp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   $0xffffff7f,%eax
  je     L49
  mov    (%rsp),%edi
L52:
  call   <putchar@plt>
  jmp    L17
L2:
  mov    $0x1,%r14d
  jmp    L49
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D16
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string

tool: D16: %2x

tool: D12: %1[0-9A-Fa-f]%1[0-9A-Fa-f]

assistant: ```c
int main(int argc, char **argv) {
  while (1) {
    int c;
    switch (c = getchar()) {
    case EOF:
      return 0;
    case '_': {
      char hex[3];
      int i;
      i = scanf("%1[0-9A-Fa-f]%1[0-9A-Fa-f]", &hex[0], &hex[1]);
      if (i != 2) {
        return 1;
      }
      i = sscanf(hex, "%2x", &c);
      if (i == EOF || i == 0) {
        return 2;
      }
    }
    default:
      putchar(c);
      break;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x5(%rsp),%rbx
  lea    0x6(%rsp),%r12
L38:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,(%rsp)
  mov    %eax,%edi
  cmp    $0x5f,%eax
  je     L5
  cmp    $0x80,%eax
  jne    L64
  xor    %r14d,%r14d
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x10,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %eax,%r14d
  cmp    $0x2,%eax
  jne    L45
  mov    %rsp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   $0xffffff7f,%eax
  je     L21
  mov    (%rsp),%edi
L64:
  call   <putchar@plt>
  jmp    L38
L45:
  mov    $0x1,%r14d
  jmp    L21
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D6
type: string

tool: D1: %2x

tool: D6: %1[0-9A-Fa-f]%1[0-9A-Fa-f]

assistant: ```c
int main(int argc, char **argv) {
  while (1) {
    int c;
    switch (c = getchar()) {
    case EOF:
      return 0;
    case '_': {
      char hex[3];
      int i;
      i = scanf("%1[0-9A-Fa-f]%1[0-9A-Fa-f]", &hex[0], &hex[1]);
      if (i != 2) {
        return 1;
      }
      i = sscanf(hex, "%2x", &c);
      if (i == EOF || i == 0) {
        return 2;
      }
    }
    default:
      putchar(c);
      break;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CharMediaList_Compare>:
  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
  mov    %al,-0x12(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  cmpb   $0x40,-0x12(%rbp)
  jle    L35
  cmpb   $0x5a,-0x12(%rbp)
  jg     L35
  movzbl -0x12(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x12(%rbp)
L35:
  cmpb   $0x40,-0x11(%rbp)
  jle    L36
  cmpb   $0x5a,-0x11(%rbp)
  jg     L36
  movzbl -0x11(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x11(%rbp)
  jmp    L36
L6:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x12(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
L36:
  movzbl -0x12(%rbp),%eax
  cmp    -0x11(%rbp),%al
  jne    L63
  cmpb   $0x0,-0x12(%rbp)
  je     L63
  cmpb   $0x0,-0x11(%rbp)
  jne    L6
L63:
  movsbl -0x12(%rbp),%eax
  movsbl -0x11(%rbp),%edx
  sub    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int CharMediaList_Compare(const void *ptr1, const void *ptr2) {
  const char *str1, *str2;
  char chr1, chr2;
  str1 = (const char *)ptr1;
  str2 = (const char *)ptr2;
  chr1 = *str1;
  chr2 = *str2;
  if (chr1 >= 'A' && chr1 <= 'Z')
    chr1 += 32;
  if (chr2 >= 'A' && chr2 <= 'Z')
    chr2 += 32;
  while (chr1 == chr2 && chr1 && chr2) {
    chr1 = *(str1++);
    chr2 = *(str2++);
  }
  return ((int)chr1 - (int)chr2);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<CharMediaList_Compare>:
  endbr64
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  lea    -0x41(%rax),%r8d
  lea    0x20(%rax),%ecx
  cmp    $0x1a,%r8b
  cmovb  %ecx,%eax
  lea    -0x41(%rdx),%r8d
  lea    0x20(%rdx),%ecx
  cmp    $0x1a,%r8b
  cmovb  %ecx,%edx
  cmp    %dl,%al
  sete   %r8b
  test   %dl,%dl
  setne  %cl
  test   %cl,%r8b
  je     L29
  test   %al,%al
  je     L29
  mov    $0x0,%ecx
L34:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%edx
  add    $0x1,%rcx
  cmp    %dl,%al
  sete   %r9b
  test   %al,%al
  setne  %r8b
  test   %r8b,%r9b
  je     L29
  test   %dl,%dl
  jne    L34
L29:
  movsbl %al,%eax
  movsbl %dl,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int CharMediaList_Compare(const void *ptr1, const void *ptr2) {
  const char *str1, *str2;
  char chr1, chr2;
  str1 = (const char *)ptr1;
  str2 = (const char *)ptr2;
  chr1 = *str1;
  chr2 = *str2;
  if (chr1 >= 'A' && chr1 <= 'Z')
    chr1 += 32;
  if (chr2 >= 'A' && chr2 <= 'Z')
    chr2 += 32;
  while (chr1 == chr2 && chr1 && chr2) {
    chr1 = *(str1++);
    chr2 = *(str2++);
  }
  return ((int)chr1 - (int)chr2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<CharMediaList_Compare>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%ecx
  lea    -0x41(%rax),%r8d
  lea    0x20(%rax),%edx
  cmp    $0x1a,%r8b
  lea    -0x41(%rcx),%r9d
  lea    0x20(%rcx),%r8d
  cmovae %eax,%edx
  cmp    $0x1a,%r9b
  cmovae %ecx,%r8d
  test   %dl,%dl
  je     L59
  cmp    %r8b,%dl
  jne    L59
  xor    %edx,%edx
  jmp    L49
  nopl   (%rax)
L38:
  test   %al,%al
  je     L7
  movsbl (%rdi,%rdx,1),%eax
  movsbl (%rsi,%rdx,1),%ecx
L49:
  add    $0x1,%rdx
  cmp    %cl,%al
  je     L38
L7:
  sub    %ecx,%eax
  ret
L59:
  movsbl %r8b,%ecx
  movsbl %dl,%eax
  jmp    L7
```

assistant: ```c
int CharMediaList_Compare(const void *ptr1, const void *ptr2) {
  const char *str1, *str2;
  char chr1, chr2;
  str1 = (const char *)ptr1;
  str2 = (const char *)ptr2;
  chr1 = *str1;
  chr2 = *str2;
  if (chr1 >= 'A' && chr1 <= 'Z')
    chr1 += 32;
  if (chr2 >= 'A' && chr2 <= 'Z')
    chr2 += 32;
  while (chr1 == chr2 && chr1 && chr2) {
    chr1 = *(str1++);
    chr2 = *(str2++);
  }
  return ((int)chr1 - (int)chr2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CharMediaList_Compare>:
  endbr64
  movzbl (%rdi),%eax
  movzbl (%rsi),%ecx
  lea    -0x41(%rax),%r8d
  lea    0x20(%rax),%edx
  cmp    $0x1a,%r8b
  lea    -0x41(%rcx),%r8d
  cmovb  %edx,%eax
  lea    0x20(%rcx),%edx
  cmp    $0x1a,%r8b
  cmovb  %edx,%ecx
  test   %al,%al
  je     L23
  cmp    %cl,%al
  jne    L23
  xor    %edx,%edx
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L47:
  test   %al,%al
  je     L23
L19:
  movzbl (%rdi,%rdx,1),%eax
  movzbl (%rsi,%rdx,1),%ecx
  add    $0x1,%rdx
  cmp    %cl,%al
  je     L47
L23:
  movsbl %cl,%ecx
  movsbl %al,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int CharMediaList_Compare(const void *ptr1, const void *ptr2) {
  const char *str1, *str2;
  char chr1, chr2;
  str1 = (const char *)ptr1;
  str2 = (const char *)ptr2;
  chr1 = *str1;
  chr2 = *str2;
  if (chr1 >= 'A' && chr1 <= 'Z')
    chr1 += 32;
  if (chr2 >= 'A' && chr2 <= 'Z')
    chr2 += 32;
  while (chr1 == chr2 && chr1 && chr2) {
    chr1 = *(str1++);
    chr2 = *(str2++);
  }
  return ((int)chr1 - (int)chr2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<imprime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L5
L7:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void imprime(char *s) {
  while (*s != 0x00) {
    putchar(*s++);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<imprime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L12
L6:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <putchar@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L6
L12:
  pop    %rbx
  ret
```

assistant: ```c
void imprime(char *s) {
  while (*s != 0x00) {
    putchar(*s++);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<imprime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L13
L4:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L4
L13:
  pop    %rbx
  ret
```

assistant: ```c
void imprime(char *s) {
  while (*s != 0x00) {
    putchar(*s++);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<imprime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L0
L18:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L18
L0:
  pop    %rbx
  ret
```

assistant: ```c
void imprime(char *s) {
  while (*s != 0x00) {
    putchar(*s++);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movq   $0xa,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x0,%r11d
  mov    -0x30(%rbp),%rax
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    -0x30(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L25:
  cmp    %rdx,%rsp
  je     L10
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L25
L10:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L35
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L35:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <overflow@plt>
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const size_t length = 10;
  int buffer[length];
  overflow(buffer, length);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0xa,%esi
  call   <overflow@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  const size_t length = 10;
  int buffer[length];
  overflow(buffer, length);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0xa,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <overflow@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  const size_t length = 10;
  int buffer[length];
  overflow(buffer, length);
  return 0;
}
```