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

```asm
<key_common_len>:
  endbr64
  cmp    %ecx,%esi
  mov    %rdi,%r10
  cmovle %esi,%ecx
  test   %ecx,%ecx
  lea    0x7(%rcx),%esi
  mov    %ecx,%r9d
  cmovns %ecx,%esi
  xor    %eax,%eax
  sar    $0x3,%esi
  cmp    $0x7,%ecx
  jg     L45
  jmp    L86
  nopl   0x0(%rax,%rax,1)
L61:
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %eax,%esi
  jle    L106
L45:
  movzbl (%r10,%rax,1),%edi
  movzbl (%rdx,%rax,1),%r8d
  cmp    %r8b,%dil
  je     L61
  lea    0x0(,%rax,8),%eax
L35:
  xor    %r8d,%edi
  cmp    %eax,%r9d
  jle    L1
  test   %dil,%dil
  js     L1
  mov    $0xffffff80,%edx
  jmp    L47
  nopl   (%rax)
L72:
  test   %dl,%dil
  jne    L1
L47:
  add    $0x1,%eax
  shr    %dl
  cmp    %eax,%r9d
  jne    L72
  mov    %r9d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %eax,%r9d
  mov    %r9d,%eax
  ret
  nop
L106:
  movslq %ecx,%rax
  movzbl (%r10,%rax,1),%edi
  movzbl (%rdx,%rax,1),%r8d
  lea    0x0(,%rcx,8),%eax
  jmp    L35
L86:
  movzbl (%rdi),%edi
  movzbl (%rdx),%r8d
  jmp    L35
```

assistant: ```c
int key_common_len(char *keyi, int keyilen, char *keyj, int keyjlen) {
  int nmatch = 0;
  int minkeylen = ((keyilen) > (keyjlen) ? (keyjlen) : (keyilen));
  int keylen = 0;
  unsigned char bitmask;
  unsigned char diff;
  nmatch = 0;
  while (nmatch < minkeylen / 8 && keyi[nmatch] == keyj[nmatch]) {
    nmatch++;
  }
  keylen = nmatch * 8;
  bitmask = 0x80;
  diff = keyi[nmatch] ^ keyj[nmatch];
  while (keylen < minkeylen && !(bitmask & diff)) {
    keylen++;
    bitmask >>= 1;
  }
  return keylen;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<key_common_len>:
  endbr64
  cmp    %ecx,%esi
  mov    %rdi,%r10
  cmovle %esi,%ecx
  test   %ecx,%ecx
  lea    0x7(%rcx),%esi
  mov    %ecx,%r9d
  cmovns %ecx,%esi
  xor    %eax,%eax
  sar    $0x3,%esi
  cmp    $0x7,%ecx
  jg     L29
  jmp    L107
  nopl   0x0(%rax,%rax,1)
L65:
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %eax,%esi
  jle    L22
L29:
  movzbl (%r10,%rax,1),%edi
  movzbl (%rdx,%rax,1),%r8d
  cmp    %r8b,%dil
  je     L65
  lea    0x0(,%rax,8),%eax
L89:
  xor    %r8d,%edi
  cmp    %eax,%r9d
  jle    L52
  test   %dil,%dil
  js     L52
  mov    $0xffffff80,%edx
  jmp    L64
  nopl   (%rax)
L71:
  test   %dl,%dil
  jne    L52
L64:
  add    $0x1,%eax
  shr    %dl
  cmp    %eax,%r9d
  jne    L71
  mov    %r9d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L52:
  mov    %eax,%r9d
  mov    %r9d,%eax
  ret
  nop
L22:
  movslq %ecx,%rax
  movzbl (%r10,%rax,1),%edi
  movzbl (%rdx,%rax,1),%r8d
  lea    0x0(,%rcx,8),%eax
  jmp    L89
L107:
  movzbl (%rdi),%edi
  movzbl (%rdx),%r8d
  jmp    L89
```

assistant: ```c
int key_common_len(char *keyi, int keyilen, char *keyj, int keyjlen) {
  int nmatch = 0;
  int minkeylen = ((keyilen) > (keyjlen) ? (keyjlen) : (keyilen));
  int keylen = 0;
  unsigned char bitmask;
  unsigned char diff;
  nmatch = 0;
  while (nmatch < minkeylen / 8 && keyi[nmatch] == keyj[nmatch]) {
    nmatch++;
  }
  keylen = nmatch * 8;
  bitmask = 0x80;
  diff = keyi[nmatch] ^ keyj[nmatch];
  while (keylen < minkeylen && !(bitmask & diff)) {
    keylen++;
    bitmask >>= 1;
  }
  return keylen;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  mov    $0x1,%esi
  lea    D14(%rip),%r13
  lea    D39(%rip),%r12
  cmpl   $0x0,(%rsp)
  jg     L21
L76:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L56:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jl     L76
  movslq %ebx,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%esi
  mov    %esi,%ecx
  lea    (%rsi,%rsi,4),%eax
  add    %eax,%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  imul   %esi,%esi
  mov    %ecx,%edx
  test   %ecx,%ecx
  je     L21
L65:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%edi
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   %eax,%eax
  add    %eax,%esi
  mov    %edi,%edx
  test   %edi,%edi
  jne    L65
L21:
  mov    %ebx,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    %esi,%eax
  jne    L56
  add    $0x1,%ebp
  movslq %ebp,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  cmp    %eax,%ebp
  je     L8
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L56
L77:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D40
datatype: string
name: disassemble_data
arguments: label: D25
datatype: string
name: disassemble_data
arguments: label: D20
datatype: string
name: disassemble_data
arguments: label: D14
datatype: string
name: disassemble_data
arguments: label: D39
datatype: string

tool: D40: Please input two integers:

tool: D25: %d %d

tool: D20: The result is:

tool: D14: %d


tool: D39: %d	

assistant: ```c
int main() {
  int i, n, m, t, sum, count = 0;
  printf("Please input two integers:");
  scanf("%d %d", &n, &m);
  printf("The result is:\n");
  for (i = 1; i <= n; i++) {
    sum = 0;
    t = i;
    sum = (t % 10) * (t % 10);
    while (t /= 10)
      sum += (t % 10) * (t % 10);
    if (i / m == sum) {
      count++;
      if (count % 5)
        printf("%d\t", i);
      else
        printf("%d\n", i);
    }
  }
  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
<main>:
  endbr64
  push   %r14
  lea    D13(%rip),%rdi
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  lea    D39(%rip),%rbp
  push   %rbx
  lea    D24(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  lea    D25(%rip),%rdi
  call   <puts@plt>
  mov    (%rsp),%eax
  mov    $0x1,%esi
  test   %eax,%eax
  jle    L36
L19:
  mov    %r14d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    %esi,%eax
  je     L48
  nopl   (%rax)
L29:
  add    $0x1,%r14d
  cmp    %r14d,(%rsp)
  jl     L36
L81:
  mov    %r14d,%esi
  imul   %r12,%rsi
  shr    $0x23,%rsi
  lea    (%rsi,%rsi,4),%eax
  mov    %esi,%edx
  mov    %r14d,%esi
  add    %eax,%eax
  sub    %eax,%esi
  imul   %esi,%esi
  test   %edx,%edx
  je     L19
L87:
  mov    %edx,%eax
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %edi,%edi
  mov    %ecx,%edx
  sub    %edi,%eax
  imul   %eax,%eax
  add    %eax,%esi
  test   %ecx,%ecx
  jne    L87
  mov    %r14d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    %esi,%eax
  jne    L29
L48:
  add    $0x1,%r13d
  mov    %r14d,%esi
  imul   $0xcccccccd,%r13d,%eax
  cmp    $0x33333333,%eax
  jbe    L89
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %r14d,(%rsp)
  jge    L81
  nopl   (%rax)
L36:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L89:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L29
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D13
datatype: string
name: disassemble_data
arguments: label: D39
datatype: string
name: disassemble_data
arguments: label: D24
datatype: string
name: disassemble_data
arguments: label: D30
datatype: string
name: disassemble_data
arguments: label: D25
datatype: string

tool: D13: Please input two integers:

tool: D39: %d


tool: D24: %d	

tool: D30: %d %d

tool: D25: The result is:

assistant: ```c
int main() {
  int i, n, m, t, sum, count = 0;
  printf("Please input two integers:");
  scanf("%d %d", &n, &m);
  printf("The result is:\n");
  for (i = 1; i <= n; i++) {
    sum = 0;
    t = i;
    sum = (t % 10) * (t % 10);
    while (t /= 10)
      sum += (t % 10) * (t % 10);
    if (i / m == sum) {
      count++;
      if (count % 5)
        printf("%d\t", i);
      else
        printf("%d\n", i);
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  lea    D20(%rip),%rdi
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  lea    D39(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    (%rsp),%esi
  mov    $0x1,%ecx
  test   %esi,%esi
  jle    L12
L63:
  mov    %r14d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    %ecx,%eax
  je     L84
  nopl   (%rax)
L94:
  add    $0x1,%r14d
  cmp    %esi,%r14d
  jg     L12
L44:
  mov    %r14d,%ecx
  imul   %r12,%rcx
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%eax
  mov    %ecx,%edx
  mov    %r14d,%ecx
  add    %eax,%eax
  sub    %eax,%ecx
  imul   %ecx,%ecx
  test   %edx,%edx
  je     L63
  nop
L50:
  mov    %edx,%eax
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r8d
  mov    %eax,%edi
  mov    %edx,%eax
  add    %r8d,%r8d
  mov    %edi,%edx
  sub    %r8d,%eax
  imul   %eax,%eax
  add    %eax,%ecx
  test   %edi,%edi
  jne    L50
  mov    %r14d,%eax
  cltd
  idivl  0x4(%rsp)
  cmp    %ecx,%eax
  jne    L94
L84:
  add    $0x1,%r13d
  mov    %r14d,%esi
  imul   $0xcccccccd,%r13d,%eax
  cmp    $0x33333333,%eax
  jbe    L52
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14d
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    %esi,%r14d
  jle    L44
  nopw   0x0(%rax,%rax,1)
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L52:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  jmp    L94
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D20
type_guessed: string
name: disassemble_data
arguments: data_label: D0
type_guessed: string
name: disassemble_data
arguments: data_label: D39
type_guessed: string
name: disassemble_data
arguments: data_label: D2
type_guessed: string
name: disassemble_data
arguments: data_label: D13
type_guessed: string

tool: D20: Please input two integers:

tool: D0: %d


tool: D39: %d	

tool: D2: %d %d

tool: D13: The result is:

assistant: ```c
int main() {
  int i, n, m, t, sum, count = 0;
  printf("Please input two integers:");
  scanf("%d %d", &n, &m);
  printf("The result is:\n");
  for (i = 1; i <= n; i++) {
    sum = 0;
    t = i;
    sum = (t % 10) * (t % 10);
    while (t /= 10)
      sum += (t % 10) * (t % 10);
    if (i / m == sum) {
      count++;
      if (count % 5)
        printf("%d\t", i);
      else
        printf("%d\n", i);
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fmatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L6:
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L7
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L7:
  movq   $0x1,-0x10(%rbp)
  jmp    L20
L1:
  mov    -0x10(%rbp),%rax
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addq   $0x1,-0x10(%rbp)
L20:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x10(%rbp)
  jne    L1
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
datatype: string
name: extract_symbolic_data
arguments: data_tag: D9
datatype: string

tool: D2: Error: could not allocate memory:%d


tool: D9: Error: could not allocate memory

assistant: ```c
float **fmatrix(int xsize, int ysize) {
  long i;
  float **m;
  if ((m = (float **)malloc(xsize * sizeof(float *))) == (float **)NULL) {
    printf("Error: could not allocate memory:%d\n", xsize);
    exit(1);
  }
  if ((m[0] = (float *)malloc(xsize * ysize * sizeof(float))) == (float *)NULL) {
    printf("Error: could not allocate memory\n");
    exit(1);
  }
  for (i = 1; i != xsize; i++) {
    m[i] = m[i - 1] + ysize;
  }
  return (m);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fmatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %esi,%ebx
  movslq %edi,%r12
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L48
L8:
  mov    %r13d,%edi
  imul   %ebx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L47
L58:
  cmp    $0x1,%r13d
  je     L27
  movslq %ebx,%rcx
  shl    $0x2,%rcx
  lea    0x8(%rbp),%rax
  lea    0x0(%rbp,%r12,8),%rsi
L55:
  mov    %rcx,%rdx
  add    -0x8(%rax),%rdx
  mov    %rdx,(%rax)
  add    $0x8,%rax
  cmp    %rsi,%rax
  jne    L55
L27:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  mov    %r13d,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L8
L47:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L58
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
type_guessed: string
name: extract_symbolic_data
arguments: label: D19
type_guessed: string

tool: D10: Error: could not allocate memory:%d


tool: D19: Error: could not allocate memory

assistant: ```c
float **fmatrix(int xsize, int ysize) {
  long i;
  float **m;
  if ((m = (float **)malloc(xsize * sizeof(float *))) == (float **)NULL) {
    printf("Error: could not allocate memory:%d\n", xsize);
    exit(1);
  }
  if ((m[0] = (float *)malloc(xsize * ysize * sizeof(float))) == (float *)NULL) {
    printf("Error: could not allocate memory\n");
    exit(1);
  }
  for (i = 1; i != xsize; i++) {
    m[i] = m[i - 1] + ysize;
  }
  return (m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fmatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  mov    %esi,%ebx
  mov    %rbp,%r13
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L41
L19:
  mov    %r13d,%edi
  imul   %ebx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L25
L22:
  cmp    $0x1,%r13d
  je     L52
  movslq %ebx,%rcx
  lea    0x8(%r12),%rax
  lea    (%r12,%rbp,8),%rsi
  shl    $0x2,%rcx
  nopw   0x0(%rax,%rax,1)
L26:
  mov    -0x8(%rax),%rdx
  add    $0x8,%rax
  add    %rcx,%rdx
  mov    %rdx,-0x8(%rax)
  cmp    %rsi,%rax
  jne    L26
L52:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %ebp,%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L19
  nopl   0x0(%rax)
L25:
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L22
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string
name: disassemble_data
arguments: data_tag: D18
type_infered: string

tool: D1: Error: could not allocate memory:%d


tool: D18: Error: could not allocate memory

assistant: ```c
float **fmatrix(int xsize, int ysize) {
  long i;
  float **m;
  if ((m = (float **)malloc(xsize * sizeof(float *))) == (float **)NULL) {
    printf("Error: could not allocate memory:%d\n", xsize);
    exit(1);
  }
  if ((m[0] = (float *)malloc(xsize * ysize * sizeof(float))) == (float *)NULL) {
    printf("Error: could not allocate memory\n");
    exit(1);
  }
  for (i = 1; i != xsize; i++) {
    m[i] = m[i - 1] + ysize;
  }
  return (m);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fmatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  mov    %esi,%ebx
  mov    %rbp,%r13
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L29
L20:
  mov    %r13d,%edi
  imul   %ebx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L50
L6:
  cmp    $0x1,%r13d
  je     L0
  movslq %ebx,%rcx
  mov    (%r12),%rdx
  lea    0x8(%r12),%rax
  lea    (%r12,%rbp,8),%rsi
  shl    $0x2,%rcx
  nopl   0x0(%rax,%rax,1)
L32:
  add    %rcx,%rdx
  add    $0x8,%rax
  mov    %rdx,-0x8(%rax)
  cmp    %rax,%rsi
  jne    L32
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L29:
  mov    %ebp,%esi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L20
  nopl   0x0(%rax)
L50:
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L6
```

tool_calls: name: read_data
arguments: data_tag: D10
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string

tool: D10: Error: could not allocate memory:%d


tool: D13: Error: could not allocate memory

assistant: ```c
float **fmatrix(int xsize, int ysize) {
  long i;
  float **m;
  if ((m = (float **)malloc(xsize * sizeof(float *))) == (float **)NULL) {
    printf("Error: could not allocate memory:%d\n", xsize);
    exit(1);
  }
  if ((m[0] = (float *)malloc(xsize * ysize * sizeof(float))) == (float *)NULL) {
    printf("Error: could not allocate memory\n");
    exit(1);
  }
  for (i = 1; i != xsize; i++) {
    m[i] = m[i - 1] + ysize;
  }
  return (m);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<avl_rotate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <avl_get_child@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x34(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <avl_get_child@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <avl_get_parent@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x34(%rbp),%eax
  neg    %eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <avl_set_child@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <avl_set_parent@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x34(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <avl_set_child@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <avl_set_parent@plt>
  cmpq   $0x0,-0x10(%rbp)
  je     L4
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <avl_set_parent@plt>
L4:
  mov    -0x18(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x8(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <avl_replace_child@plt>
  nop
  leave
  ret
```

assistant: ```c
__attribute__((always_)) void avl_rotate(struct avl_tree_node **const root_ptr, struct avl_tree_node *const A, const int sign) {
  struct avl_tree_node *const B = avl_get_child(A, -sign);
  struct avl_tree_node *const E = avl_get_child(B, +sign);
  struct avl_tree_node *const P = avl_get_parent(A);
  avl_set_child(A, -sign, E);
  avl_set_parent(A, B);
  avl_set_child(B, +sign, A);
  avl_set_parent(B, P);
  if (E)
    avl_set_parent(E, A);
  avl_replace_child(root_ptr, P, A, B);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<avl_rotate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,%rbx
  mov    %edx,%r12d
  mov    %edx,%r15d
  neg    %r15d
  mov    %r15d,%esi
  mov    %rbx,%rdi
  call   <avl_get_child@plt>
  mov    %rax,%rbp
  mov    %r12d,%esi
  mov    %rax,%rdi
  call   <avl_get_child@plt>
  mov    %rax,%r13
  mov    %rbx,%rdi
  call   <avl_get_parent@plt>
  mov    %rax,%r14
  mov    %r13,%rdx
  mov    %r15d,%esi
  mov    %rbx,%rdi
  call   <avl_set_child@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <avl_set_parent@plt>
  mov    %rbx,%rdx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <avl_set_child@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <avl_set_parent@plt>
  test   %r13,%r13
  je     L8
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <avl_set_parent@plt>
L8:
  mov    %rbp,%rcx
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    0x8(%rsp),%rdi
  call   <avl_replace_child@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
__attribute__((always_)) void avl_rotate(struct avl_tree_node **const root_ptr, struct avl_tree_node *const A, const int sign) {
  struct avl_tree_node *const B = avl_get_child(A, -sign);
  struct avl_tree_node *const E = avl_get_child(B, +sign);
  struct avl_tree_node *const P = avl_get_parent(A);
  avl_set_child(A, -sign, E);
  avl_set_parent(A, B);
  avl_set_child(B, +sign, A);
  avl_set_parent(B, P);
  if (E)
    avl_set_parent(E, A);
  avl_replace_child(root_ptr, P, A, B);
}
```