ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getTimeStamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    $0x1e,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  lea    D1(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
datatype: string

tool: D1: %ld.%ld

assistant: ```c
int getTimeStamp(char *ts) {
  memset(ts, 0x00, 30);
  struct timeval secs;
  gettimeofday(&secs, NULL);
  sprintf(ts, "%ld.%ld", secs.tv_sec, secs.tv_usec);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getTimeStamp>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x1e,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%edx
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
data_type: string

tool: D1: %ld.%ld

assistant: ```c
int getTimeStamp(char *ts) {
  memset(ts, 0x00, 30);
  struct timeval secs;
  gettimeofday(&secs, NULL);
  sprintf(ts, "%ld.%ld", secs.tv_sec, secs.tv_usec);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getTimeStamp>:
  endbr64
  push   %rbp
  mov    $0x1e,%edx
  xor    %esi,%esi
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%edx
  mov    %rbp,%rdi
  lea    D5(%rip),%rsi
  call   <sprintf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type: string

tool: D5: %ld.%ld

assistant: ```c
int getTimeStamp(char *ts) {
  memset(ts, 0x00, 30);
  struct timeval secs;
  gettimeofday(&secs, NULL);
  sprintf(ts, "%ld.%ld", secs.tv_sec, secs.tv_usec);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getTimeStamp>:
  endbr64
  push   %rbp
  mov    $0x1e,%edx
  xor    %esi,%esi
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%edx
  mov    %rbp,%rdi
  lea    D1(%rip),%rsi
  call   <sprintf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D1
type_guessed: string

tool: D1: %ld.%ld

assistant: ```c
int getTimeStamp(char *ts) {
  memset(ts, 0x00, 30);
  struct timeval secs;
  gettimeofday(&secs, NULL);
  sprintf(ts, "%ld.%ld", secs.tv_sec, secs.tv_usec);
  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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x60(%rbp),%rcx
  lea    -0xc0(%rbp),%rdx
  lea    -0x120(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x60(%rbp),%rcx
  lea    -0xc0(%rbp),%rdx
  lea    -0x120(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D13
datatype: string
name: unpack_data
arguments: tag: D17
datatype: string
name: unpack_data
arguments: tag: D6
datatype: string

tool: D13: Enter text:

tool: D17: %s%s%s

tool: D6: 
s1 = %s
s2 = %s
s3 = %s


assistant: ```c
int main(void) {
  char s1[81], s2[81], s3[81];
  printf("Enter text:\n");
  scanf("%s%s%s", s1, s2, s3);
  printf("\ns1 = %s\ns2 = %s\ns3 = %s\n", s1, s2, s3);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <puts@plt>
  lea    0xc0(%rsp),%r12
  lea    0x60(%rsp),%rbp
  mov    %rsp,%rbx
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x120,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D26
data_type: string
name: read_data
arguments: data_tag: D4
data_type: string
name: read_data
arguments: data_tag: D16
data_type: string

tool: D26: Enter text:

tool: D4: %s%s%s

tool: D16: 
s1 = %s
s2 = %s
s3 = %s


assistant: ```c
int main(void) {
  char s1[81], s2[81], s3[81];
  printf("Enter text:\n");
  scanf("%s%s%s", s1, s2, s3);
  printf("\ns1 = %s\ns2 = %s\ns3 = %s\n", s1, s2, s3);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D8(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x120,%rsp
  lea    0xc0(%rsp),%r13
  lea    0x60(%rsp),%r12
  mov    %rsp,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x120,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
datatype: string
name: extract_symbolic_data
arguments: label: D3
datatype: string
name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D8: Enter text:

tool: D3: %s%s%s

tool: D0: 
s1 = %s
s2 = %s
s3 = %s


assistant: ```c
int main(void) {
  char s1[81], s2[81], s3[81];
  printf("Enter text:\n");
  scanf("%s%s%s", s1, s2, s3);
  printf("\ns1 = %s\ns2 = %s\ns3 = %s\n", s1, s2, s3);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x120,%rsp
  lea    0xc0(%rsp),%r13
  lea    0x60(%rsp),%r12
  mov    %rsp,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x120,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D28
type_infered: string

tool: D16: Enter text:

tool: D1: %s%s%s

tool: D28: 
s1 = %s
s2 = %s
s3 = %s


assistant: ```c
int main(void) {
  char s1[81], s2[81], s3[81];
  printf("Enter text:\n");
  scanf("%s%s%s", s1, s2, s3);
  printf("\ns1 = %s\ns2 = %s\ns3 = %s\n", s1, s2, s3);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<backSubstitution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <zeros@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  imul   -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  add    %rdx,%rax
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L26
L13:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L2
L12:
  mov    -0x24(%rbp),%eax
  imul   -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L2:
  mov    -0x14(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L12
  mov    -0x24(%rbp),%eax
  imul   -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x10(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  subl   $0x1,-0x18(%rbp)
L26:
  cmpl   $0x0,-0x18(%rbp)
  jns    L13
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double *backSubstitution(int n, double *matrix, double b[]) {
  int i, j;
  double temp;
  double *x = zeros(n);
  x[n - 1] = b[n - 1] / *(matrix + n * (n - 1) + n - 1);
  for (i = (n - 2); i >= 0; i--) {
    temp = b[i];
    for (j = (i + 1); j < n; j++) {
      temp -= (*(matrix + n * i + j) * x[j]);
    }
    x[i] = temp / *(matrix + n * i + i);
  }
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<backSubstitution>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <zeros@plt>
  mov    %rax,%rdx
  movslq %ebx,%r8
  lea    0x0(,%r8,8),%rdi
  lea    -0x1(%rbx),%esi
  mov    %esi,%eax
  imul   %ebx,%eax
  movslq %eax,%rcx
  add    %r8,%rcx
  movsd  -0x8(%rbp,%rdi,1),%xmm0
  divsd  -0x8(%r12,%rcx,8),%xmm0
  movsd  %xmm0,-0x8(%rdx,%rdi,1)
  mov    %ebx,%r8d
  sub    $0x2,%r8d
  js     L37
  mov    %ebx,%ecx
  neg    %ecx
  movslq %ecx,%rcx
  lea    -0x8(,%rcx,8),%r10
  movslq %r8d,%rdi
  sub    %ebx,%eax
  cltq
  add    %rdi,%rax
  lea    (%r12,%rax,8),%r9
  imul   %ebx,%r8d
  shl    $0x3,%rdi
L6:
  movsd  0x0(%rbp,%rdi,1),%xmm1
  cmp    %esi,%ebx
  jle    L39
  movslq %esi,%rax
  movslq %r8d,%rcx
  lea    (%r12,%rcx,8),%rcx
L29:
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  (%rdx,%rax,8),%xmm0
  subsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %eax,%ebx
  jg     L29
L39:
  divsd  (%r9),%xmm1
  movsd  %xmm1,(%rdx,%rdi,1)
  add    %r10,%r9
  sub    %ebx,%r8d
  sub    $0x8,%rdi
  sub    $0x1,%esi
  jne    L6
L37:
  mov    %rdx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double *backSubstitution(int n, double *matrix, double b[]) {
  int i, j;
  double temp;
  double *x = zeros(n);
  x[n - 1] = b[n - 1] / *(matrix + n * (n - 1) + n - 1);
  for (i = (n - 2); i >= 0; i--) {
    temp = b[i];
    for (j = (i + 1); j < n; j++) {
      temp -= (*(matrix + n * i + j) * x[j]);
    }
    x[i] = temp / *(matrix + n * i + i);
  }
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<backSubstitution>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %edi,%ebx
  call   <zeros@plt>
  lea    -0x1(%rbx),%ecx
  movslq %ebx,%rsi
  mov    %rax,%r8
  mov    %ecx,%eax
  lea    0x0(,%rsi,8),%r9
  imul   %ebx,%eax
  movsd  -0x8(%rbp,%r9,1),%xmm0
  movslq %eax,%rdx
  add    %rsi,%rdx
  mov    %ebx,%esi
  divsd  -0x8(%r12,%rdx,8),%xmm0
  movsd  %xmm0,-0x8(%r8,%r9,1)
  sub    $0x2,%esi
  js     L14
  mov    %ebx,%edx
  sub    %ebx,%eax
  movslq %esi,%rsi
  neg    %edx
  cltq
  movslq %edx,%rdx
  lea    -0x8(,%rdx,8),%r10
  lea    (%rax,%rsi,1),%rdx
  shl    $0x3,%rsi
  lea    (%r12,%rdx,8),%rdi
  lea    (%r12,%rax,8),%rdx
  nopl   0x0(%rax,%rax,1)
L12:
  movsd  0x0(%rbp,%rsi,1),%xmm1
  cmp    %ecx,%ebx
  jle    L23
  movslq %ecx,%rax
  nopl   (%rax)
L32:
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  (%r8,%rax,8),%xmm0
  add    $0x1,%rax
  subsd  %xmm0,%xmm1
  cmp    %eax,%ebx
  jg     L32
L23:
  divsd  (%rdi),%xmm1
  sub    %r9,%rdx
  movsd  %xmm1,(%r8,%rsi,1)
  add    %r10,%rdi
  sub    $0x8,%rsi
  sub    $0x1,%ecx
  jne    L12
L14:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double *backSubstitution(int n, double *matrix, double b[]) {
  int i, j;
  double temp;
  double *x = zeros(n);
  x[n - 1] = b[n - 1] / *(matrix + n * (n - 1) + n - 1);
  for (i = (n - 2); i >= 0; i--) {
    temp = b[i];
    for (j = (i + 1); j < n; j++) {
      temp -= (*(matrix + n * i + j) * x[j]);
    }
    x[i] = temp / *(matrix + n * i + i);
  }
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<backSubstitution>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <zeros@plt>
  lea    -0x1(%rbp),%edi
  movslq %ebp,%rsi
  mov    %edi,%edx
  mov    %rax,%r8
  lea    0x0(,%rsi,8),%rcx
  imul   %ebp,%edx
  movsd  -0x8(%r12,%rcx,1),%xmm0
  movslq %edx,%rax
  add    %rsi,%rax
  divsd  -0x8(%rbx,%rax,8),%xmm0
  mov    %ebp,%eax
  movsd  %xmm0,-0x8(%r8,%rcx,1)
  sub    $0x2,%eax
  js     L54
  mov    %ebp,%r11d
  sub    %ebp,%edx
  cltq
  mov    $0x1,%r9d
  neg    %r11d
  movslq %edx,%rdx
  lea    0x0(,%rax,8),%rcx
  movslq %r11d,%r11
  lea    0x1(%rax,%rdx,1),%rax
  lea    (%r12,%rcx,1),%r10
  lea    (%rbx,%rax,8),%rsi
  lea    -0x8(,%r11,8),%r12
  lea    0x8(%r8,%rcx,1),%rcx
  xchg   %ax,%ax
L32:
  movsd  (%r10),%xmm0
  mov    %edi,%eax
  cmp    %edi,%ebp
  jle    L56
  cmp    $0x1,%r9d
  je     L37
  mov    %r9d,%r13d
  xor    %eax,%eax
  shr    %r13d
  shl    $0x4,%r13
  nopl   (%rax)
L45:
  movupd (%rcx,%rax,1),%xmm1
  movupd (%rsi,%rax,1),%xmm3
  add    $0x10,%rax
  mulpd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  subsd  %xmm1,%xmm0
  cmp    %rax,%r13
  jne    L45
  mov    %r9d,%r13d
  and    $0xfffffffe,%r13d
  lea    (%rdi,%r13,1),%eax
  cmp    %r13d,%r9d
  je     L56
L37:
  cltq
  lea    (%rdx,%rax,1),%r13
  movsd  (%rbx,%r13,8),%xmm1
  mulsd  (%r8,%rax,8),%xmm1
  subsd  %xmm1,%xmm0
L56:
  divsd  -0x8(%rsi),%xmm0
  sub    $0x8,%r10
  movsd  %xmm0,-0x8(%rcx)
  add    %r11,%rdx
  add    $0x1,%r9d
  add    %r12,%rsi
  sub    $0x8,%rcx
  sub    $0x1,%edi
  jne    L32
L54:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double *backSubstitution(int n, double *matrix, double b[]) {
  int i, j;
  double temp;
  double *x = zeros(n);
  x[n - 1] = b[n - 1] / *(matrix + n * (n - 1) + n - 1);
  for (i = (n - 2); i >= 0; i--) {
    temp = b[i];
    for (j = (i + 1); j < n; j++) {
      temp -= (*(matrix + n * i + j) * x[j]);
    }
    x[i] = temp / *(matrix + n * i + i);
  }
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<iread>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
L10:
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L19
  mov    $0x0,%eax
  call   <wasintr@plt>
  test   %eax,%eax
  je     L19
  jmp    L10
L19:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long iread(int f, void *a, int n) {
  long m;
  for (;;) {
    m = read(f, a, n);
    if (m >= 0 || !wasintr())
      break;
  }
  return m;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<iread>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %edx,%ebp
L9:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jns    L6
  mov    $0x0,%eax
  call   <wasintr@plt>
  test   %eax,%eax
  jne    L9
L6:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long iread(int f, void *a, int n) {
  long m;
  for (;;) {
    m = read(f, a, n);
    if (m >= 0 || !wasintr())
      break;
  }
  return m;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<iread>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L29:
  xor    %eax,%eax
  call   <wasintr@plt>
  test   %eax,%eax
  je     L28
L18:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %rax,%r12
  test   %rax,%rax
  js     L29
L28:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long iread(int f, void *a, int n) {
  long m;
  for (;;) {
    m = read(f, a, n);
    if (m >= 0 || !wasintr())
      break;
  }
  return m;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<iread>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  call   <wasintr@plt>
  test   %eax,%eax
  je     L25
L14:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %rax,%r12
  test   %rax,%rax
  js     L8
L25:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long iread(int f, void *a, int n) {
  long m;
  for (;;) {
    m = read(f, a, n);
    if (m >= 0 || !wasintr())
      break;
  }
  return m;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xxstrdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L9
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L9:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *xxstrdup(const char *s) {
  size_t size = strlen(s) + 1;
  char *p = malloc(size);
  if (p) {
    memcpy(p, s, size);
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xxstrdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L8
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L8:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *xxstrdup(const char *s) {
  size_t size = strlen(s) + 1;
  char *p = malloc(size);
  if (p) {
    memcpy(p, s, size);
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xxstrdup>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L5
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L5:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *xxstrdup(const char *s) {
  size_t size = strlen(s) + 1;
  char *p = malloc(size);
  if (p) {
    memcpy(p, s, size);
  }
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xxstrdup>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L8:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *xxstrdup(const char *s) {
  size_t size = strlen(s) + 1;
  char *p = malloc(size);
  if (p) {
    memcpy(p, s, size);
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<match>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L24
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L24
  mov    $0x0,%eax
  jmp    L19
L24:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <amatch@plt>
L19:
  leave
  ret
```

assistant: ```c
match(s, p) char *s, *p;
{
  if (*s == '.' && *p != '.')
    return (0);
  return (amatch(s, p));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<match>:
  endbr64
  cmpb   $0x2e,(%rdi)
  jne    L0
  mov    $0x0,%eax
  cmpb   $0x2e,(%rsi)
  jne    L7
L0:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <amatch@plt>
  add    $0x8,%rsp
  ret
L7:
  ret
```

assistant: ```c
match(s, p) char *s, *p;
{
  if (*s == '.' && *p != '.')
    return (0);
  return (amatch(s, p));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<match>:
  endbr64
  cmpb   $0x2e,(%rdi)
  jne    L16
  cmpb   $0x2e,(%rsi)
  je     L16
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
  jmp    1050 <amatch@plt>
```

assistant: ```c
match(s, p) char *s, *p;
{
  if (*s == '.' && *p != '.')
    return (0);
  return (amatch(s, p));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<match>:
  endbr64
  cmpb   $0x2e,(%rdi)
  jne    L1
  cmpb   $0x2e,(%rsi)
  je     L1
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  jmp    1050 <amatch@plt>
```

assistant: ```c
match(s, p) char *s, *p;
{
  if (*s == '.' && *p != '.')
    return (0);
  return (amatch(s, p));
}
```