ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string

tool: D9: Alonso Cerpa


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; ++i)
    printf("Alonso Cerpa\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<write_next_solution_with_diagnostics>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    %r8d,-0x4c(%rbp)
  mov    %r9d,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x40(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x50(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    0x10(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x48(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x9d,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x2c(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jne    L6
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
L6:
  mov    -0x2c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int write_next_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(157, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<write_next_solution_with_diagnostics>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %esi,0x8(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %r8d,0x10(%rsp)
  mov    %r9d,0x14(%rsp)
  mov    0x40(%rsp),%eax
  mov    %eax,0x18(%rsp)
  mov    0x48(%rsp),%eax
  mov    %eax,0x1c(%rsp)
  mov    (%rdi),%eax
  mov    %eax,0x20(%rsp)
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0x9d,%edi
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L1
  addl   $0x1,(%rbx)
L1:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x30,%rsp
  pop    %rbx
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(157, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<write_next_solution_with_diagnostics>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    0x40(%rsp),%eax
  mov    %esi,0x8(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %edx,0xc(%rsp)
  lea    0x8(%rsp),%rdx
  mov    %eax,0x18(%rsp)
  mov    0x48(%rsp),%eax
  mov    %r8d,0x10(%rsp)
  xor    %r8d,%r8d
  mov    %eax,0x1c(%rsp)
  mov    (%rdi),%eax
  mov    $0x9d,%edi
  mov    %r9d,0x14(%rsp)
  mov    %eax,0x20(%rsp)
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L3
  addl   $0x1,(%rbx)
L3:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x30,%rsp
  pop    %rbx
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(157, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<write_next_solution_with_diagnostics>:
  endbr64
  push   %rbx
  movd   %esi,%xmm0
  movd   %edx,%xmm2
  movd   %r9d,%xmm4
  punpckldq %xmm2,%xmm0
  mov    %rdi,%rbx
  sub    $0x30,%rsp
  movd   0x40(%rsp),%xmm1
  movd   0x48(%rsp),%xmm3
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   %xmm0,0x8(%rsp)
  movd   %r8d,%xmm0
  mov    (%rdi),%eax
  punpckldq %xmm3,%xmm1
  punpckldq %xmm4,%xmm0
  xor    %r8d,%r8d
  mov    $0x9d,%edi
  punpcklqdq %xmm1,%xmm0
  mov    %eax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L8
  addl   $0x1,(%rbx)
L8:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x30,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(157, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<prepend_srcdir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    D45(%rip),%rax
  test   %rax,%rax
  jne    L15
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <getenv@plt>
  mov    %rax,D70(%rip)
  mov    D70(%rip),%rax
  test   %rax,%rax
  jne    L15
  lea    D70(%rip),%rax
  mov    %rax,D25(%rip)
L15:
  mov    D25(%rip),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rbx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <xmalloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    D25(%rip),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%rax
  lea    D25(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D23
datatype: string
name: read_data
arguments: data_label: D70
datatype: string
name: read_data
arguments: data_label: D25
datatype: string

tool: D23: srcdir

tool: D70: .

tool: D25: /

assistant: ```c
char *prepend_srcdir(const char *fname) {
  static const char *srcdir;
  char *result;
  if (!srcdir)
    if (!(srcdir = getenv("srcdir")))
      srcdir = ".";
  result = xmalloc(strlen(srcdir) + 1 + strlen(fname) + 1);
  strcpy(result, srcdir);
  strcat(result, "/");
  strcat(result, fname);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sdsortrec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  cmpq   $0x1,-0x10(%rbp)
  jbe    L18
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L24
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  cmpq   $0x0,-0x18(%rbp)
  je     L24
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    $0x0,%ecx
  mov    %rax,%rdi
  call   <sdsortrec>
L24:
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  cmp    %rax,-0x10(%rbp)
  jbe    L55
  cmpq   $0x1,-0x20(%rbp)
  jne    L55
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    $0x1,%ecx
  mov    %rax,%rdi
  call   <sdsortrec>
  jmp    L55
L18:
  nop
L55:
  leave
  ret
```

assistant: ```c
void sdsortrec(int *a, size_t n, size_t i, size_t d) {
  if (n < 2)
    return;
  if (a[i + 1] < a[i]) {
    swap(&a[i + 1], &a[i]);
    if (i > 0)
      sdsortrec(a, n, i - 1, 0);
  }
  if (i + 2 < n && d == 1)
    sdsortrec(a, n, i + 1, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sdsortrec>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L68
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rcx,%r13
  lea    0x1(%rdx),%r14
  lea    0x0(,%r14,4),%rax
  lea    (%rdi,%rax,1),%rdi
  lea    -0x4(%r12,%rax,1),%rsi
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  jl     L6
L19:
  add    $0x2,%rbx
  cmp    %rbp,%rbx
  jae    L55
  cmp    $0x1,%r13
  je     L28
L55:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <swap@plt>
  test   %rbx,%rbx
  je     L19
  lea    -0x1(%rbx),%rdx
  mov    $0x0,%ecx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <sdsortrec>
  jmp    L19
L28:
  mov    $0x1,%ecx
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <sdsortrec>
  jmp    L55
L68:
  ret
```

assistant: ```c
void sdsortrec(int *a, size_t n, size_t i, size_t d) {
  if (n < 2)
    return;
  if (a[i + 1] < a[i]) {
    swap(&a[i + 1], &a[i]);
    if (i > 0)
      sdsortrec(a, n, i - 1, 0);
  }
  if (i + 2 < n && d == 1)
    sdsortrec(a, n, i + 1, 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sdsortrec>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    0x4(%rdi,%rdx,4),%rbx
L47:
  cmp    $0x1,%r12
  jbe    L42
  mov    -0x4(%rbx),%eax
  lea    -0x4(%rbx),%rsi
  cmp    %eax,(%rbx)
  jl     L17
L21:
  lea    0x2(%rbp),%rax
  cmp    %rax,%r12
  seta   %dl
  cmp    $0x1,%r13
  mov    $0x1,%r13d
  sete   %al
  add    $0x1,%rbp
  add    $0x4,%rbx
  test   %al,%dl
  jne    L47
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rbx,%rdi
  call   <swap@plt>
  test   %rbp,%rbp
  je     L21
  lea    -0x1(%rbp),%rdx
  xor    %ecx,%ecx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <sdsortrec>
  jmp    L21
```

assistant: ```c
void sdsortrec(int *a, size_t n, size_t i, size_t d) {
  if (n < 2)
    return;
  if (a[i + 1] < a[i]) {
    swap(&a[i + 1], &a[i]);
    if (i > 0)
      sdsortrec(a, n, i - 1, 0);
  }
  if (i + 2 < n && d == 1)
    sdsortrec(a, n, i + 1, 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sdsortrec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  lea    -0x8(%rdi,%rdx,4),%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
L2:
  cmp    $0x1,%r12
  jbe    L98
  mov    0x8(%rbp),%eax
  lea    0xc(%rbp),%rdi
  lea    0x8(%rbp),%r15
  lea    0x4(%rbp),%r14
  cmp    %eax,0xc(%rbp)
  jl     L53
L73:
  lea    0x2(%rbx),%rax
  mov    %r14,%rbp
  cmp    %rax,%r12
  seta   %dl
  cmp    $0x1,%r13
  mov    $0x1,%r13d
  sete   %al
  add    $0x1,%rbx
  test   %al,%dl
  jne    L2
L98:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  mov    %r15,%rsi
  lea    0x4(%rbp),%r14
  call   <swap@plt>
  test   %rbx,%rbx
  je     L73
  mov    0x4(%rbp),%eax
  cmp    %eax,0x8(%rbp)
  jge    L73
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <swap@plt>
  cmp    $0x1,%rbx
  je     L73
  mov    0x0(%rbp),%eax
  cmp    %eax,0x4(%rbp)
  jge    L73
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <swap@plt>
  cmp    $0x2,%rbx
  je     L73
  mov    -0x4(%rbp),%eax
  lea    -0x4(%rbp),%rsi
  cmp    %eax,0x0(%rbp)
  jge    L73
  mov    %rbp,%rdi
  call   <swap@plt>
  cmp    $0x3,%rbx
  je     L73
  mov    0x8(%rsp),%rdi
  lea    -0x4(%rbx),%rdx
  xor    %ecx,%ecx
  mov    %r12,%rsi
  call   <sdsortrec>
  jmp    L73
```

assistant: ```c
void sdsortrec(int *a, size_t n, size_t i, size_t d) {
  if (n < 2)
    return;
  if (a[i + 1] < a[i]) {
    swap(&a[i + 1], &a[i]);
    if (i > 0)
      sdsortrec(a, n, i - 1, 0);
  }
  if (i + 2 < n && d == 1)
    sdsortrec(a, n, i + 1, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L10
L4:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L10:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L4
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
  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
  jmp    L16
L14:
  call   <putchar@plt>
L16:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D1(%rip),%rax
  cmp    %edi,(%rax)
  jne    L14
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
  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   %rbx
  mov    D1(%rip),%rbx
  jmp    L6
  xchg   %ax,%ax
L16:
  call   <putchar@plt>
L6:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L16
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D2(%rip),%rbx
  jmp    L18
  xchg   %ax,%ax
L10:
  call   <putchar@plt>
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L10
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cvt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D35(%rip),%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x8(%rbp)
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  je     L18
  mov    D27(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jne    L27
L18:
  mov    D35(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L27
  movq   $0xffffffffffffffff,-0x8(%rbp)
L27:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
__attribute__((unused)) long cvt(char *str) {
  long ret;
  errno = 0;
  ret = strtol(str, NULL, 10);
  if ((ret == LONG_MAX || ret == LONG_MIN) && errno)
    ret = -1;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cvt>:
  endbr64
  sub    $0x8,%rsp
  mov    D16(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    D28(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L10
  mov    D32(%rip),%rdx
  cmp    %rax,(%rdx)
  jne    L5
L10:
  mov    D16(%rip),%rdx
  cmpq   $0x0,(%rdx)
  mov    $0xffffffffffffffff,%rdx
  cmovne %rdx,%rax
L5:
  add    $0x8,%rsp
  ret
```

assistant: ```c
__attribute__((unused)) long cvt(char *str) {
  long ret;
  errno = 0;
  ret = strtol(str, NULL, 10);
  if ((ret == LONG_MAX || ret == LONG_MIN) && errno)
    ret = -1;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cvt>:
  endbr64
  push   %rbx
  mov    D22(%rip),%rbx
  mov    $0xa,%edx
  xor    %esi,%esi
  movq   $0x0,(%rbx)
  call   <strtol@plt>
  mov    D10(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L10
  mov    D27(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L10
  pop    %rbx
  ret
  nopl   0x0(%rax)
L10:
  cmpq   $0x0,(%rbx)
  mov    $0xffffffffffffffff,%rdx
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
__attribute__((unused)) long cvt(char *str) {
  long ret;
  errno = 0;
  ret = strtol(str, NULL, 10);
  if ((ret == LONG_MAX || ret == LONG_MIN) && errno)
    ret = -1;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cvt>:
  endbr64
  push   %rbx
  mov    D20(%rip),%rbx
  mov    $0xa,%edx
  xor    %esi,%esi
  movq   $0x0,(%rbx)
  call   <strtol@plt>
  mov    D4(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L17
  mov    D11(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L17
  pop    %rbx
  ret
  nopl   0x0(%rax)
L17:
  cmpq   $0x0,(%rbx)
  mov    $0xffffffffffffffff,%rdx
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
__attribute__((unused)) long cvt(char *str) {
  long ret;
  errno = 0;
  ret = strtol(str, NULL, 10);
  if ((ret == LONG_MAX || ret == LONG_MIN) && errno)
    ret = -1;
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  movl   $0x8,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <acc_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <acc_map_data@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L6
L19:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x28(%rbp)
L6:
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L19
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <acc_deviceptr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <acc_unmap_data@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <acc_free@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int N = 8;
  float *a, *b, *c, *d;
  int i;
  a = (float *)malloc(N * sizeof(float));
  b = (float *)malloc(N * sizeof(float));
  c = (float *)malloc(N * sizeof(float));
  d = (float *)acc_malloc(N * sizeof(float));
  acc_map_data(c, d, N * sizeof(float));
#pragma acc data present(a [0:N]) present(c [0:N]) present(b [0:N])
  {
#pragma acc parallel
    {
      int ii;
      for (ii = 0; ii < N; ii++) {
        c[ii] = a[ii];
        b[ii] = c[ii];
      }
    }
  }
  d = (float *)acc_deviceptr(c);
  acc_unmap_data(c);
  acc_free(d);
  free(a);
  free(b);
  free(c);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x20,%edi
  call   <acc_malloc@plt>
  mov    %rax,%rsi
  mov    $0x20,%edx
  mov    %rbx,%rdi
  call   <acc_map_data@plt>
  lea    0x20(%rbp),%rsi
  mov    %rbp,%rax
  mov    %rbx,%rcx
  sub    %rbp,%rcx
  mov    %r12,%rdx
  sub    %rbp,%rdx
L2:
  movss  (%rax),%xmm0
  movss  %xmm0,(%rcx,%rax,1)
  movss  %xmm0,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L2
  mov    %rbx,%rdi
  call   <acc_deviceptr@plt>
  mov    %rax,%r13
  mov    %rbx,%rdi
  call   <acc_unmap_data@plt>
  mov    %r13,%rdi
  call   <acc_free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int N = 8;
  float *a, *b, *c, *d;
  int i;
  a = (float *)malloc(N * sizeof(float));
  b = (float *)malloc(N * sizeof(float));
  c = (float *)malloc(N * sizeof(float));
  d = (float *)acc_malloc(N * sizeof(float));
  acc_map_data(c, d, N * sizeof(float));
#pragma acc data present(a [0:N]) present(c [0:N]) present(b [0:N])
  {
#pragma acc parallel
    {
      int ii;
      for (ii = 0; ii < N; ii++) {
        c[ii] = a[ii];
        b[ii] = c[ii];
      }
    }
  }
  d = (float *)acc_deviceptr(c);
  acc_unmap_data(c);
  acc_free(d);
  free(a);
  free(b);
  free(c);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x20,%edi
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x20,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    $0x20,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x20,%edi
  mov    %rax,%r12
  call   <acc_malloc@plt>
  mov    $0x20,%edx
  mov    %r12,%rdi
  mov    %rax,%rsi
  call   <acc_map_data@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    0x20(%r13),%rsi
  mov    %r13,%rax
  sub    %r13,%rcx
  sub    %r13,%rdx
  nopl   0x0(%rax,%rax,1)
L3:
  movss  (%rax),%xmm0
  movss  %xmm0,(%rcx,%rax,1)
  movss  %xmm0,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L3
  mov    %r12,%rdi
  call   <acc_deviceptr@plt>
  mov    %r12,%rdi
  mov    %rax,%r14
  call   <acc_unmap_data@plt>
  mov    %r14,%rdi
  call   <acc_free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int N = 8;
  float *a, *b, *c, *d;
  int i;
  a = (float *)malloc(N * sizeof(float));
  b = (float *)malloc(N * sizeof(float));
  c = (float *)malloc(N * sizeof(float));
  d = (float *)acc_malloc(N * sizeof(float));
  acc_map_data(c, d, N * sizeof(float));
#pragma acc data present(a [0:N]) present(c [0:N]) present(b [0:N])
  {
#pragma acc parallel
    {
      int ii;
      for (ii = 0; ii < N; ii++) {
        c[ii] = a[ii];
        b[ii] = c[ii];
      }
    }
  }
  d = (float *)acc_deviceptr(c);
  acc_unmap_data(c);
  acc_free(d);
  free(a);
  free(b);
  free(c);
  return 0;
}
```