ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<pop_unsigned_int>:
  endbr64
  sub    $0x18,%rsp
  mov    (%rdi),%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  cmovns %esi,%eax
  mov    %r8,(%rsp)
  sar    $0x3,%eax
  cmp    $0x3f,%esi
  jg     L4
  mov    $0x7,%edx
  xor    %esi,%esi
  sub    %eax,%edx
  cltq
  add    $0x1,%rdx
  lea    (%rsp,%rax,1),%rdi
  call   <memset@plt>
  mov    (%rsp),%r8
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    %r8,%rax
  add    $0x18,%rsp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long pop_unsigned_int(const char *source, int num_bits) {
  int long_size = sizeof(unsigned long), i;
  unsigned long result = *(unsigned long *)(source);
  char *result_bytes = (char *)(&result);
  for (i = num_bits / 8; i < long_size; i++)
    result_bytes[i] = 0;
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<run_all_tests>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <unit_test_st_aligned_malloc@plt>
  test   %rax,%rax
  je     L29
  movl   $0xffffffff,-0x4(%rbp)
L29:
  mov    $0x0,%eax
  call   <unit_test_st_aligned_realloc@plt>
  test   %rax,%rax
  je     L19
  movl   $0xffffffff,-0x4(%rbp)
L19:
  mov    $0x0,%eax
  call   <unit_test_st_mem_usage@plt>
  test   %rax,%rax
  je     L24
  movl   $0xffffffff,-0x4(%rbp)
L24:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int run_all_tests() {
  int ret = 0;
  if (unit_test_st_aligned_malloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_aligned_realloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_mem_usage() != 0) {
    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
<run_all_tests>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <unit_test_st_aligned_malloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <unit_test_st_aligned_realloc@plt>
  mov    $0xffffffff,%ebx
  test   %rax,%rax
  jne    L0
  neg    %rbp
  sbb    %ebx,%ebx
L0:
  mov    $0x0,%eax
  call   <unit_test_st_mem_usage@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%ebx
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int run_all_tests() {
  int ret = 0;
  if (unit_test_st_aligned_malloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_aligned_realloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_mem_usage() != 0) {
    ret = -1;
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<run_all_tests>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    $0xffffffff,%r12d
  push   %rbx
  sub    $0x8,%rsp
  call   <unit_test_st_aligned_malloc@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <unit_test_st_aligned_realloc@plt>
  test   %rax,%rax
  jne    L5
  neg    %rbx
  sbb    %r12d,%r12d
L5:
  xor    %eax,%eax
  call   <unit_test_st_mem_usage@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int run_all_tests() {
  int ret = 0;
  if (unit_test_st_aligned_malloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_aligned_realloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_mem_usage() != 0) {
    ret = -1;
  }
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<run_all_tests>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    $0xffffffff,%r12d
  push   %rbx
  sub    $0x8,%rsp
  call   <unit_test_st_aligned_malloc@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <unit_test_st_aligned_realloc@plt>
  test   %rax,%rax
  jne    L0
  neg    %rbx
  sbb    %r12d,%r12d
L0:
  xor    %eax,%eax
  call   <unit_test_st_mem_usage@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int run_all_tests() {
  int ret = 0;
  if (unit_test_st_aligned_malloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_aligned_realloc() != 0) {
    ret = -1;
  }
  if (unit_test_st_mem_usage() != 0) {
    ret = -1;
  }
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fd_readline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %ecx,%rcx
  lea    -0x1(%rdx,%rcx,1),%r15
  mov    %rdx,%rbx
  lea    0x13(%rsp),%r14
L78:
  mov    $0x1,%edx
  mov    %r14,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L43
  movzbl 0x13(%rsp),%eax
  cmp    $0x8,%al
  je     L86
  cmp    $0xd,%al
  je     L45
  cmp    %r15,%rbx
  jae    L5
  lea    0x1(%rbx),%rcx
  mov    %rcx,0x8(%rsp)
  mov    %al,(%rbx)
  movzbl %al,%eax
  mov    %eax,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  call   <write@plt>
  mov    0x8(%rsp),%rbx
  jmp    L78
L43:
  sub    $0x1,%eax
  jmp    L52
L86:
  cmp    %rbp,%rbx
  jbe    L105
  sub    $0x1,%rbx
  movl   $0x8,0x14(%rsp)
  lea    0x14(%rsp),%rax
  mov    $0x1,%edx
  mov    %rax,0x8(%rsp)
  mov    %rax,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  movl   $0x20,0x14(%rsp)
  mov    $0x1,%edx
  mov    0x8(%rsp),%rsi
  mov    %r13d,%edi
  call   <write@plt>
  movzbl 0x13(%rsp),%eax
  mov    %eax,0x14(%rsp)
  mov    $0x1,%edx
  mov    0x8(%rsp),%rsi
  mov    %r13d,%edi
  call   <write@plt>
  jmp    L78
L105:
  movl   $0x7,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  call   <write@plt>
  jmp    L78
L45:
  movb   $0x0,(%rbx)
  mov    %ebx,%eax
  sub    %ebp,%eax
L52:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L106
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  movl   $0x7,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  call   <write@plt>
  jmp    L78
L106:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fd_readline(int fdi, int fdo, char *b, int bsz) {
  int r;
  unsigned char c;
  unsigned char *bp, *bpe;
  bp = b;
  bpe = b + bsz - 1;
  while (1) {
    r = read(fdi, &c, 1);
    if (r <= 0) {
      r--;
      goto out;
    }
    switch (c) {
    case '\b':
      if (bp > (unsigned char *)b) {
        bp--;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = ' ';
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    case '\r':
      *bp = '\0';
      r = bp - (unsigned char *)b;
      goto out;
    default:
      if (bp < bpe) {
        *bp++ = c;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    }
  }
out:
  return r;
}
```
user: What is the c source code of the assembly code below:

```asm
<fd_readline>:
  endbr64
  push   %r15
  movslq %ecx,%rcx
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    -0x1(%rdx,%rcx,1),%rax
  lea    0x13(%rsp),%r14
  mov    %rax,(%rsp)
  lea    0x14(%rsp),%r15
L28:
  mov    $0x1,%edx
  mov    %r14,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L55
  movzbl 0x13(%rsp),%eax
  cmp    $0x8,%al
  je     L35
  cmp    $0xd,%al
  je     L33
  cmp    (%rsp),%rbx
  jae    L85
  mov    %al,(%rbx)
  lea    0x1(%rbx),%rcx
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    %rcx,0x8(%rsp)
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  mov    0x8(%rsp),%rbx
  jmp    L28
  nopl   (%rax)
L35:
  cmp    %rbp,%rbx
  ja     L19
L85:
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %r13d,%edi
  movl   $0x7,0x14(%rsp)
  call   <write@plt>
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L33:
  movb   $0x0,(%rbx)
  mov    %ebx,%eax
  sub    %ebp,%eax
L80:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L50
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  sub    $0x1,%rbx
  movl   $0x8,0x14(%rsp)
  call   <write@plt>
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  movl   $0x20,0x14(%rsp)
  call   <write@plt>
  movzbl 0x13(%rsp),%eax
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L55:
  sub    $0x1,%eax
  jmp    L80
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fd_readline(int fdi, int fdo, char *b, int bsz) {
  int r;
  unsigned char c;
  unsigned char *bp, *bpe;
  bp = b;
  bpe = b + bsz - 1;
  while (1) {
    r = read(fdi, &c, 1);
    if (r <= 0) {
      r--;
      goto out;
    }
    switch (c) {
    case '\b':
      if (bp > (unsigned char *)b) {
        bp--;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = ' ';
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    case '\r':
      *bp = '\0';
      r = bp - (unsigned char *)b;
      goto out;
    default:
      if (bp < bpe) {
        *bp++ = c;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    }
  }
out:
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fd_readline>:
  endbr64
  push   %r15
  movslq %ecx,%rcx
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    -0x1(%rdx,%rcx,1),%rax
  lea    0x13(%rsp),%r14
  mov    %rax,(%rsp)
  lea    0x14(%rsp),%r15
L66:
  mov    $0x1,%edx
  mov    %r14,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L45
  movzbl 0x13(%rsp),%eax
  cmp    $0x8,%al
  je     L46
  cmp    $0xd,%al
  je     L26
  cmp    (%rsp),%rbx
  jae    L24
  mov    %al,(%rbx)
  lea    0x1(%rbx),%rcx
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    %rcx,0x8(%rsp)
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  mov    0x8(%rsp),%rbx
  jmp    L66
  nopl   (%rax)
L46:
  cmp    %rbp,%rbx
  ja     L57
L24:
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %r13d,%edi
  movl   $0x7,0x14(%rsp)
  call   <write@plt>
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L26:
  movb   $0x0,(%rbx)
  mov    %ebx,%eax
  sub    %ebp,%eax
L34:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L90
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L57:
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  sub    $0x1,%rbx
  movl   $0x8,0x14(%rsp)
  call   <write@plt>
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  movl   $0x20,0x14(%rsp)
  call   <write@plt>
  movzbl 0x13(%rsp),%eax
  mov    %r15,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L45:
  sub    $0x1,%eax
  jmp    L34
L90:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fd_readline(int fdi, int fdo, char *b, int bsz) {
  int r;
  unsigned char c;
  unsigned char *bp, *bpe;
  bp = b;
  bpe = b + bsz - 1;
  while (1) {
    r = read(fdi, &c, 1);
    if (r <= 0) {
      r--;
      goto out;
    }
    switch (c) {
    case '\b':
      if (bp > (unsigned char *)b) {
        bp--;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = ' ';
          write((fdo), &(cl), 1);
        } while (0);
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    case '\r':
      *bp = '\0';
      r = bp - (unsigned char *)b;
      goto out;
    default:
      if (bp < bpe) {
        *bp++ = c;
        do {
          int cl = c;
          write((fdo), &(cl), 1);
        } while (0);
      } else {
        do {
          int cl = '\x07';
          write((fdo), &(cl), 1);
        } while (0);
      }
      break;
    }
  }
out:
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0x3c(%rbp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L76:
  cmp    %rcx,%rsp
  je     L51
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L76
L51:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L39
  orq    $0x0,-0x8(%rsp,%rax,1)
L39:
  lea    0x3(%rsp),%r14
  mov    %r14,%r15
  shr    $0x2,%r15
  and    $0xfffffffffffffffc,%r14
  test   %edx,%edx
  jle    L10
  mov    %r14,%r12
  mov    $0x0,%ebx
  lea    D21(%rip),%r13
L55:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    %ebx,-0x3c(%rbp)
  jg     L55
L10:
  mov    0x0(,%r15,4),%r12d
  cmpl   $0x1,-0x3c(%rbp)
  jle    L69
  mov    %r12d,%ebx
  mov    $0x1,%r13d
L54:
  mov    (%r14,%r13,4),%edi
  lea    (%rdi,%rbx,1),%esi
  call   <FindMax@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  mov    %r12d,%edi
  call   <FindMax@plt>
  mov    %eax,%r12d
  add    $0x1,%r13
  cmp    %r13d,-0x3c(%rbp)
  jg     L54
L69:
  mov    %r12d,%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D21
type: string

tool: D21: %d

assistant: ```c
int main() {
  int n, i, MAX = 0, maxVal = 0;
  scanf("%d", &n);
  int arr[n];
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  maxVal = MAX = arr[0];
  for (i = 1; i < n; i++) {
    maxVal = FindMax(arr[i], maxVal + arr[i]);
    MAX = FindMax(MAX, maxVal);
  }
  printf("%d", MAX);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L99
L29:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L29
L99:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L23
L25:
  lea    0x3(%rsp),%r13
  mov    %r13,%r15
  and    $0xfffffffffffffffc,%r13
  shr    $0x2,%r15
  test   %ecx,%ecx
  jle    L19
  mov    %r13,%r14
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L84:
  mov    %r14,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r14
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L84
  mov    0x0(,%r15,4),%ebx
  mov    %ebx,%r14d
  cmp    $0x1,%eax
  jle    L72
  mov    $0x1,%r15d
  nopw   0x0(%rax,%rax,1)
L36:
  mov    0x0(%r13,%r15,4),%edi
  add    $0x1,%r15
  lea    (%rdi,%rbx,1),%esi
  call   <FindMax@plt>
  mov    %r14d,%edi
  mov    %eax,%esi
  mov    %eax,%ebx
  call   <FindMax@plt>
  mov    %eax,%r14d
  cmp    %r15d,-0x3c(%rbp)
  jg     L36
L72:
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L23:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L25
L19:
  mov    0x0(,%r15,4),%r14d
  jmp    L72
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
datatype: string

tool: D7: %d

assistant: ```c
int main() {
  int n, i, MAX = 0, maxVal = 0;
  scanf("%d", &n);
  int arr[n];
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  maxVal = MAX = arr[0];
  for (i = 1; i < n; i++) {
    maxVal = FindMax(arr[i], maxVal + arr[i]);
    MAX = FindMax(MAX, maxVal);
  }
  printf("%d", MAX);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L74
L20:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L20
L74:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L12
L37:
  lea    0x3(%rsp),%r13
  mov    %r13,%r15
  and    $0xfffffffffffffffc,%r13
  shr    $0x2,%r15
  test   %ecx,%ecx
  jle    L77
  mov    %r13,%r14
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r14,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r14
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L0
  mov    0x0(,%r15,4),%ebx
  mov    %ebx,%r14d
  cmp    $0x1,%eax
  jle    L86
  mov    $0x1,%r15d
  nopw   0x0(%rax,%rax,1)
L31:
  mov    0x0(%r13,%r15,4),%edi
  add    $0x1,%r15
  lea    (%rdi,%rbx,1),%esi
  call   <FindMax@plt>
  mov    %r14d,%edi
  mov    %eax,%esi
  mov    %eax,%ebx
  call   <FindMax@plt>
  mov    %eax,%r14d
  cmp    %r15d,-0x3c(%rbp)
  jg     L31
L86:
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L12:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L37
L77:
  mov    0x0(,%r15,4),%r14d
  jmp    L86
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string

tool: D9: %d

assistant: ```c
int main() {
  int n, i, MAX = 0, maxVal = 0;
  scanf("%d", &n);
  int arr[n];
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  maxVal = MAX = arr[0];
  for (i = 1; i < n; i++) {
    maxVal = FindMax(arr[i], maxVal + arr[i]);
    MAX = FindMax(MAX, maxVal);
  }
  printf("%d", MAX);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<closeWithMsg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <sendMsgToClient@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <shutdown@plt>
  nop
  leave
  ret
```

assistant: ```c
void closeWithMsg(char *str, int clientFD) {
  sendMsgToClient(clientFD, str);
  shutdown(clientFD, SHUT_RDWR);
}
```
user: What is the c source code of the assembly code below?

```asm
<closeWithMsg>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  mov    %rdi,%rsi
  mov    %ebx,%edi
  call   <sendMsgToClient@plt>
  mov    D6(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <shutdown@plt>
  pop    %rbx
  ret
```

assistant: ```c
void closeWithMsg(char *str, int clientFD) {
  sendMsgToClient(clientFD, str);
  shutdown(clientFD, SHUT_RDWR);
}
```
user: What is the c source code of the assembly code below:

```asm
<closeWithMsg>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %rdi,%rsi
  mov    %ebp,%edi
  call   <sendMsgToClient@plt>
  mov    D5(%rip),%rax
  mov    %ebp,%edi
  pop    %rbp
  mov    (%rax),%esi
  jmp    1060 <shutdown@plt>
```

assistant: ```c
void closeWithMsg(char *str, int clientFD) {
  sendMsgToClient(clientFD, str);
  shutdown(clientFD, SHUT_RDWR);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<closeWithMsg>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %rdi,%rsi
  mov    %ebp,%edi
  call   <sendMsgToClient@plt>
  mov    D5(%rip),%rax
  mov    %ebp,%edi
  pop    %rbp
  mov    (%rax),%esi
  jmp    1060 <shutdown@plt>
```

assistant: ```c
void closeWithMsg(char *str, int clientFD) {
  sendMsgToClient(clientFD, str);
  shutdown(clientFD, SHUT_RDWR);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<downlo>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%r9
  mov    %rdx,%rdi
  mov    %ecx,%r11d
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  mov    %ecx,%edx
  shr    $0x1f,%edx
  add    %ecx,%edx
  sar    %edx
  mov    %ebx,%eax
  sub    %edx,%eax
  lea    -0x1(%rcx,%rax,2),%edx
  cmp    %esi,%edx
  setge  %dl
  movzbl %dl,%edx
  sub    %edx,%eax
  mov    $0xffffffff,%r10d
  cmovns %eax,%r10d
  mov    $0x0,%ebp
  mov    %ecx,%ecx
  jns    L18
L6:
  add    $0x1,%r10d
  cmp    %r10d,%ebx
  jle    L37
  movslq %r10d,%rbp
  lea    -0x1(%r11),%eax
  lea    0x8(%rdi,%rax,8),%r10
L53:
  lea    0x0(%rbp,%rbp,1),%eax
  test   %r11d,%r11d
  jle    L55
  mov    %rdi,%rdx
  pxor   %xmm1,%xmm1
L73:
  mov    %eax,%ecx
  sub    %esi,%ecx
  cmp    %eax,%esi
  cmovle %ecx,%eax
  movslq %eax,%rcx
  movsd  (%r9,%rcx,8),%xmm0
  mulsd  (%rdx),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x1,%eax
  add    $0x8,%rdx
  cmp    %r10,%rdx
  jne    L73
L36:
  movsd  %xmm1,(%r8,%rbp,8)
  add    $0x1,%rbp
  cmp    %ebp,%ebx
  jg     L53
L37:
  pop    %rbx
  pop    %rbp
  ret
L97:
  lea    0x0(%rbp,%rbp,1),%eax
  cltq
  lea    (%r9,%rax,8),%rdx
  mov    $0x0,%eax
  pxor   %xmm1,%xmm1
L39:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rdx,%rax,8),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L39
L16:
  movsd  %xmm1,(%r8,%rbp,8)
  add    $0x1,%rbp
  cmp    %ebp,%r10d
  jl     L6
L18:
  pxor   %xmm1,%xmm1
  test   %r11d,%r11d
  jg     L97
  jmp    L16
L55:
  pxor   %xmm1,%xmm1
  jmp    L36
```

assistant: ```c
void downlo(x, n, lpf, m, y) double x[], lpf[], y[];
int n, m;
{
  int n2, mlo, mhi, i, h, j;
  double s;
  n2 = n / 2;
  mlo = m / 2;
  mhi = n2 - mlo;
  if (2 * mhi + (m - 1) >= n)
    --mhi;
  if (mhi < 0)
    mhi = -1;
  for (i = 0; i <= mhi; i++) {
    s = 0.;
    for (h = 0; h < m; h++)
      s += lpf[h] * x[2 * i + h];
    y[i] = s;
  }
  for (i = mhi + 1; i < n2; i++) {
    s = 0.;
    j = 2 * i;
    for (h = 0; h < m; h++) {
      if (j >= n)
        j -= n;
      s += lpf[h] * x[j];
      j++;
    }
    y[i] = s;
  }
}
```