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

```asm
<toml_rtoi>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L48
  lea    0x4(%rsp),%rax
  mov    %rsi,%rbx
  test   %rsi,%rsi
  cmove  %rax,%rbx
  movzbl (%rdi),%eax
  cmp    $0x2b,%al
  je     L16
  lea    0x10(%rsp),%rdx
  cmp    $0x2d,%al
  je     L152
L108:
  cmp    $0x30,%al
  je     L95
L22:
  lea    0x74(%rsp),%rcx
  test   %al,%al
  jne    L28
  jmp    L11
L138:
  cmp    %rcx,%rdx
  jae    L48
L28:
  add    $0x1,%rdi
  cmp    $0x5f,%al
  je     L97
  mov    %al,(%rdx)
  add    $0x1,%rdx
L97:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L138
  cmp    %rcx,%rdx
  je     L48
L11:
  movb   $0x0,(%rdx)
  mov    D1(%rip),%rbp
  xor    %edx,%edx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movq   $0x0,0x0(%rbp)
  call   <strtoll@plt>
  cmpq   $0x0,0x0(%rbp)
  mov    %eax,(%rbx)
  jne    L48
  mov    0x8(%rsp),%rax
  movzbl (%rax),%eax
  neg    %al
  sbb    %eax,%eax
L14:
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L152:
  movzbl 0x1(%rdi),%eax
  movb   $0x2d,0x10(%rsp)
  add    $0x1,%rdi
  lea    0x11(%rsp),%rdx
  cmp    $0x30,%al
  jne    L22
L95:
  cmpb   $0x0,0x1(%rdi)
  jne    L48
  movl   $0x0,(%rbx)
  xor    %eax,%eax
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L48:
  mov    $0xffffffff,%eax
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L16:
  movzbl 0x1(%rdi),%eax
  movb   $0x2b,0x10(%rsp)
  add    $0x1,%rdi
  lea    0x11(%rsp),%rdx
  jmp    L108
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toml_rtoi(const char *src, int64_t *ret_) {
  if (!src)
    return -1;
  char buf[100];
  char *p = buf;
  char *q = p + sizeof(buf);
  const char *s = src;
  int64_t dummy;
  int64_t *ret = ret_ ? ret_ : &dummy;
  if (*s == '+')
    *p++ = *s++;
  else if (*s == '-')
    *p++ = *s++;
  if ('0' == s[0]) {
    if (s[1])
      return -1;
    return *ret = 0, 0;
  }
  while (*s && p < q) {
    int ch = *s++;
    if (ch == '_')
      ;
    else
      *p++ = ch;
  }
  if (*s || p == q)
    return -1;
  *p = 0;
  char *endp;
  errno = 0;
  *ret = strtoll(buf, &endp, 0);
  return (errno || *endp) ? -1 : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subst_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x50,%rsp
  mov    %rdi,-0x2048(%rbp)
  mov    %rsi,-0x2050(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L34
L37:
  lea    -0x2020(%rbp),%rax
  mov    %rax,-0x2038(%rbp)
  lea    -0x1010(%rbp),%rax
  mov    %rax,-0x2030(%rbp)
  jmp    L28
L49:
  mov    -0x2038(%rbp),%rax
  mov    %rax,-0x2028(%rbp)
  mov    -0x2030(%rbp),%rax
  mov    %rax,-0x2038(%rbp)
  mov    -0x2028(%rbp),%rax
  mov    %rax,-0x2030(%rbp)
L28:
  mov    -0x2030(%rbp),%rdx
  mov    -0x2038(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <subst_line@plt>
  test   %rax,%rax
  jne    L49
  mov    -0x2038(%rbp),%rdx
  mov    -0x2050(%rbp),%rax
  lea    D4(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
L34:
  mov    -0x2048(%rbp),%rdx
  lea    -0x2020(%rbp),%rax
  mov    $0x1000,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L37
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D4
type: string

tool: D4: %s

assistant: ```c
int subst_file(FILE *in, FILE *out) {
  char linebuf1[4096 + 1];
  char linebuf2[4096 + 1];
  char *origline, *newline, *tmpptr;
  char *atptr, *nextptr;
  char *varname, *value;
  int dump;
  while (fgets(linebuf1, 4096, in)) {
    origline = linebuf1;
    newline = linebuf2;
    while (subst_line(origline, newline)) {
      tmpptr = origline;
      origline = newline;
      newline = tmpptr;
    }
    fprintf(out, "%s", origline);
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<subst_file>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x20,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x2018(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%r14
  jmp    L47
L17:
  mov    %rdx,%rbx
L4:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <subst_line@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rbp
  test   %rax,%rax
  jne    L17
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <fprintf@plt>
L47:
  mov    %rsp,%rdi
  mov    %r12,%rdx
  mov    $0x1000,%esi
  call   <fgets@plt>
  test   %rax,%rax
  je     L27
  lea    0x1010(%rsp),%rbp
  mov    %rsp,%rbx
  jmp    L4
L27:
  mov    0x2018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x1,%eax
  add    $0x2020,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_guessed: string

tool: D6: %s

assistant: ```c
int subst_file(FILE *in, FILE *out) {
  char linebuf1[4096 + 1];
  char linebuf2[4096 + 1];
  char *origline, *newline, *tmpptr;
  char *atptr, *nextptr;
  char *varname, *value;
  int dump;
  while (fgets(linebuf1, 4096, in)) {
    origline = linebuf1;
    newline = linebuf2;
    while (subst_line(origline, newline)) {
      tmpptr = origline;
      origline = newline;
      newline = tmpptr;
    }
    fprintf(out, "%s", origline);
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<subst_file>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  lea    D6(%rip),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x2018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %rsp,%rbp
  nopl   (%rax)
L0:
  mov    %r13,%rdx
  mov    $0x1000,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L44
  lea    0x1010(%rsp),%rbx
  mov    %rbp,%r12
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %rdx,%r12
L5:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <subst_line@plt>
  mov    %rbx,%rdx
  mov    %r12,%rbx
  test   %rax,%rax
  jne    L12
  mov    %r12,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <fprintf@plt>
  jmp    L0
  nopl   0x0(%rax)
L44:
  mov    0x2018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x2028,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: %s

assistant: ```c
int subst_file(FILE *in, FILE *out) {
  char linebuf1[4096 + 1];
  char linebuf2[4096 + 1];
  char *origline, *newline, *tmpptr;
  char *atptr, *nextptr;
  char *varname, *value;
  int dump;
  while (fgets(linebuf1, 4096, in)) {
    origline = linebuf1;
    newline = linebuf2;
    while (subst_line(origline, newline)) {
      tmpptr = origline;
      origline = newline;
      newline = tmpptr;
    }
    fprintf(out, "%s", origline);
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subst_file>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  lea    D5(%rip),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x2018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %rsp,%rbp
  nopl   (%rax)
L42:
  mov    %r13,%rdx
  mov    $0x1000,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L21
  lea    0x1010(%rsp),%rbx
  mov    %rbp,%r12
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L46:
  mov    %rdx,%r12
L19:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <subst_line@plt>
  mov    %rbx,%rdx
  mov    %r12,%rbx
  test   %rax,%rax
  jne    L46
  mov    %r12,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <fprintf@plt>
  jmp    L42
  nopl   0x0(%rax)
L21:
  mov    0x2018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x2028,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type_infered: string

tool: D5: %s

assistant: ```c
int subst_file(FILE *in, FILE *out) {
  char linebuf1[4096 + 1];
  char linebuf2[4096 + 1];
  char *origline, *newline, *tmpptr;
  char *atptr, *nextptr;
  char *varname, *value;
  int dump;
  while (fgets(linebuf1, 4096, in)) {
    origline = linebuf1;
    newline = linebuf2;
    while (subst_line(origline, newline)) {
      tmpptr = origline;
      origline = newline;
      newline = tmpptr;
    }
    fprintf(out, "%s", origline);
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<decode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x1,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,(%rax)
  movl   $0x0,-0x2c(%rbp)
  jmp    L38
L40:
  addl   $0x1,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L38
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x38(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x28(%rbp)
  addl   $0x1,-0x2c(%rbp)
L38:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
  mov    -0x40(%rbp),%rax
  mov    -0x28(%rbp),%edx
  mov    %edx,(%rax)
  movl   $0x0,-0x2c(%rbp)
  jmp    L46
L32:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <iohtoi@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x2c(%rbp)
L46:
  mov    -0x2c(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L32
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned char *decode(char *input, int *size) {
  int lsize = strlen(input);
  unsigned char *result = malloc(sizeof(unsigned char) * lsize);
  int i;
  char **mots = malloc(sizeof(char *) * lsize);
  int nb_mots = 1;
  mots[0] = input;
  i = 0;
  while (input[i] != 0) {
    i++;
    if (input[i] == ',') {
      mots[nb_mots] = input + i + 1;
      input[i] = 0;
      nb_mots++;
      i++;
    }
  }
  *size = nb_mots;
  for (i = 0; i < nb_mots; i++)
    result[i] = iohtoi(mots[i]);
  free(mots);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<decode>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r12,(%rax)
  mov    $0x1,%r8d
  mov    $0x0,%edx
L47:
  mov    %r8d,%r9d
  movslq %edx,%rdx
  lea    0x1(%r12,%rdx,1),%rax
L23:
  mov    %edx,%edi
  mov    %rax,%rcx
  cmpb   $0x0,-0x1(%rax)
  je     L24
  lea    0x1(%rdx),%rsi
  add    $0x1,%rax
  cmpb   $0x2c,(%rcx)
  je     L37
  mov    %rsi,%rdx
  jmp    L23
L37:
  lea    0x2(%r12,%rdx,1),%rax
  mov    %rax,(%rbx,%r8,8)
  movb   $0x0,(%rcx)
  lea    0x2(%rdi),%edx
  add    $0x1,%r8
  jmp    L47
L24:
  mov    %r9d,0x0(%r13)
  test   %r9d,%r9d
  jle    L12
  mov    %r8d,%r13d
  mov    $0x0,%r12d
L59:
  mov    (%rbx,%r12,8),%rdi
  call   <iohtoi@plt>
  mov    %al,0x0(%rbp,%r12,1)
  add    $0x1,%r12
  cmp    %r13,%r12
  jne    L59
L12:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned char *decode(char *input, int *size) {
  int lsize = strlen(input);
  unsigned char *result = malloc(sizeof(unsigned char) * lsize);
  int i;
  char **mots = malloc(sizeof(char *) * lsize);
  int nb_mots = 1;
  mots[0] = input;
  i = 0;
  while (input[i] != 0) {
    i++;
    if (input[i] == ',') {
      mots[nb_mots] = input + i + 1;
      input[i] = 0;
      nb_mots++;
      i++;
    }
  }
  *size = nb_mots;
  for (i = 0; i < nb_mots; i++)
    result[i] = iohtoi(mots[i]);
  free(mots);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<decode>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <malloc@plt>
  lea    0x0(,%rbx,8),%edi
  mov    $0x1,%ebx
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %rbp,(%rax)
  mov    %rax,%r13
  xor    %eax,%eax
L14:
  movslq %eax,%rdx
  add    $0x1,%eax
  mov    %ebx,%edi
  movzbl 0x0(%rbp,%rdx,1),%edx
  cltq
  jmp    L41
  nopl   (%rax)
L10:
  lea    0x0(%rbp,%rax,1),%rsi
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%edx
  cmp    $0x2c,%dl
  je     L40
L41:
  mov    %eax,%ecx
  test   %dl,%dl
  jne    L10
  mov    %edi,(%r14)
  xor    %ebp,%ebp
  xchg   %ax,%ax
L48:
  mov    0x0(%r13,%rbp,8),%rdi
  call   <iohtoi@plt>
  mov    %al,(%r12,%rbp,1)
  add    $0x1,%rbp
  cmp    %rbx,%rbp
  jne    L48
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L40:
  add    %rbp,%rax
  mov    %rax,0x0(%r13,%rbx,8)
  lea    0x1(%rcx),%eax
  add    $0x1,%rbx
  movb   $0x0,(%rsi)
  jmp    L14
```

assistant: ```c
unsigned char *decode(char *input, int *size) {
  int lsize = strlen(input);
  unsigned char *result = malloc(sizeof(unsigned char) * lsize);
  int i;
  char **mots = malloc(sizeof(char *) * lsize);
  int nb_mots = 1;
  mots[0] = input;
  i = 0;
  while (input[i] != 0) {
    i++;
    if (input[i] == ',') {
      mots[nb_mots] = input + i + 1;
      input[i] = 0;
      nb_mots++;
      i++;
    }
  }
  *size = nb_mots;
  for (i = 0; i < nb_mots; i++)
    result[i] = iohtoi(mots[i]);
  free(mots);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decode>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <malloc@plt>
  lea    0x0(,%rbp,8),%edi
  mov    $0x1,%ebp
  mov    %rax,%r12
  call   <malloc@plt>
  xor    %edx,%edx
  mov    %rbx,(%rax)
  mov    %rax,%r13
L23:
  movslq %edx,%rdx
  mov    %ebp,%r8d
  lea    0x1(%rbx,%rdx,1),%rax
  jmp    L45
  nopl   0x0(%rax)
L38:
  add    $0x1,%rax
  cmpb   $0x2c,(%rcx)
  lea    0x1(%rdx),%rsi
  je     L5
  mov    %rsi,%rdx
L45:
  cmpb   $0x0,-0x1(%rax)
  mov    %edx,%edi
  mov    %rax,%rcx
  jne    L38
  mov    %r8d,(%r14)
  xor    %ebx,%ebx
L27:
  mov    0x0(%r13,%rbx,8),%rdi
  call   <iohtoi@plt>
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L27
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  lea    0x2(%rbx,%rdx,1),%rax
  lea    0x2(%rdi),%edx
  mov    %rax,0x0(%r13,%rbp,8)
  add    $0x1,%rbp
  movb   $0x0,(%rcx)
  jmp    L23
```

assistant: ```c
unsigned char *decode(char *input, int *size) {
  int lsize = strlen(input);
  unsigned char *result = malloc(sizeof(unsigned char) * lsize);
  int i;
  char **mots = malloc(sizeof(char *) * lsize);
  int nb_mots = 1;
  mots[0] = input;
  i = 0;
  while (input[i] != 0) {
    i++;
    if (input[i] == ',') {
      mots[nb_mots] = input + i + 1;
      input[i] = 0;
      nb_mots++;
      i++;
    }
  }
  *size = nb_mots;
  for (i = 0; i < nb_mots; i++)
    result[i] = iohtoi(mots[i]);
  free(mots);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
L7:
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  jne    L7
  mov    %r14,%rdi
  call   <Mem_Free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r13,%rdi
  call   <Mem_Free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r12,%rdi
  call   <Mem_Free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <Mem_Free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <Mem_Free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x9c4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  void *ptr[7];
  ptr[0] = Mem_Alloc(500);
  assert(ptr[0] != NULL);
  ptr[1] = Mem_Alloc(500);
  assert(ptr[1] != NULL);
  ptr[2] = Mem_Alloc(500);
  assert(ptr[2] != NULL);
  ptr[3] = Mem_Alloc(500);
  assert(ptr[3] != NULL);
  ptr[4] = Mem_Alloc(500);
  assert(ptr[4] != NULL);
  ptr[5] = Mem_Alloc(500);
  assert(ptr[5] != NULL);
  ptr[6] = Mem_Alloc(500);
  assert(ptr[6] != NULL);
  while (Mem_Alloc(500) != NULL)
    ;
  assert(Mem_Free(ptr[1]) == 0);
  assert(Mem_Free(ptr[5]) == 0);
  assert(Mem_Free(ptr[2]) == 0);
  assert(Mem_Free(ptr[4]) == 0);
  assert(Mem_Free(ptr[3]) == 0);
  ptr[2] = Mem_Alloc(2500);
  assert(ptr[2] != NULL);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1000,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r15
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r13
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r14
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  nopl   0x0(%rax)
L1:
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  jne    L1
  mov    %r15,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r14,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r13,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x9c4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  xor    %edi,%edi
  call   <exit@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  void *ptr[7];
  ptr[0] = Mem_Alloc(500);
  assert(ptr[0] != NULL);
  ptr[1] = Mem_Alloc(500);
  assert(ptr[1] != NULL);
  ptr[2] = Mem_Alloc(500);
  assert(ptr[2] != NULL);
  ptr[3] = Mem_Alloc(500);
  assert(ptr[3] != NULL);
  ptr[4] = Mem_Alloc(500);
  assert(ptr[4] != NULL);
  ptr[5] = Mem_Alloc(500);
  assert(ptr[5] != NULL);
  ptr[6] = Mem_Alloc(500);
  assert(ptr[6] != NULL);
  while (Mem_Alloc(500) != NULL)
    ;
  assert(Mem_Free(ptr[1]) == 0);
  assert(Mem_Free(ptr[5]) == 0);
  assert(Mem_Free(ptr[2]) == 0);
  assert(Mem_Free(ptr[4]) == 0);
  assert(Mem_Free(ptr[3]) == 0);
  ptr[2] = Mem_Alloc(2500);
  assert(ptr[2] != NULL);
  exit(0);
}
```