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
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L1
L5:
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long strlen(const char *s) {
  unsigned long count = 0;
  for (; *s != '\0'; ++count) {
    ;
  }
  return count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copiar_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int copiar_string(void *destino, const void *origen) {
  char **d = (char **)destino;
  char **o = (char **)origen;
  int len = strlen(*o);
  *d = malloc((len + 1) * sizeof(char));
  strcpy(*d, *o);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<copiar_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    (%rsi),%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(%rbp)
  mov    (%rbx),%rsi
  call   <strcpy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int copiar_string(void *destino, const void *origen) {
  char **d = (char **)destino;
  char **o = (char **)origen;
  int len = strlen(*o);
  *d = malloc((len + 1) * sizeof(char));
  strcpy(*d, *o);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copiar_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  mov    (%rbx),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int copiar_string(void *destino, const void *origen) {
  char **d = (char **)destino;
  char **o = (char **)origen;
  int len = strlen(*o);
  *d = malloc((len + 1) * sizeof(char));
  strcpy(*d, *o);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<child>:
  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    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sethostname@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <uname@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x0,%edx
  lea    D16(%rip),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <execlp@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D29
type_infered: string
name: disassemble_data
arguments: label: D16
type_infered: string

tool: D29: uts.nodename in child: %s


tool: D16: bash

assistant: ```c
int child(void *arg) {
  struct utsname uts;
  sethostname(arg, strlen(arg));
  uname(&uts);
  printf("uts.nodename in child: %s\n", uts.nodename);
  sleep(1);
  execlp("bash", "bash", (char *)NULL);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<child>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <sethostname@plt>
  mov    %rsp,%rdi
  call   <uname@plt>
  mov    (%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x0,%edx
  lea    D2(%rip),%rdi
  mov    %rdi,%rsi
  call   <execlp@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D7
data_type: string
name: parse_data
arguments: data_tag: D2
data_type: string

tool: D7: uts.nodename in child: %s


tool: D2: bash

assistant: ```c
int child(void *arg) {
  struct utsname uts;
  sethostname(arg, strlen(arg));
  uname(&uts);
  printf("uts.nodename in child: %s\n", uts.nodename);
  sleep(1);
  execlp("bash", "bash", (char *)NULL);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<child>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <sethostname@plt>
  mov    %rsp,%rdi
  call   <uname@plt>
  mov    (%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D10(%rip),%rdi
  xor    %edx,%edx
  mov    %rdi,%rsi
  call   <execlp@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D16
type_infered: string
name: read_data
arguments: data_tag: D10
type_infered: string

tool: D16: uts.nodename in child: %s


tool: D10: bash

assistant: ```c
int child(void *arg) {
  struct utsname uts;
  sethostname(arg, strlen(arg));
  uname(&uts);
  printf("uts.nodename in child: %s\n", uts.nodename);
  sleep(1);
  execlp("bash", "bash", (char *)NULL);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<child>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <sethostname@plt>
  mov    %rsp,%rdi
  call   <uname@plt>
  mov    (%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D9(%rip),%rdi
  xor    %edx,%edx
  mov    %rdi,%rsi
  call   <execlp@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
type: string
name: parse_data
arguments: data_tag: D9
type: string

tool: D3: uts.nodename in child: %s


tool: D9: bash

assistant: ```c
int child(void *arg) {
  struct utsname uts;
  sethostname(arg, strlen(arg));
  uname(&uts);
  printf("uts.nodename in child: %s\n", uts.nodename);
  sleep(1);
  execlp("bash", "bash", (char *)NULL);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D35(%rip),%rdi
  lea    D69(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D74(%rip),%rdi
  call   <puts@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L39
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  sub    $0x1,%eax
  je     L12
L39:
  lea    D82(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L30:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x40,%rsp
  pop    %rbp
  ret
L12:
  movsd  0x8(%rsp),%xmm0
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  movsd  D31(%rip),%xmm1
  lea    0x30(%rsp),%rsi
  lea    0x28(%rsp),%rdi
  mulsd  %xmm1,%xmm0
  mulsd  0x10(%rsp),%xmm1
  call   <exactJ@plt>
  lea    D49(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movsd  0x28(%rsp),%xmm0
  mov    $0x2,%eax
  movsd  D114(%rip),%xmm1
  lea    D111(%rip),%rdi
  mulsd  0x30(%rsp),%xmm1
  divsd  D119(%rip),%xmm1
  call   <printf@plt>
  movsd  D119(%rip),%xmm0
  movsd  D119(%rip),%xmm1
  mulsd  0x18(%rsp),%xmm0
  lea    D12(%rip),%rdi
  mov    $0x2,%eax
  mulsd  0x20(%rsp),%xmm1
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L30
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D35
type: string
name: retrieve_data
arguments: label: D69
type: string
name: retrieve_data
arguments: label: D74
type: string
name: retrieve_data
arguments: label: D82
type: string
name: retrieve_data
arguments: label: D31
type: f64
name: retrieve_data
arguments: label: D49
type: string
name: retrieve_data
arguments: label: D114
type: byte[8]
name: retrieve_data
arguments: label: D111
type: string
name: retrieve_data
arguments: label: D119
type: f64
name: retrieve_data
arguments: label: D12
type: string

tool: D35: Enter  x, y  separated by space (or newline);

tool: D69: %lf

tool: D74:     (units: km, km; e.g. 100 100):

tool: D82: ... input error; exiting

tool: D31: 1000

tool: D49: 
Results from Test J (returned %d):


tool: {"D114": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D111:   H   = %10.3f (m)     nu    = %10.3f (MPa a)


tool: D119: 31556926

tool: D12:   u   = %10.5f (m/a)   v     = %10.5f (m/a)


assistant: ```c
int main() {
  double x, y, u, v, H, nu;
  int scanret;
  const double secpera = 31556926.0;
  printf("Enter  x, y  separated by space (or newline);\n");
  printf("    (units: km, km; e.g. 100 100):\n");
  scanret = scanf("%lf", &x);
  if (scanret != 1) {
    printf("... input error; exiting\n");
    return 1;
  }
  scanret = scanf("%lf", &y);
  if (scanret != 1) {
    printf("... input error; exiting\n");
    return 1;
  }
  printf("\nResults from Test J (returned %d):\n", exactJ(x * 1000.0, y * 1000.0, &H, &nu, &u, &v));
  printf("  H   = %10.3f (m)     nu    = %10.3f (MPa a)\n", H, (nu * 1.0e-6) / secpera);
  printf("  u   = %10.5f (m/a)   v     = %10.5f (m/a)\n", u * secpera, v * secpera);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D28(%rip),%rdi
  lea    D117(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D103(%rip),%rdi
  call   <puts@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L36
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  sub    $0x1,%eax
  je     L13
L36:
  lea    D9(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L19:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  add    $0x40,%rsp
  pop    %rbp
  ret
L13:
  movsd  0x8(%rsp),%xmm0
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  movsd  D93(%rip),%xmm1
  lea    0x30(%rsp),%rsi
  lea    0x28(%rsp),%rdi
  mulsd  %xmm1,%xmm0
  mulsd  0x10(%rsp),%xmm1
  call   <exactJ@plt>
  lea    D40(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movsd  0x28(%rsp),%xmm0
  mov    $0x2,%eax
  movsd  D43(%rip),%xmm1
  lea    D27(%rip),%rdi
  mulsd  0x30(%rsp),%xmm1
  divsd  D12(%rip),%xmm1
  call   <printf@plt>
  movsd  D12(%rip),%xmm0
  movsd  D12(%rip),%xmm1
  mulsd  0x18(%rsp),%xmm0
  lea    D17(%rip),%rdi
  mov    $0x2,%eax
  mulsd  0x20(%rsp),%xmm1
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L19
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D28
type_infered: string
name: read_data
arguments: label: D117
type_infered: string
name: read_data
arguments: label: D103
type_infered: string
name: read_data
arguments: label: D9
type_infered: string
name: read_data
arguments: label: D93
type_infered: f64
name: read_data
arguments: label: D40
type_infered: string
name: read_data
arguments: label: D43
type_infered: byte[8]
name: read_data
arguments: label: D27
type_infered: string
name: read_data
arguments: label: D12
type_infered: f64
name: read_data
arguments: label: D17
type_infered: string

tool: D28: Enter  x, y  separated by space (or newline);

tool: D117: %lf

tool: D103:     (units: km, km; e.g. 100 100):

tool: D9: ... input error; exiting

tool: D93: 1000

tool: D40: 
Results from Test J (returned %d):


tool: {"D43": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D27:   H   = %10.3f (m)     nu    = %10.3f (MPa a)


tool: D12: 31556926

tool: D17:   u   = %10.5f (m/a)   v     = %10.5f (m/a)


assistant: ```c
int main() {
  double x, y, u, v, H, nu;
  int scanret;
  const double secpera = 31556926.0;
  printf("Enter  x, y  separated by space (or newline);\n");
  printf("    (units: km, km; e.g. 100 100):\n");
  scanret = scanf("%lf", &x);
  if (scanret != 1) {
    printf("... input error; exiting\n");
    return 1;
  }
  scanret = scanf("%lf", &y);
  if (scanret != 1) {
    printf("... input error; exiting\n");
    return 1;
  }
  printf("\nResults from Test J (returned %d):\n", exactJ(x * 1000.0, y * 1000.0, &H, &nu, &u, &v));
  printf("  H   = %10.3f (m)     nu    = %10.3f (MPa a)\n", H, (nu * 1.0e-6) / secpera);
  printf("  u   = %10.5f (m/a)   v     = %10.5f (m/a)\n", u * secpera, v * secpera);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rnd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L3
  movl   $0x1,-0x4(%rbp)
L3:
  mov    $0x0,%eax
  call   <ya_random@plt>
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  leave
  ret
```

assistant: ```c
int rnd(int q) {
  if (q < 1)
    q = 1;
  return ya_random() % q;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rnd>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <ya_random@plt>
  test   %ebx,%ebx
  mov    $0x1,%edx
  cmovle %edx,%ebx
  cltd
  idiv   %ebx
  mov    %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int rnd(int q) {
  if (q < 1)
    q = 1;
  return ya_random() % q;
}
```
user: What is the c source code of the assembly code below:

```asm
<rnd>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <ya_random@plt>
  test   %ebx,%ebx
  mov    $0x1,%edx
  cmovg  %ebx,%edx
  pop    %rbx
  mov    %edx,%edi
  cltd
  idiv   %edi
  mov    %edx,%eax
  ret
```

assistant: ```c
int rnd(int q) {
  if (q < 1)
    q = 1;
  return ya_random() % q;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rnd>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <ya_random@plt>
  test   %ebx,%ebx
  mov    $0x1,%edx
  cmovg  %ebx,%edx
  pop    %rbx
  mov    %edx,%edi
  cltd
  idiv   %edi
  mov    %edx,%eax
  ret
```

assistant: ```c
int rnd(int q) {
  if (q < 1)
    q = 1;
  return ya_random() % q;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sumVeryLongIntegers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L58
L43:
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L60
L56:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L18
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  imul   $0x198,%rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  add    %eax,-0xc(%rbp)
L18:
  addl   $0x1,-0x10(%rbp)
L60:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L56
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,(%rsi)
  addl   $0x1,-0x14(%rbp)
L58:
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jle    L43
  jmp    L7
L46:
  subl   $0x1,-0x14(%rbp)
L7:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L46
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sumVeryLongIntegers(int n, int veryLongIntegers[102][102], int integersLengths[102], int result[102], int maxLength) {
  int loop, inner_loop, partialSum, valueToSend, numberIndex;
  valueToSend = 0;
  loop = 0;
  for (loop = 0; loop < maxLength + 2; loop++) {
    partialSum = 0;
    for (inner_loop = 0; inner_loop < n; inner_loop++) {
      if (loop < integersLengths[inner_loop]) {
        numberIndex = integersLengths[inner_loop] - loop - 1;
        partialSum = partialSum + veryLongIntegers[inner_loop][numberIndex];
      }
    }
    partialSum = partialSum + valueToSend;
    valueToSend = partialSum / 10;
    result[loop] = partialSum % 10;
  }
  while (result[loop - 1] == 0)
    loop--;
  return loop;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sumVeryLongIntegers>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  cmp    $0xffffffff,%r8d
  jl     L41
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %rdx,%r13
  lea    0x1(%r8),%r15d
  lea    -0x1(%rdi),%eax
  lea    0x4(%rdx,%rax,4),%r10
  mov    $0x0,%edi
  mov    $0x0,%r9d
  mov    $0x0,%r14d
  jmp    L8
L72:
  add    $0x4,%rdx
  add    $0x198,%rcx
  cmp    %r10,%rdx
  je     L34
L1:
  mov    (%rdx),%eax
  cmp    %edi,%eax
  jle    L72
  sub    %r11d,%eax
  sub    $0x1,%eax
  cltq
  add    (%rcx,%rax,4),%esi
  jmp    L72
L34:
  add    %r9d,%esi
  movslq %esi,%r9
  imul   $0x66666667,%r9,%r9
  sar    $0x22,%r9
  mov    %esi,%eax
  sar    $0x1f,%eax
  sub    %eax,%r9d
  lea    (%r9,%r9,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %esi,(%rbx,%rdi,4)
  lea    0x1(%rdi),%rax
  cmp    %r15,%rdi
  je     L16
  mov    %rax,%rdi
L8:
  mov    %edi,%r11d
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %r14d,%esi
  test   %ebp,%ebp
  jg     L1
  jmp    L34
L16:
  add    $0x2,%r8d
L4:
  movslq %r8d,%rax
  cmpl   $0x0,-0x4(%rbx,%rax,4)
  jne    L48
  sub    $0x1,%r8d
  movslq %r8d,%rax
L10:
  mov    %eax,%r8d
  sub    $0x1,%rax
  cmpl   $0x0,(%rbx,%rax,4)
  je     L10
L48:
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L41:
  mov    $0x0,%r8d
  jmp    L4
```

assistant: ```c
int sumVeryLongIntegers(int n, int veryLongIntegers[102][102], int integersLengths[102], int result[102], int maxLength) {
  int loop, inner_loop, partialSum, valueToSend, numberIndex;
  valueToSend = 0;
  loop = 0;
  for (loop = 0; loop < maxLength + 2; loop++) {
    partialSum = 0;
    for (inner_loop = 0; inner_loop < n; inner_loop++) {
      if (loop < integersLengths[inner_loop]) {
        numberIndex = integersLengths[inner_loop] - loop - 1;
        partialSum = partialSum + veryLongIntegers[inner_loop][numberIndex];
      }
    }
    partialSum = partialSum + valueToSend;
    valueToSend = partialSum / 10;
    result[loop] = partialSum % 10;
  }
  while (result[loop - 1] == 0)
    loop--;
  return loop;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sumVeryLongIntegers>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  cmp    $0xffffffff,%r8d
  jl     L29
  mov    %rsi,%r12
  lea    -0x1(%rdi),%eax
  mov    %edi,%ebp
  mov    %rdx,%r13
  xor    %edi,%edi
  lea    0x4(%rdx,%rax,4),%r10
  lea    0x1(%r8),%r14d
  xor    %r9d,%r9d
  mov    %edi,%r11d
  mov    %r12,%rcx
  mov    %r13,%rdx
  xor    %esi,%esi
  test   %ebp,%ebp
  jle    L89
  nopl   0x0(%rax)
L95:
  mov    (%rdx),%eax
  cmp    %edi,%eax
  jle    L40
  sub    %r11d,%eax
  sub    $0x1,%eax
  cltq
  add    (%rcx,%rax,4),%esi
L40:
  add    $0x4,%rdx
  add    $0x198,%rcx
  cmp    %rdx,%r10
  jne    L95
  add    %r9d,%esi
L19:
  movslq %esi,%r9
  mov    %esi,%eax
  imul   $0x66666667,%r9,%r9
  sar    $0x1f,%eax
  sar    $0x22,%r9
  sub    %eax,%r9d
  lea    (%r9,%r9,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  lea    0x1(%rdi),%rax
  mov    %esi,(%rbx,%rdi,4)
  cmp    %rdi,%r14
  je     L66
  mov    %rax,%rdi
  mov    %r12,%rcx
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %edi,%r11d
  test   %ebp,%ebp
  jg     L95
L89:
  mov    %r9d,%esi
  jmp    L19
  nopl   (%rax)
L66:
  add    $0x2,%r8d
  movslq %r8d,%rax
  lea    -0x4(,%rax,4),%rax
L32:
  mov    (%rbx,%rax,1),%ecx
  test   %ecx,%ecx
  jne    L41
  sub    $0x1,%r8d
  movslq %r8d,%rax
  nopl   (%rax)
L37:
  mov    %eax,%r8d
  sub    $0x1,%rax
  mov    (%rbx,%rax,4),%edx
  test   %edx,%edx
  je     L37
L41:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L29:
  mov    $0xfffffffffffffffc,%rax
  xor    %r8d,%r8d
  jmp    L32
```

assistant: ```c
int sumVeryLongIntegers(int n, int veryLongIntegers[102][102], int integersLengths[102], int result[102], int maxLength) {
  int loop, inner_loop, partialSum, valueToSend, numberIndex;
  valueToSend = 0;
  loop = 0;
  for (loop = 0; loop < maxLength + 2; loop++) {
    partialSum = 0;
    for (inner_loop = 0; inner_loop < n; inner_loop++) {
      if (loop < integersLengths[inner_loop]) {
        numberIndex = integersLengths[inner_loop] - loop - 1;
        partialSum = partialSum + veryLongIntegers[inner_loop][numberIndex];
      }
    }
    partialSum = partialSum + valueToSend;
    valueToSend = partialSum / 10;
    result[loop] = partialSum % 10;
  }
  while (result[loop - 1] == 0)
    loop--;
  return loop;
}
```