ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  movl   $0x80000000,-0x28(%rsp)
  mov    -0x28(%rsp),%eax
  mov    -0x28(%rsp),%edx
  neg    %eax
  and    %edx,%eax
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x28(%rsp)
  movabs $0x8000000000000000,%rdx
  mov    %rdx,-0x20(%rsp)
  mov    -0x20(%rsp),%rax
  mov    -0x20(%rsp),%rcx
  neg    %rax
  and    %rcx,%rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x20(%rsp)
  mov    %rdx,-0x10(%rsp)
  mov    -0x10(%rsp),%rax
  mov    -0x10(%rsp),%rdx
  neg    %rax
  and    %rdx,%rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x10(%rsp)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int e = 1, f = -1;
  volatile int i = (-0x7fffffff - 1);
  volatile int i2 = i & (((((((-i) + 1) - 1) + 1) - 1) + 1) - 1);
  i2 = -(i + e + f);
  i = -i;
  volatile long int li = (-0x7fffffffffffffffL - 1L);
  volatile long int li2 = li & (((((((-li) + 1) - 1) + 1) - 1) + 1) - 1);
  li2 = -(li + e + f);
  li = -li;
  volatile long long int lli = (-0x7fffffffffffffffLL - 1LL);
  volatile long long int lli2 = lli & (((((((-lli) + 1) - 1) + 1) - 1) + 1) - 1);
  lli2 = -(lli + e + f);
  lli = -lli;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  movl   $0x80000000,-0x28(%rsp)
  mov    -0x28(%rsp),%eax
  mov    -0x28(%rsp),%edx
  neg    %eax
  and    %edx,%eax
  movabs $0x8000000000000000,%rdx
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x28(%rsp)
  mov    %rdx,-0x20(%rsp)
  mov    -0x20(%rsp),%rax
  mov    -0x20(%rsp),%rcx
  neg    %rax
  and    %rcx,%rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x20(%rsp)
  mov    %rdx,-0x10(%rsp)
  mov    -0x10(%rsp),%rax
  mov    -0x10(%rsp),%rdx
  neg    %rax
  and    %rdx,%rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x10(%rsp)
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int e = 1, f = -1;
  volatile int i = (-0x7fffffff - 1);
  volatile int i2 = i & (((((((-i) + 1) - 1) + 1) - 1) + 1) - 1);
  i2 = -(i + e + f);
  i = -i;
  volatile long int li = (-0x7fffffffffffffffL - 1L);
  volatile long int li2 = li & (((((((-li) + 1) - 1) + 1) - 1) + 1) - 1);
  li2 = -(li + e + f);
  li = -li;
  volatile long long int lli = (-0x7fffffffffffffffLL - 1LL);
  volatile long long int lli2 = lli & (((((((-lli) + 1) - 1) + 1) - 1) + 1) - 1);
  lli2 = -(lli + e + f);
  lli = -lli;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  movl   $0x80000000,-0x28(%rsp)
  mov    -0x28(%rsp),%eax
  mov    -0x28(%rsp),%edx
  neg    %eax
  and    %edx,%eax
  movabs $0x8000000000000000,%rdx
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x24(%rsp)
  mov    -0x28(%rsp),%eax
  neg    %eax
  mov    %eax,-0x28(%rsp)
  mov    %rdx,-0x20(%rsp)
  mov    -0x20(%rsp),%rax
  mov    -0x20(%rsp),%rcx
  neg    %rax
  and    %rcx,%rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x18(%rsp)
  mov    -0x20(%rsp),%rax
  neg    %rax
  mov    %rax,-0x20(%rsp)
  mov    %rdx,-0x10(%rsp)
  mov    -0x10(%rsp),%rax
  mov    -0x10(%rsp),%rdx
  neg    %rax
  and    %rdx,%rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x8(%rsp)
  mov    -0x10(%rsp),%rax
  neg    %rax
  mov    %rax,-0x10(%rsp)
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int e = 1, f = -1;
  volatile int i = (-0x7fffffff - 1);
  volatile int i2 = i & (((((((-i) + 1) - 1) + 1) - 1) + 1) - 1);
  i2 = -(i + e + f);
  i = -i;
  volatile long int li = (-0x7fffffffffffffffL - 1L);
  volatile long int li2 = li & (((((((-li) + 1) - 1) + 1) - 1) + 1) - 1);
  li2 = -(li + e + f);
  li = -li;
  volatile long long int lli = (-0x7fffffffffffffffLL - 1LL);
  volatile long long int lli2 = lli & (((((((-lli) + 1) - 1) + 1) - 1) + 1) - 1);
  lli2 = -(lli + e + f);
  lli = -lli;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<fn1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  shl    $0x2,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    0x14(%rcx),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  shl    $0x2,%eax
  sub    %eax,%edx
  mov    %edx,%ecx
  lea    0x1e(%rcx),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

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

tool: D4: fn1

assistant: ```c
char *fn1() {
  char *result;
  result = malloc(20 + rand() % 20);
  result = realloc(result, 30 + rand() % 20);
  strcpy((char *)result, "fn1");
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<fn1>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x14(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x1e(%rax),%esi
  mov    %rbx,%rdi
  call   <realloc@plt>
  mov    %rax,%rbx
  lea    D3(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D3: fn1

assistant: ```c
char *fn1() {
  char *result;
  result = malloc(20 + rand() % 20);
  result = realloc(result, 30 + rand() % 20);
  strcpy((char *)result, "fn1");
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<fn1>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x14(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    %r12,%rdi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x1e(%rax),%esi
  call   <realloc@plt>
  lea    D4(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
datatype: string

tool: D4: fn1

assistant: ```c
char *fn1() {
  char *result;
  result = malloc(20 + rand() % 20);
  result = realloc(result, 30 + rand() % 20);
  strcpy((char *)result, "fn1");
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fn1>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x14(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    %r12,%rdi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  lea    0x1e(%rax),%esi
  call   <realloc@plt>
  lea    D5(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

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

tool: D5: fn1

assistant: ```c
char *fn1() {
  char *result;
  result = malloc(20 + rand() % 20);
  result = realloc(result, 30 + rand() % 20);
  strcpy((char *)result, "fn1");
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L9
L24:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L15
L11:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D10(%rip),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L15:
  cmpl   $0x2,-0x14(%rbp)
  jle    L11
  addl   $0x1,-0x18(%rbp)
L9:
  cmpl   $0x6,-0x18(%rbp)
  jle    L24
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D10
type_guessed: string
name: read_data
arguments: data_tag: D4
type_guessed: string
name: read_data
arguments: data_tag: D20
type_guessed: string

tool: D10: smskcbrt

tool: D4: # c::%s:len=%d


tool: D20: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "smskcbrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D27(%rip),%r15
  jmp    L31
L7:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L0
L31:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L37:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L37
  jmp    L7
L0:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: label: D27
type_guessed: string
name: retrieve_data
arguments: label: D13
type_guessed: string
name: retrieve_data
arguments: label: D9
type_guessed: string

tool: D27: smskcbrt

tool: D13: # c::%s:len=%d


tool: D9: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "smskcbrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D24(%rip),%r15
  push   %r14
  lea    D17(%rip),%r14
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L9:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L16:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L16
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L9
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D24
data_type: string
name: parse_data
arguments: data_label: D17
data_type: string
name: parse_data
arguments: data_label: D15
data_type: string

tool: D24: smskcbrt

tool: D17: # c::%s:len=%d


tool: D15: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "smskcbrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D1(%rip),%r15
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  lea    D3(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L16:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L0:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L0
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L16
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string
name: read_data
arguments: data_label: D23
type_guessed: string
name: read_data
arguments: data_label: D3
type_guessed: string

tool: D1: smskcbrt

tool: D23: # c::%s:len=%d


tool: D3: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "smskcbrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<subtract_image_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L48:
  movl   $0x0,-0x4(%rbp)
  jmp    L49
L12:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rcx,%rax
  mov    %dx,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  test   %ax,%ax
  jns    L32
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movw   $0x0,(%rax)
L32:
  addl   $0x1,-0x4(%rbp)
L49:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L12
  addl   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L48
  nop
  pop    %rbp
  ret
```

assistant: ```c
int subtract_image_array(the_image, out_image, rows, cols)
int rows, cols;
short **the_image, **out_image;
{
  int i, j, length, width;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      out_image[i][j] = the_image[i][j] - out_image[i][j];
      if (out_image[i][j] < 0)
        out_image[i][j] = 0;
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<subtract_image_array>:
  endbr64
  mov    %ecx,%r9d
  test   %edx,%edx
  jle    L73
  mov    %rsi,%r8
  lea    -0x1(%rdx),%eax
  lea    0x8(%rsi,%rax,8),%r10
  lea    -0x1(%rcx),%eax
  lea    0x2(%rax,%rax,1),%rsi
  jmp    L74
L13:
  add    $0x2,%rax
  cmp    %rsi,%rax
  je     L22
L65:
  mov    %rax,%rcx
  add    (%r8),%rcx
  mov    (%rdi),%rdx
  movzwl (%rdx,%rax,1),%edx
  sub    (%rcx),%dx
  mov    %dx,(%rcx)
  mov    %rax,%rdx
  add    (%r8),%rdx
  cmpw   $0x0,(%rdx)
  jns    L13
  movw   $0x0,(%rdx)
  jmp    L13
L22:
  add    $0x8,%rdi
  add    $0x8,%r8
  cmp    %r10,%r8
  je     L73
L74:
  mov    $0x0,%eax
  test   %r9d,%r9d
  jg     L65
  jmp    L22
L73:
  ret
```

assistant: ```c
int subtract_image_array(the_image, out_image, rows, cols)
int rows, cols;
short **the_image, **out_image;
{
  int i, j, length, width;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      out_image[i][j] = the_image[i][j] - out_image[i][j];
      if (out_image[i][j] < 0)
        out_image[i][j] = 0;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<subtract_image_array>:
  endbr64
  mov    %ecx,%r11d
  test   %edx,%edx
  jle    L13
  lea    -0x1(%rdx),%eax
  mov    %rdi,%r8
  lea    0x8(%rdi,%rax,8),%r10
  lea    -0x1(%rcx),%eax
  lea    0x2(%rax,%rax,1),%r9
  xchg   %ax,%ax
L27:
  xor    %edx,%edx
  test   %r11d,%r11d
  jle    L0
  nopw   0x0(%rax,%rax,1)
L7:
  mov    (%rsi),%rcx
  mov    (%r8),%rax
  mov    $0x0,%edi
  add    %rdx,%rcx
  movzwl (%rax,%rdx,1),%eax
  sub    (%rcx),%ax
  cmovs  %edi,%eax
  add    $0x2,%rdx
  mov    %ax,(%rcx)
  cmp    %r9,%rdx
  jne    L7
L0:
  add    $0x8,%r8
  add    $0x8,%rsi
  cmp    %r8,%r10
  jne    L27
L13:
  ret
```

assistant: ```c
int subtract_image_array(the_image, out_image, rows, cols)
int rows, cols;
short **the_image, **out_image;
{
  int i, j, length, width;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      out_image[i][j] = the_image[i][j] - out_image[i][j];
      if (out_image[i][j] < 0)
        out_image[i][j] = 0;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<subtract_image_array>:
  endbr64
  test   %edx,%edx
  jle    L4
  test   %ecx,%ecx
  jle    L4
  lea    -0x1(%rdx),%eax
  movslq %ecx,%rcx
  mov    %rdi,%r8
  lea    0x8(%rdi,%rax,8),%r10
  lea    (%rcx,%rcx,1),%r9
  xchg   %ax,%ax
L7:
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L10:
  mov    (%rsi),%rcx
  mov    (%r8),%rax
  mov    $0x0,%edi
  add    %rdx,%rcx
  movzwl (%rax,%rdx,1),%eax
  sub    (%rcx),%ax
  cmovs  %edi,%eax
  add    $0x2,%rdx
  mov    %ax,(%rcx)
  cmp    %r9,%rdx
  jne    L10
  add    $0x8,%r8
  add    $0x8,%rsi
  cmp    %r8,%r10
  jne    L7
L4:
  ret
```

assistant: ```c
int subtract_image_array(the_image, out_image, rows, cols)
int rows, cols;
short **the_image, **out_image;
{
  int i, j, length, width;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      out_image[i][j] = the_image[i][j] - out_image[i][j];
      if (out_image[i][j] < 0)
        out_image[i][j] = 0;
    }
  }
}
```