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

```asm
<findKthLargest>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  sub    %edx,%r13d
  mov    %rdi,%r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L32:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <partition@plt>
  cmp    %r13d,%eax
  jge    L1
  lea    0x1(%rax),%ebx
  jmp    L32
  cs nopw 0x0(%rax,%rax,1)
L1:
  jle    L28
  lea    -0x1(%rax),%ebp
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L28:
  cltq
  mov    (%r12,%rax,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int findKthLargest(int *nums, int numsSize, int k) {
  int lo = 0, hi = numsSize - 1;
  for (;;) {
    int p = partition(nums, lo, hi);
    if (p < numsSize - k) {
      lo = p + 1;
    } else if (p > numsSize - k) {
      hi = p - 1;
    } else {
      lo = p;
      break;
    }
  }
  return nums[lo];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<findKthLargest>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  sub    %edx,%r13d
  mov    %rdi,%r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L4:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <partition@plt>
  cmp    %r13d,%eax
  jge    L1
  lea    0x1(%rax),%ebx
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L1:
  jle    L18
  lea    -0x1(%rax),%ebp
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L18:
  cltq
  mov    (%r12,%rax,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int findKthLargest(int *nums, int numsSize, int k) {
  int lo = 0, hi = numsSize - 1;
  for (;;) {
    int p = partition(nums, lo, hi);
    if (p < numsSize - k) {
      lo = p + 1;
    } else if (p > numsSize - k) {
      hi = p - 1;
    } else {
      lo = p;
      break;
    }
  }
  return nums[lo];
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  movl   $0xa00000,-0x18(%rbp)
  movl   $0x3938700,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L14
L17:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,-0x20(%rbp)
  addl   $0x10,-0x24(%rbp)
L14:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L17
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_guessed: string

tool: D2: Sum is %d


assistant: ```c
int main(int argc, char **argv) {
  int MEM_SIZE = 10 * 1024 * 1024;
  int DURATION = 60 * 1000 * 1000;
  int DELAY_OPS = 1;
  int count = 0;
  int *mem = (int *)malloc(sizeof(int) * MEM_SIZE);
  int elapsed = 0;
  int tmp = 0;
  while (elapsed < DURATION) {
    int read_addr = count % MEM_SIZE;
    tmp += mem[read_addr];
    elapsed += DELAY_OPS;
    count += 16;
  }
  printf("Sum is %d\n", tmp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2800000,%edi
  call   <malloc@plt>
  mov    $0x0,%edi
  mov    $0x0,%ecx
L7:
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x36,%rdx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  shl    $0x15,%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  movslq %edx,%rdx
  add    (%rax,%rdx,4),%edi
  add    $0x10,%ecx
  cmp    $0x39387000,%ecx
  jne    L7
  mov    %edi,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: string

tool: D2: Sum is %d


assistant: ```c
int main(int argc, char **argv) {
  int MEM_SIZE = 10 * 1024 * 1024;
  int DURATION = 60 * 1000 * 1000;
  int DELAY_OPS = 1;
  int count = 0;
  int *mem = (int *)malloc(sizeof(int) * MEM_SIZE);
  int elapsed = 0;
  int tmp = 0;
  while (elapsed < DURATION) {
    int read_addr = count % MEM_SIZE;
    tmp += mem[read_addr];
    elapsed += DELAY_OPS;
    count += 16;
  }
  printf("Sum is %d\n", tmp);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x2800000,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    $0xcccccccd,%edi
  nopl   0x0(%rax)
L4:
  mov    %ecx,%edx
  imul   %rdi,%rdx
  shr    $0x37,%rdx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    $0x10,%ecx
  shl    $0x15,%esi
  sub    %esi,%edx
  movslq %edx,%rdx
  add    (%rax,%rdx,4),%r8d
  cmp    $0x39387000,%ecx
  jne    L4
  mov    %r8d,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: string

tool: D8: Sum is %d


assistant: ```c
int main(int argc, char **argv) {
  int MEM_SIZE = 10 * 1024 * 1024;
  int DURATION = 60 * 1000 * 1000;
  int DELAY_OPS = 1;
  int count = 0;
  int *mem = (int *)malloc(sizeof(int) * MEM_SIZE);
  int elapsed = 0;
  int tmp = 0;
  while (elapsed < DURATION) {
    int read_addr = count % MEM_SIZE;
    tmp += mem[read_addr];
    elapsed += DELAY_OPS;
    count += 16;
  }
  printf("Sum is %d\n", tmp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  mov    $0x2800000,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    $0xcccccccd,%edi
  nopl   0x0(%rax)
L9:
  mov    %ecx,%edx
  imul   %rdi,%rdx
  shr    $0x37,%rdx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    $0x10,%ecx
  shl    $0x15,%esi
  sub    %esi,%edx
  movslq %edx,%rdx
  add    (%rax,%rdx,4),%r8d
  cmp    $0x39387000,%ecx
  jne    L9
  mov    %r8d,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
data_type: string

tool: D9: Sum is %d


assistant: ```c
int main(int argc, char **argv) {
  int MEM_SIZE = 10 * 1024 * 1024;
  int DURATION = 60 * 1000 * 1000;
  int DELAY_OPS = 1;
  int count = 0;
  int *mem = (int *)malloc(sizeof(int) * MEM_SIZE);
  int elapsed = 0;
  int tmp = 0;
  while (elapsed < DURATION) {
    int read_addr = count % MEM_SIZE;
    tmp += mem[read_addr];
    elapsed += DELAY_OPS;
    count += 16;
  }
  printf("Sum is %d\n", tmp);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L45:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L11
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L11
  cmpl   $0xa,-0x8(%rbp)
  jne    L45
L11:
  cmpl   $0xa,-0x8(%rbp)
  jne    L35
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getl(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getl>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L32
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L3:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D0(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L6
  cmp    $0xa,%eax
  je     L6
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L3
  mov    %r12d,%ebx
  jmp    L14
L32:
  mov    $0x0,%ebx
L6:
  cmp    $0xa,%eax
  je     L64
L14:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L64:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  add    $0x1,%ebx
  jmp    L14
```

assistant: ```c
int getl(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<getl>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L51
  mov    D9(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L16:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L9
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L14
  mov    %edx,%r14d
L32:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L16
  cmp    $0xa,%eax
  je     L54
L51:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L54:
  lea    0x1(%r14),%edx
L9:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L14:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getl(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getl>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L0
  mov    D5(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L2:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L35
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L6
  mov    %edx,%r14d
L8:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L2
  cmp    $0xa,%eax
  je     L38
L0:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L38:
  lea    0x1(%r14),%edx
L35:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L6:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getl(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<copia>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    D1(%rip),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    D24(%rip),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    D24(%rip),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  addl   $0x1,-0x4(%rbp)
  jmp    L8
L11:
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copia(void) {
  int i;
  extern char linea[], linea_max[];
  i = 0;
  while (1) {
    *(linea_max + i) = *(linea + i);
    if (*(linea_max + i) == '\0')
      break;
    i++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<copia>:
  endbr64
  mov    D2(%rip),%rax
  movzbl (%rax),%eax
  mov    D12(%rip),%rdx
  mov    %al,(%rdx)
  test   %al,%al
  je     L16
  mov    $0x1,%eax
  mov    D2(%rip),%rsi
  mov    %rdx,%rcx
L15:
  movzbl (%rax,%rsi,1),%edx
  mov    %dl,(%rax,%rcx,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L15
L16:
  ret
```

assistant: ```c
void copia(void) {
  int i;
  extern char linea[], linea_max[];
  i = 0;
  while (1) {
    *(linea_max + i) = *(linea + i);
    if (*(linea_max + i) == '\0')
      break;
    i++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<copia>:
  endbr64
  mov    D5(%rip),%rcx
  mov    D15(%rip),%rsi
  movzbl (%rcx),%eax
  mov    %al,(%rsi)
  test   %al,%al
  je     L18
  mov    $0x1,%eax
L8:
  movzbl (%rax,%rcx,1),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax,%rsi,1)
  test   %dl,%dl
  jne    L8
L18:
  ret
```

assistant: ```c
void copia(void) {
  int i;
  extern char linea[], linea_max[];
  i = 0;
  while (1) {
    *(linea_max + i) = *(linea + i);
    if (*(linea_max + i) == '\0')
      break;
    i++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<copia>:
  endbr64
  mov    D12(%rip),%rcx
  mov    D7(%rip),%rsi
  movzbl (%rcx),%eax
  mov    %al,(%rsi)
  test   %al,%al
  je     L1
  mov    $0x1,%eax
L11:
  movzbl (%rax,%rcx,1),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax,%rsi,1)
  test   %dl,%dl
  jne    L11
L1:
  ret
```

assistant: ```c
void copia(void) {
  int i;
  extern char linea[], linea_max[];
  i = 0;
  while (1) {
    *(linea_max + i) = *(linea + i);
    if (*(linea_max + i) == '\0')
      break;
    i++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <instance1@plt>
  mov    $0x0,%edi
  call   <instance2@plt>
  mov    $0x0,%edi
  call   <instance3@plt>
  mov    $0x0,%edi
  call   <instance4@plt>
  mov    $0x0,%edi
  call   <instance5@plt>
  mov    $0x0,%edi
  call   <instance6@plt>
  mov    $0x0,%edi
  call   <instance7@plt>
  mov    $0x0,%edi
  call   <instance8@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  instance1(0);
__CPROVER_ASYNC_1:
  instance2(0);
__CPROVER_ASYNC_2:
  instance3(0);
__CPROVER_ASYNC_3:
  instance4(0);
__CPROVER_ASYNC_4:
  instance5(0);
__CPROVER_ASYNC_5:
  instance6(0);
__CPROVER_ASYNC_6:
  instance7(0);
__CPROVER_ASYNC_7:
  instance8(0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <instance1@plt>
  mov    $0x0,%edi
  call   <instance2@plt>
  mov    $0x0,%edi
  call   <instance3@plt>
  mov    $0x0,%edi
  call   <instance4@plt>
  mov    $0x0,%edi
  call   <instance5@plt>
  mov    $0x0,%edi
  call   <instance6@plt>
  mov    $0x0,%edi
  call   <instance7@plt>
  mov    $0x0,%edi
  call   <instance8@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  instance1(0);
__CPROVER_ASYNC_1:
  instance2(0);
__CPROVER_ASYNC_2:
  instance3(0);
__CPROVER_ASYNC_3:
  instance4(0);
__CPROVER_ASYNC_4:
  instance5(0);
__CPROVER_ASYNC_5:
  instance6(0);
__CPROVER_ASYNC_6:
  instance7(0);
__CPROVER_ASYNC_7:
  instance8(0);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <instance1@plt>
  xor    %edi,%edi
  call   <instance2@plt>
  xor    %edi,%edi
  call   <instance3@plt>
  xor    %edi,%edi
  call   <instance4@plt>
  xor    %edi,%edi
  call   <instance5@plt>
  xor    %edi,%edi
  call   <instance6@plt>
  xor    %edi,%edi
  call   <instance7@plt>
  xor    %edi,%edi
  call   <instance8@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  instance1(0);
__CPROVER_ASYNC_1:
  instance2(0);
__CPROVER_ASYNC_2:
  instance3(0);
__CPROVER_ASYNC_3:
  instance4(0);
__CPROVER_ASYNC_4:
  instance5(0);
__CPROVER_ASYNC_5:
  instance6(0);
__CPROVER_ASYNC_6:
  instance7(0);
__CPROVER_ASYNC_7:
  instance8(0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <instance1@plt>
  xor    %edi,%edi
  call   <instance2@plt>
  xor    %edi,%edi
  call   <instance3@plt>
  xor    %edi,%edi
  call   <instance4@plt>
  xor    %edi,%edi
  call   <instance5@plt>
  xor    %edi,%edi
  call   <instance6@plt>
  xor    %edi,%edi
  call   <instance7@plt>
  xor    %edi,%edi
  call   <instance8@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  instance1(0);
__CPROVER_ASYNC_1:
  instance2(0);
__CPROVER_ASYNC_2:
  instance3(0);
__CPROVER_ASYNC_3:
  instance4(0);
__CPROVER_ASYNC_4:
  instance5(0);
__CPROVER_ASYNC_5:
  instance6(0);
__CPROVER_ASYNC_6:
  instance7(0);
__CPROVER_ASYNC_7:
  instance8(0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<multiply1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L12
L5:
  movl   $0x0,-0x3c(%rbp)
  jmp    L27
L1:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L27:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L1
  addl   $0x1,-0x40(%rbp)
L12:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L5
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply1(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<multiply1>:
  endbr64
  test   %edi,%edi
  jle    L24
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L4:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L10:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L10
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  ret
```

assistant: ```c
void multiply1(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<multiply1>:
  endbr64
  test   %edi,%edi
  jle    L3
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L24:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L24
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L1
  pop    %rbx
  ret
L3:
  ret
```

assistant: ```c
void multiply1(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<multiply1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L8
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L23:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L33
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L33
  nopl   0x0(%rax)
L3:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L3
  cmp    -0x24(%rsp),%r11d
  je     L94
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L94
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L94
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L94:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L23
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L33:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L56:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L56
  mov    -0x18(%rsp),%r8
  jmp    L94
```

assistant: ```c
void multiply1(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```