ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<idamax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  addq   $0x8,-0x10(%rbp)
  lea    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  lea    -0x4(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <BLAS_idamax@plt>
  leave
  ret
```

assistant: ```c
int idamax(int n, double *x, int is) {
  x++;
  return (BLAS_idamax(&n, x, &is));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<idamax>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  mov    %edx,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  add    $0x8,%rsi
  lea    0xc(%rsp),%rdi
  call   <BLAS_idamax@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int idamax(int n, double *x, int is) {
  x++;
  return (BLAS_idamax(&n, x, &is));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<idamax>:
  endbr64
  sub    $0x18,%rsp
  add    $0x8,%rsi
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rdi
  mov    %edx,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  call   <BLAS_idamax@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int idamax(int n, double *x, int is) {
  x++;
  return (BLAS_idamax(&n, x, &is));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<idamax>:
  endbr64
  sub    $0x18,%rsp
  add    $0x8,%rsi
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rdi
  mov    %edx,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  call   <BLAS_idamax@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int idamax(int n, double *x, int is) {
  x++;
  return (BLAS_idamax(&n, x, &is));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_transmit_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L9
L22:
  movl   $0x7,-0x4(%rbp)
  jmp    L15
L1:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  mov    -0x20(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_transmit_to_gpio_with_increased_bit@plt>
  subl   $0x1,-0x4(%rbp)
L15:
  cmpl   $0x0,-0x4(%rbp)
  jns    L1
  addl   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L22
  nop
  nop
  leave
  ret
```

assistant: ```c
void _transmit_bytes(const unsigned char *data, const int length, const int delay_per_bit) {
  unsigned char bit;
  int i, j;
  for (i = 0; i < length; i++) {
    for (j = 7; j >= 0; j--) {
      bit = (data[i] >> j) & 1;
      _transmit_to_gpio_with_increased_bit(bit, delay_per_bit);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_transmit_bytes>:
  endbr64
  test   %esi,%esi
  jle    L4
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%r12d
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r13
  jmp    L39
L37:
  add    $0x1,%rbp
  cmp    %r13,%rbp
  je     L24
L39:
  mov    $0x7,%ebx
L20:
  movzbl 0x0(%rbp),%edi
  mov    %ebx,%ecx
  sar    %cl,%edi
  and    $0x1,%edi
  mov    %r12d,%esi
  call   <_transmit_to_gpio_with_increased_bit@plt>
  sub    $0x1,%ebx
  cmp    $0xffffffff,%ebx
  jne    L20
  jmp    L37
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  ret
```

assistant: ```c
void _transmit_bytes(const unsigned char *data, const int length, const int delay_per_bit) {
  unsigned char bit;
  int i, j;
  for (i = 0; i < length; i++) {
    for (j = 7; j >= 0; j--) {
      bit = (data[i] >> j) & 1;
      _transmit_to_gpio_with_increased_bit(bit, delay_per_bit);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_transmit_bytes>:
  endbr64
  test   %esi,%esi
  jle    L18
  push   %r13
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nop
L19:
  mov    $0x7,%r13d
  cs nopw 0x0(%rax,%rax,1)
L8:
  movzbl (%rbx),%edi
  mov    %r13d,%ecx
  mov    %ebp,%esi
  sar    %cl,%edi
  and    $0x1,%edi
  call   <_transmit_to_gpio_with_increased_bit@plt>
  sub    $0x1,%r13d
  jae    L8
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L19
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  ret
```

assistant: ```c
void _transmit_bytes(const unsigned char *data, const int length, const int delay_per_bit) {
  unsigned char bit;
  int i, j;
  for (i = 0; i < length; i++) {
    for (j = 7; j >= 0; j--) {
      bit = (data[i] >> j) & 1;
      _transmit_to_gpio_with_increased_bit(bit, delay_per_bit);
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<_transmit_bytes>:
  endbr64
  test   %esi,%esi
  jle    L4
  push   %r13
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nop
L10:
  mov    $0x7,%r13d
  cs nopw 0x0(%rax,%rax,1)
L9:
  movzbl (%rbx),%edi
  mov    %r13d,%ecx
  mov    %ebp,%esi
  sar    %cl,%edi
  and    $0x1,%edi
  call   <_transmit_to_gpio_with_increased_bit@plt>
  sub    $0x1,%r13d
  jae    L9
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L10
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  ret
```

assistant: ```c
void _transmit_bytes(const unsigned char *data, const int length, const int delay_per_bit) {
  unsigned char bit;
  int i, j;
  for (i = 0; i < length; i++) {
    for (j = 7; j >= 0; j--) {
      bit = (data[i] >> j) & 1;
      _transmit_to_gpio_with_increased_bit(bit, delay_per_bit);
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<somma_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int somma_n(int a, int b) {
  int somma = a + b;
  return somma;
}
```
user: What is the c source code of the assembly code below:

```asm
<somma_n>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int somma_n(int a, int b) {
  int somma = a + b;
  return somma;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<somma_n>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int somma_n(int a, int b) {
  int somma = a + b;
  return somma;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<somma_n>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int somma_n(int a, int b) {
  int somma = a + b;
  return somma;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<yy_1_TableBlockHeader>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void yy_1_TableBlockHeader(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<yy_1_TableBlockHeader>:
  endbr64
  ret
```

assistant: ```c
void yy_1_TableBlockHeader(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<yy_1_TableBlockHeader>:
  endbr64
  ret
```

assistant: ```c
void yy_1_TableBlockHeader(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<yy_1_TableBlockHeader>:
  endbr64
  ret
```

assistant: ```c
void yy_1_TableBlockHeader(char *yytext, int yyleng) {
  ;
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<qcount_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x38(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x2c(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jns    L37
  mov    $0x0,%eax
  jmp    L31
L37:
  lea    -0x20(%rbp),%rcx
  mov    -0x2c(%rbp),%eax
  mov    $0x14,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  movl   $0x0,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L38
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <atoll@plt>
  mov    %eax,-0x30(%rbp)
L38:
  mov    -0x30(%rbp),%eax
L31:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
int64_t qcount_read(const char *filepath) {
  int fd = open(filepath, O_RDONLY, 0);
  if (fd < 0)
    return 0;
  char buf[20 + 1];
  ssize_t readed = read(fd, buf, (sizeof(buf) - 1));
  close(fd);
  int64_t num = 0;
  if (readed > 0) {
    buf[readed] = '\0';
    num = atoll(buf);
  }
  return num;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<qcount_read>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edx
  mov    D1(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  test   %ebx,%ebx
  js     L9
  mov    %rsp,%rsi
  mov    $0x14,%edx
  mov    %ebx,%edi
  call   <read@plt>
  mov    %rax,%rbp
  mov    %ebx,%edi
  call   <close@plt>
  mov    $0x0,%eax
  test   %rbp,%rbp
  jne    L34
L9:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L37
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L34:
  movb   $0x0,(%rsp,%rbp,1)
  mov    %rsp,%rdi
  call   <atoll@plt>
  jmp    L9
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t qcount_read(const char *filepath) {
  int fd = open(filepath, O_RDONLY, 0);
  if (fd < 0)
    return 0;
  char buf[20 + 1];
  ssize_t readed = read(fd, buf, (sizeof(buf) - 1));
  close(fd);
  int64_t num = 0;
  if (readed > 0) {
    buf[readed] = '\0';
    num = atoll(buf);
  }
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<qcount_read>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  test   %eax,%eax
  js     L26
  mov    %rsp,%r12
  mov    %eax,%edi
  mov    $0x14,%edx
  mov    %eax,%ebp
  mov    %r12,%rsi
  call   <read@plt>
  mov    %ebp,%edi
  mov    %rax,%rbx
  call   <close@plt>
  test   %rbx,%rbx
  jne    L22
L26:
  xor    %eax,%eax
L17:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L22:
  mov    %r12,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <atoll@plt>
  jmp    L17
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t qcount_read(const char *filepath) {
  int fd = open(filepath, O_RDONLY, 0);
  if (fd < 0)
    return 0;
  char buf[20 + 1];
  ssize_t readed = read(fd, buf, (sizeof(buf) - 1));
  close(fd);
  int64_t num = 0;
  if (readed > 0) {
    buf[readed] = '\0';
    num = atoll(buf);
  }
  return num;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<qcount_read>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  test   %eax,%eax
  js     L12
  mov    %rsp,%r12
  mov    %eax,%edi
  mov    $0x14,%edx
  mov    %eax,%ebp
  mov    %r12,%rsi
  call   <read@plt>
  mov    %ebp,%edi
  mov    %rax,%rbx
  call   <close@plt>
  test   %rbx,%rbx
  jne    L36
L12:
  xor    %eax,%eax
L11:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L36:
  mov    %r12,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <atoll@plt>
  jmp    L11
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t qcount_read(const char *filepath) {
  int fd = open(filepath, O_RDONLY, 0);
  if (fd < 0)
    return 0;
  char buf[20 + 1];
  ssize_t readed = read(fd, buf, (sizeof(buf) - 1));
  close(fd);
  int64_t num = 0;
  if (readed > 0) {
    buf[readed] = '\0';
    num = atoll(buf);
  }
  return num;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L19:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %eax,%eax
  jne    L16
  cmpl   $0x0,-0x4(%rbp)
  jns    L19
L16:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int Trim(char *buffer) {
  int n = strlen(buffer) - 1;
  while (!isalnum(buffer[n]) && n >= 0)
    buffer[n--] = '\0';
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Trim>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
L7:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isalnum@plt>
  test   %ebx,%ebx
  js     L3
  test   %eax,%eax
  jne    L3
  movb   $0x0,0x0(%rbp,%rbx,1)
  sub    $0x1,%rbx
  jmp    L7
L3:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Trim(char *buffer) {
  int n = strlen(buffer) - 1;
  while (!isalnum(buffer[n]) && n >= 0)
    buffer[n--] = '\0';
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Trim>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L25:
  test   %eax,%eax
  jne    L29
  movb   $0x0,0x0(%rbp,%rbx,1)
  sub    $0x1,%rbx
L11:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isalnum@plt>
  test   %ebx,%ebx
  jns    L25
L29:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Trim(char *buffer) {
  int n = strlen(buffer) - 1;
  while (!isalnum(buffer[n]) && n >= 0)
    buffer[n--] = '\0';
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Trim>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  jmp    L8
  nopw   0x0(%rax,%rax,1)
L5:
  test   %eax,%eax
  jne    L23
  movb   $0x0,0x0(%rbp,%rbx,1)
  sub    $0x1,%rbx
L8:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isalnum@plt>
  test   %ebx,%ebx
  jns    L5
L23:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Trim(char *buffer) {
  int n = strlen(buffer) - 1;
  while (!isalnum(buffer[n]) && n >= 0)
    buffer[n--] = '\0';
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x32,%edi
  call   <echo@plt>
  mov    $0x78,%edi
  call   <echo@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
main() {
  echo(2 * 10 + 10 * 3);
  echo((2 * (10 + 10)) * 3);
}
```
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
  mov    $0x32,%edi
  call   <echo@plt>
  mov    $0x78,%edi
  call   <echo@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  echo(2 * 10 + 10 * 3);
  echo((2 * (10 + 10)) * 3);
}
```
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
  mov    $0x32,%edi
  call   <echo@plt>
  mov    $0x78,%edi
  call   <echo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  echo(2 * 10 + 10 * 3);
  echo((2 * (10 + 10)) * 3);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%edi
  call   <echo@plt>
  mov    $0x78,%edi
  call   <echo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  echo(2 * 10 + 10 * 3);
  echo((2 * (10 + 10)) * 3);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<adjust_coeffs_to_mod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L21
L24:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rcx
  cqto
  idivq  -0x28(%rbp)
  mov    %rdx,%rax
  mov    %rax,(%rcx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jns    L25
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
L25:
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L24
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long *adjust_coeffs_to_mod(int degree, long *coeffs, long mod) {
  long *adjustedCoeffs = calloc(degree + 1, sizeof(long));
  int i;
  for (i = 0; i <= degree; i++) {
    adjustedCoeffs[i] = coeffs[i] % mod;
    if (adjustedCoeffs[i] < 0) {
      adjustedCoeffs[i] += mod;
    }
  }
  return adjustedCoeffs;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<adjust_coeffs_to_mod>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  lea    0x1(%rdi),%edi
  mov    $0x8,%esi
  call   <calloc@plt>
  mov    %rax,%rsi
  test   %r12d,%r12d
  js     L15
  lea    0x1(%r12),%edi
  shl    $0x3,%rdi
  mov    $0x0,%ecx
L16:
  mov    0x0(%rbp,%rcx,1),%rax
  cqto
  idiv   %rbx
  lea    (%rdx,%rbx,1),%rax
  test   %rdx,%rdx
  cmovs  %rax,%rdx
  mov    %rdx,(%rsi,%rcx,1)
  add    $0x8,%rcx
  cmp    %rdi,%rcx
  jne    L16
L15:
  mov    %rsi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long *adjust_coeffs_to_mod(int degree, long *coeffs, long mod) {
  long *adjustedCoeffs = calloc(degree + 1, sizeof(long));
  int i;
  for (i = 0; i <= degree; i++) {
    adjustedCoeffs[i] = coeffs[i] % mod;
    if (adjustedCoeffs[i] < 0) {
      adjustedCoeffs[i] += mod;
    }
  }
  return adjustedCoeffs;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<adjust_coeffs_to_mod>:
  endbr64
  push   %r12
  movslq %edi,%r12
  push   %rbp
  lea    0x1(%r12),%edi
  mov    %rsi,%rbp
  mov    $0x8,%esi
  push   %rbx
  mov    %rdx,%rbx
  call   <calloc@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  js     L6
  lea    0x8(,%r12,8),%rsi
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L19:
  mov    0x0(%rbp,%rcx,1),%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  lea    (%rdx,%rbx,1),%rax
  cmovs  %rax,%rdx
  mov    %rdx,(%r8,%rcx,1)
  add    $0x8,%rcx
  cmp    %rcx,%rsi
  jne    L19
L6:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long *adjust_coeffs_to_mod(int degree, long *coeffs, long mod) {
  long *adjustedCoeffs = calloc(degree + 1, sizeof(long));
  int i;
  for (i = 0; i <= degree; i++) {
    adjustedCoeffs[i] = coeffs[i] % mod;
    if (adjustedCoeffs[i] < 0) {
      adjustedCoeffs[i] += mod;
    }
  }
  return adjustedCoeffs;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<adjust_coeffs_to_mod>:
  endbr64
  push   %r12
  movslq %edi,%r12
  push   %rbp
  lea    0x1(%r12),%edi
  mov    %rsi,%rbp
  mov    $0x8,%esi
  push   %rbx
  mov    %rdx,%rbx
  call   <calloc@plt>
  mov    %rax,%r8
  test   %r12d,%r12d
  js     L8
  lea    0x8(,%r12,8),%rsi
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L0:
  mov    0x0(%rbp,%rcx,1),%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  lea    (%rbx,%rdx,1),%rax
  cmovs  %rax,%rdx
  mov    %rdx,(%r8,%rcx,1)
  add    $0x8,%rcx
  cmp    %rcx,%rsi
  jne    L0
L8:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long *adjust_coeffs_to_mod(int degree, long *coeffs, long mod) {
  long *adjustedCoeffs = calloc(degree + 1, sizeof(long));
  int i;
  for (i = 0; i <= degree; i++) {
    adjustedCoeffs[i] = coeffs[i] % mod;
    if (adjustedCoeffs[i] < 0) {
      adjustedCoeffs[i] += mod;
    }
  }
  return adjustedCoeffs;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rb_assert_delete>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rb_delete@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *(rb_assert_delete)(struct rb_table *table, void *item) {
  void *p = rb_delete(table, item);
  assert(p != NULL);
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rb_assert_delete>:
  endbr64
  push   %rbx
  call   <rb_delete@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
void *(rb_assert_delete)(struct rb_table *table, void *item) {
  void *p = rb_delete(table, item);
  assert(p != NULL);
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rb_assert_delete>:
  endbr64
  push   %r12
  call   <rb_delete@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *(rb_assert_delete)(struct rb_table *table, void *item) {
  void *p = rb_delete(table, item);
  assert(p != NULL);
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rb_assert_delete>:
  endbr64
  push   %r12
  call   <rb_delete@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *(rb_assert_delete)(struct rb_table *table, void *item) {
  void *p = rb_delete(table, item);
  assert(p != NULL);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x55,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x48(%rbp),%rdi
  mov    -0x48(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <string_is@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void test_str(const char *file, unsigned int line, const char *s, size_t n, const char *expected) {
  char buffer[16];
setup:
  memset(&buffer[0], 0x55, sizeof buffer);
exercise:
  (void)strncpy(&buffer[0], s, n);
verify:
  string_is(file, line, &buffer[0], expected, expected);
teardown:
  return;
}
```