ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<C_a_i_bytevector>:
  endbr64
  mov    (%rdi),%rcx
  sar    %rdx
  lea    0x0(,%rdx,8),%eax
  cltq
  movabs $0x5000000000000000,%rsi
  or     %rsi,%rax
  mov    %rax,(%rcx)
  movslq %edx,%rdx
  lea    0x8(%rcx,%rdx,8),%rax
  mov    %rax,(%rdi)
  mov    %rcx,%rax
  ret
```

assistant: ```c
long C_a_i_bytevector(long **ptr, int c, long num) {
  long *p = *ptr, *p0;
  int n = ((num) >> 1);
  p0 = p;
  *(p++) = (0x00000000 | 0x4000000000000000L | 0x1000000000000000L) | ((n) << 3);
  *ptr = p + n;
  return (long)p0;
}
```
user: What is the c source code of the assembly code below:

```asm
<C_a_i_bytevector>:
  endbr64
  movabs $0x5000000000000000,%rcx
  sar    %rdx
  mov    (%rdi),%r8
  lea    0x0(,%rdx,8),%eax
  movslq %edx,%rdx
  cltq
  or     %rcx,%rax
  mov    %rax,(%r8)
  lea    0x8(%r8,%rdx,8),%rax
  mov    %rax,(%rdi)
  mov    %r8,%rax
  ret
```

assistant: ```c
long C_a_i_bytevector(long **ptr, int c, long num) {
  long *p = *ptr, *p0;
  int n = ((num) >> 1);
  p0 = p;
  *(p++) = (0x00000000 | 0x4000000000000000L | 0x1000000000000000L) | ((n) << 3);
  *ptr = p + n;
  return (long)p0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<C_a_i_bytevector>:
  endbr64
  movabs $0x5000000000000000,%rcx
  sar    %rdx
  mov    (%rdi),%r8
  lea    0x0(,%rdx,8),%eax
  movslq %edx,%rdx
  cltq
  or     %rcx,%rax
  mov    %rax,(%r8)
  lea    0x8(%r8,%rdx,8),%rax
  mov    %rax,(%rdi)
  mov    %r8,%rax
  ret
```

assistant: ```c
long C_a_i_bytevector(long **ptr, int c, long num) {
  long *p = *ptr, *p0;
  int n = ((num) >> 1);
  p0 = p;
  *(p++) = (0x00000000 | 0x4000000000000000L | 0x1000000000000000L) | ((n) << 3);
  *ptr = p + n;
  return (long)p0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_coverage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %rdi,-0x408(%rbp)
  mov    %esi,-0x40c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x408(%rbp),%rcx
  mov    -0x408(%rbp),%rdx
  lea    -0x3f0(%rbp),%rax
  lea    D19(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    %eax,-0x3f4(%rbp)
  mov    -0x408(%rbp),%rsi
  mov    -0x40c(%rbp),%ecx
  mov    -0x408(%rbp),%rdx
  lea    -0x3f0(%rbp),%rax
  mov    %rsi,%r8
  lea    D10(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    %eax,-0x3f4(%rbp)
  mov    -0x3f4(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D19
data_type: string
name: retrieve_data
arguments: label: D10
data_type: string

tool: D19: samtools mpileup %s > %s.pileup

tool: D10: cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp

assistant: ```c
int get_coverage(char *input_filename, int depth) {
  int return_value;
  char system_command[1000];
  sprintf(system_command, "samtools mpileup %s > %s.pileup", input_filename, input_filename);
  return_value = system(system_command);
  sprintf(system_command, "cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp", input_filename, depth, input_filename);
  return_value = system(system_command);
  return return_value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_coverage>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  lea    D10(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    %rbx,%r8
  mov    %r12d,%ecx
  mov    %rbx,%rdx
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x3f0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D10
datatype: string
name: parse_data
arguments: data_tag: D5
datatype: string

tool: D10: samtools mpileup %s > %s.pileup

tool: D5: cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp

assistant: ```c
int get_coverage(char *input_filename, int depth) {
  int return_value;
  char system_command[1000];
  sprintf(system_command, "samtools mpileup %s > %s.pileup", input_filename, input_filename);
  return_value = system(system_command);
  sprintf(system_command, "cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp", input_filename, depth, input_filename);
  return_value = system(system_command);
  return return_value;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<get_coverage>:
  endbr64
  push   %r13
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  mov    %esi,%r13d
  push   %r12
  lea    D13(%rip),%rsi
  mov    %rdi,%r12
  push   %rbp
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    %r12,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %r12,%r8
  mov    %r13d,%ecx
  lea    D17(%rip),%rsi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x3f0,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
datatype: string
name: parse_data
arguments: data_label: D17
datatype: string

tool: D13: samtools mpileup %s > %s.pileup

tool: D17: cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp

assistant: ```c
int get_coverage(char *input_filename, int depth) {
  int return_value;
  char system_command[1000];
  sprintf(system_command, "samtools mpileup %s > %s.pileup", input_filename, input_filename);
  return_value = system(system_command);
  sprintf(system_command, "cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp", input_filename, depth, input_filename);
  return_value = system(system_command);
  return return_value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_coverage>:
  endbr64
  push   %r13
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  mov    %esi,%r13d
  push   %r12
  lea    D5(%rip),%rsi
  mov    %rdi,%r12
  push   %rbp
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    %r12,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %r12,%r8
  mov    %r13d,%ecx
  lea    D18(%rip),%rsi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x3f0,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D5
type: string
name: retrieve_data
arguments: tag: D18
type: string

tool: D5: samtools mpileup %s > %s.pileup

tool: D18: cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp

assistant: ```c
int get_coverage(char *input_filename, int depth) {
  int return_value;
  char system_command[1000];
  sprintf(system_command, "samtools mpileup %s > %s.pileup", input_filename, input_filename);
  return_value = system(system_command);
  sprintf(system_command, "cut -f1,2,4 %s.pileup | awk '{ if ($3 > %d) print }' > %s.tmp", input_filename, depth, input_filename);
  return_value = system(system_command);
  return return_value;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cuentachar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L22:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  cmpb   $0x40,-0xd(%rbp)
  jle    L38
  cmpb   $0x5a,-0xd(%rbp)
  jle    L25
L38:
  cmpb   $0x60,-0xd(%rbp)
  jle    L1
  cmpb   $0x7a,-0xd(%rbp)
  jle    L25
L1:
  cmpb   $0xd1,-0xd(%rbp)
  je     L25
  cmpb   $0xf1,-0xd(%rbp)
  jne    L55
L25:
  addl   $0x1,-0x8(%rbp)
L55:
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L22
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int cuentachar(char *cadena) {
  int tamano;
  int i;
  char tmp;
  int numCaracteres = 0;
  tamano = strlen(cadena);
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -47) || (tmp == -15))) {
      numCaracteres++;
    }
  }
  return numCaracteres;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cuentachar>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L29
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rdi
  mov    $0x0,%esi
  jmp    L36
L22:
  add    $0x1,%esi
L58:
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  je     L25
L36:
  movzbl (%rdx),%ecx
  mov    %ecx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  jbe    L22
  and    $0xffffffdf,%ecx
  cmp    $0xd1,%cl
  jne    L58
  jmp    L22
L29:
  mov    $0x0,%esi
L25:
  mov    %esi,%eax
  pop    %rbx
  ret
```

assistant: ```c
int cuentachar(char *cadena) {
  int tamano;
  int i;
  char tmp;
  int numCaracteres = 0;
  tamano = strlen(cadena);
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -47) || (tmp == -15))) {
      numCaracteres++;
    }
  }
  return numCaracteres;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cuentachar>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L37
  sub    $0x1,%eax
  mov    %rbx,%rdi
  xor    %r8d,%r8d
  lea    0x1(%rbx,%rax,1),%rcx
  nop
L32:
  movzbl (%rdi),%eax
  and    $0xffffffdf,%eax
  lea    -0x41(%rax),%edx
  cmp    $0x19,%dl
  jbe    L25
  cmp    $0xd1,%al
  jne    L5
L25:
  add    $0x1,%r8d
L5:
  add    $0x1,%rdi
  cmp    %rdi,%rcx
  jne    L32
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L37:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int cuentachar(char *cadena) {
  int tamano;
  int i;
  char tmp;
  int numCaracteres = 0;
  tamano = strlen(cadena);
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -47) || (tmp == -15))) {
      numCaracteres++;
    }
  }
  return numCaracteres;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<validateISBN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0xd,-0x1c(%rbp)
  je     L91
  mov    $0xffffffff,%eax
  jmp    L24
L91:
  movl   $0x0,-0x2c(%rbp)
  jmp    L79
L74:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x2d(%rbp)
  lea    -0x2d(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x28(%rbp),%eax
  add    %eax,-0x24(%rbp)
  cmpl   $0x3,-0x28(%rbp)
  jne    L80
  movl   $0x1,-0x28(%rbp)
  jmp    L34
L80:
  movl   $0x3,-0x28(%rbp)
L34:
  addl   $0x1,-0x2c(%rbp)
L79:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L74
  mov    -0x24(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x2d(%rbp)
  lea    -0x2d(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0xa,%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L37
  movl   $0x0,-0x20(%rbp)
L37:
  mov    -0x20(%rbp),%eax
  cltq
  cmp    %rax,-0x10(%rbp)
  jne    L90
  mov    $0x1,%eax
  jmp    L24
L90:
  mov    $0x0,%eax
L24:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L65
  call   <__stack_chk_fail@plt>
L65:
  leave
  ret
```

assistant: ```c
int validateISBN(char isbn[]) {
  long checksum = 0;
  int i, mul = 1, sum = 0, m10, j;
  int len = strlen(isbn);
  if (len != 13)
    return -1;
  for (i = 0; i < len - 1; i++) {
    char c = isbn[i];
    int val = atoi(&c);
    sum += val * mul;
    if (mul == 3)
      mul = 1;
    else
      mul = 3;
  }
  m10 = sum % 10;
  char checkDigit = isbn[len - 1];
  checksum = atoi(&checkDigit);
  int sub = 10 - m10;
  if (m10 == 0)
    sub = 0;
  if (sub == checksum)
    return 1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<validateISBN>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0xffffffff,%eax
  cmp    $0xd,%edx
  jne    L17
  mov    %r13,%rbp
  lea    0xc(%r13),%r15
  mov    $0x0,%r12d
  mov    $0x1,%ebx
  lea    0x7(%rsp),%r14
L26:
  movzbl 0x0(%rbp),%eax
  mov    %al,0x7(%rsp)
  mov    %r14,%rdi
  call   <atoi@plt>
  imul   %ebx,%eax
  add    %eax,%r12d
  cmp    $0x3,%ebx
  setne  %bl
  movzbl %bl,%ebx
  lea    0x1(%rbx,%rbx,1),%ebx
  add    $0x1,%rbp
  cmp    %r15,%rbp
  jne    L26
  movslq %r12d,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  mov    %r12d,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  mov    %r12d,%ebx
  sub    %eax,%ebx
  movzbl 0xc(%r13),%eax
  mov    %al,0x7(%rsp)
  lea    0x7(%rsp),%rdi
  call   <atoi@plt>
  mov    %rax,%rdx
  mov    $0xa,%eax
  sub    %ebx,%eax
  test   %ebx,%ebx
  cmovne %eax,%ebx
  movslq %ebx,%rbx
  cmp    %rdx,%rbx
  sete   %al
  movzbl %al,%eax
L17:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int validateISBN(char isbn[]) {
  long checksum = 0;
  int i, mul = 1, sum = 0, m10, j;
  int len = strlen(isbn);
  if (len != 13)
    return -1;
  for (i = 0; i < len - 1; i++) {
    char c = isbn[i];
    int val = atoi(&c);
    sum += val * mul;
    if (mul == 3)
      mul = 1;
    else
      mul = 3;
  }
  m10 = sum % 10;
  char checkDigit = isbn[len - 1];
  checksum = atoi(&checkDigit);
  int sub = 10 - m10;
  if (m10 == 0)
    sub = 0;
  if (sub == checksum)
    return 1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<validateISBN>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    $0xffffffff,%eax
  cmp    $0xd,%r8d
  jne    L10
  mov    %r14,%rbp
  lea    0xc(%r14),%r13
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  lea    0x7(%rsp),%r15
L38:
  movzbl 0x0(%rbp),%eax
  mov    %r15,%rdi
  mov    %al,0x7(%rsp)
  call   <atoi@plt>
  imul   %ebx,%eax
  add    %eax,%r12d
  cmp    $0x3,%ebx
  setne  %bl
  add    $0x1,%rbp
  movzbl %bl,%ebx
  lea    0x1(%rbx,%rbx,1),%ebx
  cmp    %r13,%rbp
  jne    L38
  movslq %r12d,%rbx
  mov    %r12d,%eax
  mov    %r15,%rdi
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  mov    %r12d,%ebx
  add    %eax,%eax
  sub    %eax,%ebx
  movzbl 0xc(%r14),%eax
  mov    %al,0x7(%rsp)
  call   <atoi@plt>
  xor    %edx,%edx
  test   %ebx,%ebx
  jne    L30
L3:
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  mov    $0xa,%edx
  sub    %ebx,%edx
  movslq %edx,%rdx
  jmp    L3
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int validateISBN(char isbn[]) {
  long checksum = 0;
  int i, mul = 1, sum = 0, m10, j;
  int len = strlen(isbn);
  if (len != 13)
    return -1;
  for (i = 0; i < len - 1; i++) {
    char c = isbn[i];
    int val = atoi(&c);
    sum += val * mul;
    if (mul == 3)
      mul = 1;
    else
      mul = 3;
  }
  m10 = sum % 10;
  char checkDigit = isbn[len - 1];
  checksum = atoi(&checkDigit);
  int sub = 10 - m10;
  if (m10 == 0)
    sub = 0;
  if (sub == checksum)
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<validateISBN>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    $0xffffffff,%eax
  cmp    $0xd,%r8d
  jne    L29
  mov    %r14,%rbp
  lea    0xc(%r14),%r13
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  lea    0x7(%rsp),%r15
L5:
  movzbl 0x0(%rbp),%eax
  mov    %r15,%rdi
  mov    %al,0x7(%rsp)
  call   <atoi@plt>
  imul   %ebx,%eax
  add    %eax,%r12d
  cmp    $0x3,%ebx
  setne  %bl
  add    $0x1,%rbp
  movzbl %bl,%ebx
  lea    0x1(%rbx,%rbx,1),%ebx
  cmp    %r13,%rbp
  jne    L5
  movslq %r12d,%rbx
  mov    %r12d,%eax
  mov    %r15,%rdi
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  mov    %r12d,%ebx
  add    %eax,%eax
  sub    %eax,%ebx
  movzbl 0xc(%r14),%eax
  mov    %al,0x7(%rsp)
  call   <atoi@plt>
  xor    %edx,%edx
  test   %ebx,%ebx
  jne    L38
L18:
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
L29:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  mov    $0xa,%edx
  sub    %ebx,%edx
  movslq %edx,%rdx
  jmp    L18
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int validateISBN(char isbn[]) {
  long checksum = 0;
  int i, mul = 1, sum = 0, m10, j;
  int len = strlen(isbn);
  if (len != 13)
    return -1;
  for (i = 0; i < len - 1; i++) {
    char c = isbn[i];
    int val = atoi(&c);
    sum += val * mul;
    if (mul == 3)
      mul = 1;
    else
      mul = 3;
  }
  m10 = sum % 10;
  char checkDigit = isbn[len - 1];
  checksum = atoi(&checkDigit);
  int sub = 10 - m10;
  if (m10 == 0)
    sub = 0;
  if (sub == checksum)
    return 1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<HashString2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L19:
  mov    -0x4(%rbp),%eax
  imul   $0x2d9,%eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x4(%rbp),%eax
  sar    %eax
  xor    %edx,%eax
  mov    %eax,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int HashString2(const char *s) {
  int hval = 0;
  while (*s) {
    hval = (hval * 729 + *s * 37) ^ (hval >> 1);
    s++;
  }
  return hval;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<HashString2>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L8
  mov    $0x0,%eax
L18:
  imul   $0x2d9,%eax,%ecx
  movsbl %dl,%edx
  lea    (%rdx,%rdx,8),%esi
  lea    (%rdx,%rsi,4),%edx
  add    %ecx,%edx
  sar    %eax
  xor    %edx,%eax
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L18
  ret
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int HashString2(const char *s) {
  int hval = 0;
  while (*s) {
    hval = (hval * 729 + *s * 37) ^ (hval >> 1);
    s++;
  }
  return hval;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<HashString2>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L19
  nopl   (%rax)
L11:
  imul   $0x2d9,%eax,%ecx
  lea    (%rdx,%rdx,8),%esi
  sar    %eax
  add    $0x1,%rdi
  lea    (%rdx,%rsi,4),%edx
  add    %ecx,%edx
  xor    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L11
  ret
  xchg   %ax,%ax
L19:
  ret
```

assistant: ```c
int HashString2(const char *s) {
  int hval = 0;
  while (*s) {
    hval = (hval * 729 + *s * 37) ^ (hval >> 1);
    s++;
  }
  return hval;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<HashString2>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L18
  nopl   (%rax)
L0:
  imul   $0x2d9,%eax,%ecx
  lea    (%rdx,%rdx,8),%esi
  sar    %eax
  add    $0x1,%rdi
  lea    (%rdx,%rsi,4),%edx
  add    %ecx,%edx
  xor    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L0
  ret
  xchg   %ax,%ax
L18:
  ret
```

assistant: ```c
int HashString2(const char *s) {
  int hval = 0;
  while (*s) {
    hval = (hval * 729 + *s * 37) ^ (hval >> 1);
    s++;
  }
  return hval;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fcomp_preg32_dword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0xd8,%edi
  call   <put8@plt>
  mov    -0x4(%rbp),%eax
  add    $0x18,%eax
  mov    %eax,%edi
  call   <put8@plt>
  nop
  leave
  ret
```

assistant: ```c
void fcomp_preg32_dword(int reg32) {
  put8(0xD8);
  put8(0x18 + reg32);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fcomp_preg32_dword>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xd8,%edi
  call   <put8@plt>
  lea    0x18(%rbx),%edi
  call   <put8@plt>
  pop    %rbx
  ret
```

assistant: ```c
void fcomp_preg32_dword(int reg32) {
  put8(0xD8);
  put8(0x18 + reg32);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fcomp_preg32_dword>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xd8,%edi
  call   <put8@plt>
  lea    0x18(%rbx),%edi
  pop    %rbx
  jmp    1050 <put8@plt>
```

assistant: ```c
void fcomp_preg32_dword(int reg32) {
  put8(0xD8);
  put8(0x18 + reg32);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fcomp_preg32_dword>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xd8,%edi
  call   <put8@plt>
  lea    0x18(%rbx),%edi
  pop    %rbx
  jmp    1050 <put8@plt>
```

assistant: ```c
void fcomp_preg32_dword(int reg32) {
  put8(0xD8);
  put8(0x18 + reg32);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x13,-0x4(%rbp)
  jle    L16
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: string

tool: D0: %d

assistant: ```c
void read(int *array) {
  int i;
  for (i = 0; i < 20; i++)
    scanf("%d", array + i);
}
```