ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_ucRcAT1608_AuthEx>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x64,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x2527,%edx
  lea    D16(%rip),%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %r12,%rsi
  lea    0x9(%rbp),%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    $0x29,%eax
  mov    %rbp,%rdi
  mov    $0x2,%edx
  mov    %ax,0x19(%rbp)
  mov    $0x1a,%esi
  call   <iDoRemoteFunc@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <free@plt>
  test   %r12d,%r12d
  jne    L5
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbp
  pop    %r12
  jmp    10c0 <iGetParaInt@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r12d,%esi
  lea    D5(%rip),%rdi
  call   <vWriteLogTxt@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D16
type_guessed: string
name: parse_data
arguments: tag: D5
type_guessed: string

tool: D16: %u(b016

tool: D5: _AT1608_AuthEx iDoRemoteFunc ret = %d

assistant: ```c
unsigned char _ucRcAT1608_AuthEx(unsigned char *Gc) {
  unsigned int uiDataLen, uiRet;
  char *psFunCall = malloc(100);
  uiDataLen = sprintf(psFunCall, "%u(b016", 9511);
  memcpy(psFunCall + uiDataLen, Gc, 16);
  uiDataLen += 16;
  uiDataLen += sprintf(psFunCall + uiDataLen, ")");
  uiRet = iDoRemoteFunc((unsigned char *)psFunCall, uiDataLen, 2);
  free(psFunCall);
  if (uiRet) {
    vWriteLogTxt("_AT1608_AuthEx iDoRemoteFunc ret = %d", uiRet);
    return 1;
  }
  return iGetParaInt(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_ucRcAT1608_AuthEx>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x64,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x2527,%edx
  lea    D17(%rip),%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %r12,%rsi
  lea    0x9(%rbp),%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    $0x29,%eax
  mov    %rbp,%rdi
  mov    $0x2,%edx
  mov    %ax,0x19(%rbp)
  mov    $0x1a,%esi
  call   <iDoRemoteFunc@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <free@plt>
  test   %r12d,%r12d
  jne    L4
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbp
  pop    %r12
  jmp    10c0 <iGetParaInt@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %r12d,%esi
  lea    D10(%rip),%rdi
  call   <vWriteLogTxt@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
data_type: string
name: extract_symbolic_data
arguments: data_tag: D10
data_type: string

tool: D17: %u(b016

tool: D10: _AT1608_AuthEx iDoRemoteFunc ret = %d

assistant: ```c
unsigned char _ucRcAT1608_AuthEx(unsigned char *Gc) {
  unsigned int uiDataLen, uiRet;
  char *psFunCall = malloc(100);
  uiDataLen = sprintf(psFunCall, "%u(b016", 9511);
  memcpy(psFunCall + uiDataLen, Gc, 16);
  uiDataLen += 16;
  uiDataLen += sprintf(psFunCall + uiDataLen, ")");
  uiRet = iDoRemoteFunc((unsigned char *)psFunCall, uiDataLen, 2);
  free(psFunCall);
  if (uiRet) {
    vWriteLogTxt("_AT1608_AuthEx iDoRemoteFunc ret = %d", uiRet);
    return 1;
  }
  return iGetParaInt(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  movl   $0x6,-0x30(%rbp)
  movl   $0x2,-0x2c(%rbp)
  movl   $0x5,-0x28(%rbp)
  movl   $0x5,-0x24(%rbp)
  movl   $0x4,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  movl   $0x6,-0x18(%rbp)
  movl   $0x3,-0x14(%rbp)
  movl   $0x7,-0x10(%rbp)
  movl   $0x6,-0xc(%rbp)
  jmp    L17
L9:
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  mov    -0x30(%rbp,%rax,4),%eax
  add    %eax,-0x34(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x44(%rbp)
L17:
  mov    -0x44(%rbp),%eax
  add    $0x9,%eax
  cmp    $0x12,%eax
  ja     L9
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  add    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
int fun(int x) {
  int num = 0;
  int f[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  while (x / 10 != 0) {
    num = num + f[x % 10];
    x = x / 10;
  }
  num = num + f[x];
  return num;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fun>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x6,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x5,0x8(%rsp)
  movl   $0x5,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x3,0x1c(%rsp)
  movl   $0x7,0x20(%rsp)
  movl   $0x6,0x24(%rsp)
  lea    0x9(%rdi),%eax
  cmp    $0x12,%eax
  jbe    L4
  mov    $0x0,%eax
L33:
  movslq %edi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %edi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edi
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    %edx,%edi
  add    $0x9,%edx
  cmp    $0x12,%edx
  ja     L33
L29:
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x38,%rsp
  ret
L4:
  mov    $0x0,%eax
  jmp    L29
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fun(int x) {
  int num = 0;
  int f[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  while (x / 10 != 0) {
    num = num + f[x % 10];
    x = x / 10;
  }
  num = num + f[x];
  return num;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fun>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x200000006,%rax
  mov    %rax,(%rsp)
  movabs $0x500000005,%rax
  mov    %rax,0x8(%rsp)
  sub    $0x1,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x300000006,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x600000007,%rax
  mov    %rax,0x20(%rsp)
  lea    0x9(%rdi),%eax
  cmp    $0x12,%eax
  mov    $0x0,%eax
  jbe    L22
  nopw   0x0(%rax,%rax,1)
L12:
  movslq %edi,%rdx
  mov    %edi,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edi
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    %edx,%edi
  lea    0x9(%rdx),%edx
  cmp    $0x12,%edx
  ja     L12
L22:
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fun(int x) {
  int num = 0;
  int f[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  while (x / 10 != 0) {
    num = num + f[x % 10];
    x = x / 10;
  }
  num = num + f[x];
  return num;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fun>:
  endbr64
  sub    $0x38,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D4(%rip),%rax
  movaps %xmm0,(%rsp)
  movdqa D8(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  lea    0x9(%rdi),%eax
  cmp    $0x12,%eax
  movaps %xmm0,0x10(%rsp)
  mov    $0x0,%eax
  jbe    L25
  nopw   0x0(%rax,%rax,1)
L28:
  movslq %edi,%rdx
  mov    %edi,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edi
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    %edx,%edi
  lea    0x9(%rdx),%edx
  cmp    $0x12,%edx
  ja     L28
L25:
  movslq %edi,%rdi
  add    (%rsp,%rdi,4),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x38,%rsp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D15
type_infered: byte[16]
name: disassemble_data
arguments: data_label: D4
type_infered: byte[8]
name: disassemble_data
arguments: data_label: D8
type_infered: byte[16]

tool: {"D15": [0x06,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D4": [0x07,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D8": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

assistant: ```c
int fun(int x) {
  int num = 0;
  int f[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
  while (x / 10 != 0) {
    num = num + f[x % 10];
    x = x / 10;
  }
  num = num + f[x];
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x8(%rbp)
  movl   $0x2,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bar() {
  struct A {
    int x;
  };
  struct A a;
  a.x = 1;
  {
    struct A {
      int y;
    };
    struct A b;
    b.y = 2;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  struct A {
    int x;
  };
  struct A a;
  a.x = 1;
  {
    struct A {
      int y;
    };
    struct A b;
    b.y = 2;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  struct A {
    int x;
  };
  struct A a;
  a.x = 1;
  {
    struct A {
      int y;
    };
    struct A b;
    b.y = 2;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bar>:
  endbr64
  ret
```

assistant: ```c
void bar() {
  struct A {
    int x;
  };
  struct A a;
  a.x = 1;
  {
    struct A {
      int y;
    };
    struct A b;
    b.y = 2;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<upper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L5
L4:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
L15:
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  nop
  nop
  leave
  ret
```

assistant: ```c
void upper(char *s) {
  while (*s != '\0') {
    if (islower(*s))
      *s = toupper(*s);
    s++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<upper>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  jne    L0
L18:
  pop    %rbx
  ret
L34:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L18
L0:
  movsbl %dil,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L34
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  jmp    L34
```

assistant: ```c
void upper(char *s) {
  while (*s != '\0') {
    if (islower(*s))
      *s = toupper(*s);
    s++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<upper>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L5
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L46:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L19
L5:
  call   <islower@plt>
  test   %rax,%rax
  je     L46
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L5
L19:
  pop    %rbx
  ret
```

assistant: ```c
void upper(char *s) {
  while (*s != '\0') {
    if (islower(*s))
      *s = toupper(*s);
    s++;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<upper>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L23
L32:
  call   <islower@plt>
  test   %rax,%rax
  jne    L6
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L32
L23:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L6:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L32
  pop    %rbx
  ret
```

assistant: ```c
void upper(char *s) {
  while (*s != '\0') {
    if (islower(*s))
      *s = toupper(*s);
    s++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<String_cat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *String_cat(const char *s1, const char *s2) {
  int l1 = strlen(s1);
  int l2 = strlen(s2);
  char *out = malloc(l1 + l2 + 1);
  strncpy(out, s1, l1);
  strncpy(out + l1, s2, l2 + 1);
  return out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<String_cat>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r14
  mov    %rsi,%r13
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  lea    0x1(%rbp),%edx
  movslq %ebx,%rdi
  add    %r12,%rdi
  mov    %r13,%rsi
  call   <strncpy@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *String_cat(const char *s1, const char *s2) {
  int l1 = strlen(s1);
  int l2 = strlen(s2);
  char *out = malloc(l1 + l2 + 1);
  strncpy(out, s1, l1);
  strncpy(out + l1, s2, l2 + 1);
  return out;
}
```
user: What is the c source code of the assembly code below:

```asm
<String_cat>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strncpy@plt>
  movslq %ebx,%rdi
  lea    0x1(%rbp),%edx
  mov    %r13,%rsi
  add    %r12,%rdi
  call   <strncpy@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *String_cat(const char *s1, const char *s2) {
  int l1 = strlen(s1);
  int l2 = strlen(s2);
  char *out = malloc(l1 + l2 + 1);
  strncpy(out, s1, l1);
  strncpy(out + l1, s2, l2 + 1);
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<String_cat>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strncpy@plt>
  movslq %ebx,%rdi
  lea    0x1(%rbp),%edx
  mov    %r13,%rsi
  add    %r12,%rdi
  call   <strncpy@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *String_cat(const char *s1, const char *s2) {
  int l1 = strlen(s1);
  int l2 = strlen(s2);
  char *out = malloc(l1 + l2 + 1);
  strncpy(out, s1, l1);
  strncpy(out + l1, s2, l2 + 1);
  return out;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<not>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L4
  mov    $0x0,%eax
  jmp    L1
L4:
  mov    $0x1,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
unsigned not(unsigned a) {
  if (a)
    return 0;
  else
    return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<not>:
  endbr64
  test   %edi,%edi
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
unsigned not(unsigned a) {
  if (a)
    return 0;
  else
    return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<not>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  sete   %al
  ret
```

assistant: ```c
unsigned not(unsigned a) {
  if (a)
    return 0;
  else
    return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<not>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  sete   %al
  ret
```

assistant: ```c
unsigned not(unsigned a) {
  if (a)
    return 0;
  else
    return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fvec_arg_min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movq   $0x1,-0x10(%rbp)
  jmp    L15
L8:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  movss  -0x14(%rbp),%xmm0
  comiss %xmm1,%xmm0
  jbe    L17
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L17:
  addq   $0x1,-0x10(%rbp)
L15:
  mov    -0x10(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jl     L8
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int fvec_arg_min(const float *f, long n) {
  float m = f[0];
  long i, i0 = 0;
  assert(n > 0);
  for (i = 1; i < n; i++)
    if (f[i] < m) {
      m = f[i];
      i0 = i;
    }
  return i0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fvec_arg_min>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  movss  (%rdi),%xmm2
  movss  %xmm2,0xc(%rsp)
  test   %rsi,%rsi
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%rbx
  jle    L18
  mov    $0x0,%eax
  mov    $0x1,%edx
L8:
  movss  0x0(%rbp,%rdx,4),%xmm0
  movss  0xc(%rsp),%xmm1
  comiss %xmm0,%xmm1
  cmova  %rdx,%rax
  minss  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jne    L8
L5:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0x0,%eax
  jmp    L5
```

assistant: ```c
int fvec_arg_min(const float *f, long n) {
  float m = f[0];
  long i, i0 = 0;
  assert(n > 0);
  for (i = 1; i < n; i++)
    if (f[i] < m) {
      m = f[i];
      i0 = i;
    }
  return i0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fvec_arg_min>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movss  (%rdi),%xmm0
  xor    %edi,%edi
  test   %rsi,%rsi
  setg   %dil
  movss  %xmm0,0xc(%rsp)
  call   <assert@plt>
  cmp    $0x1,%rbx
  jle    L17
  movss  0xc(%rsp),%xmm0
  xor    %eax,%eax
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L14:
  movss  0x0(%rbp,%rdx,4),%xmm1
  comiss %xmm1,%xmm0
  minss  %xmm0,%xmm1
  cmova  %rdx,%rax
  add    $0x1,%rdx
  movaps %xmm1,%xmm0
  cmp    %rdx,%rbx
  jne    L14
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L17:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fvec_arg_min(const float *f, long n) {
  float m = f[0];
  long i, i0 = 0;
  assert(n > 0);
  for (i = 1; i < n; i++)
    if (f[i] < m) {
      m = f[i];
      i0 = i;
    }
  return i0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fvec_arg_min>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movss  (%rdi),%xmm0
  xor    %edi,%edi
  test   %rsi,%rsi
  setg   %dil
  movss  %xmm0,0xc(%rsp)
  call   <assert@plt>
  cmp    $0x1,%rbx
  jle    L9
  movss  0xc(%rsp),%xmm0
  xor    %eax,%eax
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L19:
  movss  0x0(%rbp,%rdx,4),%xmm1
  comiss %xmm1,%xmm0
  minss  %xmm0,%xmm1
  cmova  %rdx,%rax
  add    $0x1,%rdx
  movaps %xmm1,%xmm0
  cmp    %rdx,%rbx
  jne    L19
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fvec_arg_min(const float *f, long n) {
  float m = f[0];
  long i, i0 = 0;
  assert(n > 0);
  for (i = 1; i < n; i++)
    if (f[i] < m) {
      m = f[i];
      i0 = i;
    }
  return i0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vector_madd>:
  endbr64
  mov    %rdx,%r9
  mov    %r8d,%r10d
  and    $0x3,%r10d
  and    $0xfffffffc,%r8d
  je     L4
  mov    $0x0,%eax
L43:
  mov    %eax,%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x1(%rax),%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x2(%rax),%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x3(%rax),%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  add    $0x4,%eax
  cmp    %r8d,%eax
  jb     L43
L25:
  test   %r10d,%r10d
  je     L35
  add    %eax,%r10d
L21:
  mov    %eax,%r8d
  mov    (%rsi,%r8,4),%edx
  imul   (%r9,%r8,4),%edx
  add    (%rcx,%r8,4),%edx
  mov    %edx,(%rdi,%r8,4)
  add    $0x1,%eax
  cmp    %eax,%r10d
  jne    L21
L35:
  ret
L4:
  mov    %r8d,%eax
  jmp    L25
```

assistant: ```c
void vector_madd(int *res, int *a, int *b, int *c, unsigned int length) {
  unsigned int i, j;
  unsigned int num_blocks = length / 4;
  unsigned int rem = length % 4;
  for (i = 0; i < num_blocks * 4; i += 4) {
    res[i + 0] = (a[i + 0] * b[i + 0] + c[i + 0]);
    res[i + 1] = (a[i + 1] * b[i + 1] + c[i + 1]);
    res[i + 2] = (a[i + 2] * b[i + 2] + c[i + 2]);
    res[i + 3] = (a[i + 3] * b[i + 3] + c[i + 3]);
  }
  for (j = 0; j < rem; j++)
    res[i + j] = (a[i + j] * b[i + j] + c[i + j]);
}
```
user: What is the c source code of the assembly code below:

```asm
<vector_madd>:
  endbr64
  mov    %r8d,%r10d
  mov    %rdx,%r9
  xor    %eax,%eax
  and    $0x3,%r10d
  and    $0xfffffffc,%r8d
  je     L31
  cs nopw 0x0(%rax,%rax,1)
L28:
  mov    %eax,%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x1(%rax),%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x2(%rax),%r11d
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  lea    0x3(%rax),%r11d
  add    $0x4,%eax
  mov    (%rsi,%r11,4),%edx
  imul   (%r9,%r11,4),%edx
  add    (%rcx,%r11,4),%edx
  mov    %edx,(%rdi,%r11,4)
  cmp    %r8d,%eax
  jb     L28
L31:
  test   %r10d,%r10d
  je     L5
  add    %eax,%r10d
L37:
  mov    %eax,%r8d
  add    $0x1,%eax
  mov    (%rsi,%r8,4),%edx
  imul   (%r9,%r8,4),%edx
  add    (%rcx,%r8,4),%edx
  mov    %edx,(%rdi,%r8,4)
  cmp    %eax,%r10d
  jne    L37
L5:
  ret
```

assistant: ```c
void vector_madd(int *res, int *a, int *b, int *c, unsigned int length) {
  unsigned int i, j;
  unsigned int num_blocks = length / 4;
  unsigned int rem = length % 4;
  for (i = 0; i < num_blocks * 4; i += 4) {
    res[i + 0] = (a[i + 0] * b[i + 0] + c[i + 0]);
    res[i + 1] = (a[i + 1] * b[i + 1] + c[i + 1]);
    res[i + 2] = (a[i + 2] * b[i + 2] + c[i + 2]);
    res[i + 3] = (a[i + 3] * b[i + 3] + c[i + 3]);
  }
  for (j = 0; j < rem; j++)
    res[i + j] = (a[i + j] * b[i + j] + c[i + j]);
}
```