ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ecc200placementcornerA>:
  endbr64
  push   %r14
  lea    -0x1(%rsi),%r14d
  mov    %ecx,%r9d
  xor    %r8d,%r8d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %ecx,%ebx
  mov    %r14d,%ecx
  sub    $0x8,%rsp
  push   $0x7
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x1,%r8d
  movl   $0x6,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  lea    -0x1(%r12),%r14d
  mov    $0x2,%r8d
  movl   $0x5,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  lea    -0x2(%r12),%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ecx,%ecx
  movl   $0x4,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ecx,%ecx
  movl   $0x3,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x1,%ecx
  movl   $0x2,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x2,%ecx
  movl   $0x1,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %r12d,%edx
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x3,%ecx
  movl   $0x0,(%rsp)
  call   <ecc200placementbit@plt>
  pop    %rax
  pop    %rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void ecc200placementcornerA(int *array, const int NR, const int NC, const int p) {
  ecc200placementbit(array, NR, NC, NR - 1, 0, p, 7);
  ecc200placementbit(array, NR, NC, NR - 1, 1, p, 6);
  ecc200placementbit(array, NR, NC, NR - 1, 2, p, 5);
  ecc200placementbit(array, NR, NC, 0, NC - 2, p, 4);
  ecc200placementbit(array, NR, NC, 0, NC - 1, p, 3);
  ecc200placementbit(array, NR, NC, 1, NC - 1, p, 2);
  ecc200placementbit(array, NR, NC, 2, NC - 1, p, 1);
  ecc200placementbit(array, NR, NC, 3, NC - 1, p, 0);
}
```
user: What is the c source code of the assembly code below?

```asm
<ecc200placementcornerA>:
  endbr64
  push   %r14
  lea    -0x1(%rsi),%r14d
  mov    %ecx,%r9d
  xor    %r8d,%r8d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %ecx,%ebx
  mov    %r14d,%ecx
  sub    $0x8,%rsp
  push   $0x7
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x1,%r8d
  movl   $0x6,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  lea    -0x1(%r12),%r14d
  mov    $0x2,%r8d
  movl   $0x5,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  lea    -0x2(%r12),%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ecx,%ecx
  movl   $0x4,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ecx,%ecx
  movl   $0x3,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x1,%ecx
  movl   $0x2,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x2,%ecx
  movl   $0x1,(%rsp)
  call   <ecc200placementbit@plt>
  mov    %r12d,%edx
  mov    %ebx,%r9d
  mov    %r14d,%r8d
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x3,%ecx
  movl   $0x0,(%rsp)
  call   <ecc200placementbit@plt>
  pop    %rax
  pop    %rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void ecc200placementcornerA(int *array, const int NR, const int NC, const int p) {
  ecc200placementbit(array, NR, NC, NR - 1, 0, p, 7);
  ecc200placementbit(array, NR, NC, NR - 1, 1, p, 6);
  ecc200placementbit(array, NR, NC, NR - 1, 2, p, 5);
  ecc200placementbit(array, NR, NC, 0, NC - 2, p, 4);
  ecc200placementbit(array, NR, NC, 0, NC - 1, p, 3);
  ecc200placementbit(array, NR, NC, 1, NC - 1, p, 2);
  ecc200placementbit(array, NR, NC, 2, NC - 1, p, 1);
  ecc200placementbit(array, NR, NC, 3, NC - 1, p, 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_unaligned_uint32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int get_unaligned_uint32(void *p) {
  struct {
    unsigned int d;
  } __attribute__((packed)) *pp = (void *)p;
  return pp->d;
}
```
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_unaligned_uint32>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
unsigned int get_unaligned_uint32(void *p) {
  struct {
    unsigned int d;
  } __attribute__((packed)) *pp = (void *)p;
  return pp->d;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_unaligned_uint32>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
unsigned int get_unaligned_uint32(void *p) {
  struct {
    unsigned int d;
  } __attribute__((packed)) *pp = (void *)p;
  return pp->d;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<is_head>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L43
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L43
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L43
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L43
  mov    $0x1,%eax
  jmp    L21
L43:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int is_head(unsigned char *b) {
  if (b[0] == '*' && b[1] == '*' && b[2] == '*' && b[3] == '*')
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<is_head>:
  endbr64
  mov    $0x0,%eax
  cmpb   $0x2a,(%rdi)
  je     L38
L7:
  ret
L38:
  cmpb   $0x2a,0x1(%rdi)
  jne    L7
  cmpb   $0x2a,0x2(%rdi)
  jne    L7
  cmpb   $0x2a,0x3(%rdi)
  sete   %al
  movzbl %al,%eax
  jmp    L7
```

assistant: ```c
int is_head(unsigned char *b) {
  if (b[0] == '*' && b[1] == '*' && b[2] == '*' && b[3] == '*')
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_head>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x2a,(%rdi)
  je     L20
L1:
  ret
  nopl   0x0(%rax)
L20:
  cmpb   $0x2a,0x1(%rdi)
  jne    L1
  cmpb   $0x2a,0x2(%rdi)
  jne    L1
  xor    %eax,%eax
  cmpb   $0x2a,0x3(%rdi)
  sete   %al
  ret
```

assistant: ```c
int is_head(unsigned char *b) {
  if (b[0] == '*' && b[1] == '*' && b[2] == '*' && b[3] == '*')
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<is_head>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x2a,(%rdi)
  je     L15
L19:
  ret
  nopl   0x0(%rax)
L15:
  cmpb   $0x2a,0x1(%rdi)
  jne    L19
  cmpb   $0x2a,0x2(%rdi)
  jne    L19
  xor    %eax,%eax
  cmpb   $0x2a,0x3(%rdi)
  sete   %al
  ret
```

assistant: ```c
int is_head(unsigned char *b) {
  if (b[0] == '*' && b[1] == '*' && b[2] == '*' && b[3] == '*')
    return 1;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<practice2D11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x55,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x3,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  sar    $0x2,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
datatype: string

tool: D5: 0x%x


assistant: ```c
void practice2D11() {
  int x = 0x55;
  printf("0x%x\n", x << 3);
  printf("0x%x\n", x >> 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<practice2D11>:
  endbr64
  push   %rbx
  mov    $0x2a8,%esi
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x15,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
type: string

tool: D3: 0x%x


assistant: ```c
void practice2D11() {
  int x = 0x55;
  printf("0x%x\n", x << 3);
  printf("0x%x\n", x >> 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<practice2D11>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  mov    $0x2a8,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x15,%esi
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type_guessed: string

tool: D6: 0x%x


assistant: ```c
void practice2D11() {
  int x = 0x55;
  printf("0x%x\n", x << 3);
  printf("0x%x\n", x >> 2);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<practice2D11>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  mov    $0x2a8,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x15,%esi
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
type_infered: string

tool: D5: 0x%x


assistant: ```c
void practice2D11() {
  int x = 0x55;
  printf("0x%x\n", x << 3);
  printf("0x%x\n", x >> 2);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<next>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,%eax
  mov    %al,-0x3c(%rbp)
  cmpb   $0x1,-0x3c(%rbp)
  jg     L31
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  movb   $0x0,-0x21(%rbp)
  jmp    L25
L15:
  movsbq -0x21(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsbq -0x21(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  movzbl -0x21(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x21(%rbp)
L25:
  movsbq -0x21(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  sub    $0x1,%rdx
  cmp    %rdx,%rax
  jb     L15
  mov    -0x18(%rbp),%rax
  jmp    L38
L31:
  movzbl -0x3c(%rbp),%eax
  sub    $0x1,%eax
  movsbl %al,%ebx
  mov    -0x38(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <next>
  mov    %ebx,%esi
  mov    %rax,%rdi
  call   <next>
L38:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *next(const char *original, char times) {
  if (times <= 1) {
    char i, *transformed;
    size_t len;
    len = strlen(original);
    transformed = (char *)malloc(len + 1);
    transformed[len] = '\0';
    transformed[len - 1] = original[0];
    for (i = 0; i < len - 1; i += 1) {
      transformed[i] = original[i + 1];
    }
    return transformed;
  }
  return next(next(original, 1), times - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<next>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  cmp    $0x1,%sil
  jle    L36
  mov    %esi,%ebx
  mov    $0x1,%esi
  call   <next>
  mov    %rax,%rdi
  lea    -0x1(%rbx),%esi
  movsbl %sil,%esi
  call   <next>
L27:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L36:
  call   <strlen@plt>
  mov    %rax,%rbx
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbx,1)
  movzbl 0x0(%rbp),%edx
  mov    %dl,-0x1(%rax,%rbx,1)
  sub    $0x1,%rbx
  mov    %rbx,%rsi
  je     L27
  mov    $0x0,%ecx
  mov    $0x0,%edx
L25:
  movzbl 0x1(%rbp,%rdx,1),%edi
  mov    %dil,(%rax,%rdx,1)
  add    $0x1,%ecx
  movsbq %cl,%rdx
  cmp    %rdx,%rsi
  ja     L25
  jmp    L27
```

assistant: ```c
char *next(const char *original, char times) {
  if (times <= 1) {
    char i, *transformed;
    size_t len;
    len = strlen(original);
    transformed = (char *)malloc(len + 1);
    transformed[len] = '\0';
    transformed[len - 1] = original[0];
    for (i = 0; i < len - 1; i += 1) {
      transformed[i] = original[i + 1];
    }
    return transformed;
  }
  return next(next(original, 1), times - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<next>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%sil
  jle    L17
  mov    %esi,%ebx
  mov    $0x1,%esi
  call   <next>
  lea    -0x1(%rbx),%esi
  add    $0x8,%rsp
  pop    %rbx
  mov    %rax,%rdi
  movzbl %sil,%esi
  pop    %rbp
  jmp    1140 <next>
  nopl   0x0(%rax)
L17:
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbx,1)
  movzbl 0x0(%rbp),%edx
  mov    %dl,-0x1(%rax,%rbx,1)
  sub    $0x1,%rbx
  mov    %rbx,%rsi
  je     L16
  xor    %ecx,%ecx
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L14:
  movzbl 0x1(%rbp,%rdx,1),%edi
  add    $0x1,%ecx
  mov    %dil,(%rax,%rdx,1)
  movsbq %cl,%rdx
  cmp    %rdx,%rsi
  ja     L14
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *next(const char *original, char times) {
  if (times <= 1) {
    char i, *transformed;
    size_t len;
    len = strlen(original);
    transformed = (char *)malloc(len + 1);
    transformed[len] = '\0';
    transformed[len - 1] = original[0];
    for (i = 0; i < len - 1; i += 1) {
      transformed[i] = original[i + 1];
    }
    return transformed;
  }
  return next(next(original, 1), times - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<IACK>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%eax
  call   <readTCR@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <writeTCR@plt>
  mov    $0xbfffffc1,%eax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <writeTCR@plt>
  nop
  leave
  ret
```

assistant: ```c
void IACK(unsigned long ipl) {
  unsigned long tcr;
  char dummy;
  tcr = readTCR();
  writeTCR(ipl);
  dummy = *((char *)0xbfffffc1);
  writeTCR(tcr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<IACK>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%eax
  call   <readTCR@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <writeTCR@plt>
  mov    %rbx,%rdi
  call   <writeTCR@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void IACK(unsigned long ipl) {
  unsigned long tcr;
  char dummy;
  tcr = readTCR();
  writeTCR(ipl);
  dummy = *((char *)0xbfffffc1);
  writeTCR(tcr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<IACK>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <readTCR@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <writeTCR@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <writeTCR@plt>
```

assistant: ```c
void IACK(unsigned long ipl) {
  unsigned long tcr;
  char dummy;
  tcr = readTCR();
  writeTCR(ipl);
  dummy = *((char *)0xbfffffc1);
  writeTCR(tcr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<IACK>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <readTCR@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <writeTCR@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <writeTCR@plt>
```

assistant: ```c
void IACK(unsigned long ipl) {
  unsigned long tcr;
  char dummy;
  tcr = readTCR();
  writeTCR(ipl);
  dummy = *((char *)0xbfffffc1);
  writeTCR(tcr);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<assign_text>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L10
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L5
L10:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void assign_text(char **dest, const char *src) {
  if (src == NULL)
    *dest = NULL;
  else {
    *dest = malloc(strlen(src) + 1);
    strcpy(*dest, src);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<assign_text>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L8
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(%rbp)
  mov    %rbx,%rsi
  call   <strcpy@plt>
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  movq   $0x0,(%rdi)
  jmp    L2
```

assistant: ```c
void assign_text(char **dest, const char *src) {
  if (src == NULL)
    *dest = NULL;
  else {
    *dest = malloc(strlen(src) + 1);
    strcpy(*dest, src);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<assign_text>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L6
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,(%rbx)
  add    $0x8,%rsp
  mov    %rax,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <strcpy@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  movq   $0x0,(%rdi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void assign_text(char **dest, const char *src) {
  if (src == NULL)
    *dest = NULL;
  else {
    *dest = malloc(strlen(src) + 1);
    strcpy(*dest, src);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<assign_text>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L7
  mov    %rsi,%rdi
  mov    %rsi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,(%rbx)
  add    $0x8,%rsp
  mov    %rax,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1070 <strcpy@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  movq   $0x0,(%rdi)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void assign_text(char **dest, const char *src) {
  if (src == NULL)
    *dest = NULL;
  else {
    *dest = malloc(strlen(src) + 1);
    strcpy(*dest, src);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Quaternion_Scalar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  movss  %xmm0,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0x14(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mulss  -0x14(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mulss  -0x14(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  mulss  -0x14(%rbp),%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Quaternion_Scalar(float *r, float *q, float scalar) {
  r[0] = q[0] * scalar;
  r[1] = q[1] * scalar;
  r[2] = q[2] * scalar;
  r[3] = q[3] * scalar;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Quaternion_Scalar>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rsi),%xmm1
  movss  %xmm1,(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x4(%rsi),%xmm1
  movss  %xmm1,0x4(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x8(%rsi),%xmm1
  movss  %xmm1,0x8(%rdi)
  mulss  0xc(%rsi),%xmm0
  movss  %xmm0,0xc(%rdi)
  ret
```

assistant: ```c
void Quaternion_Scalar(float *r, float *q, float scalar) {
  r[0] = q[0] * scalar;
  r[1] = q[1] * scalar;
  r[2] = q[2] * scalar;
  r[3] = q[3] * scalar;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Quaternion_Scalar>:
  endbr64
  movss  (%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi)
  movss  0x4(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,0x4(%rdi)
  movss  0x8(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,0x8(%rdi)
  mulss  0xc(%rsi),%xmm0
  movss  %xmm0,0xc(%rdi)
  ret
```

assistant: ```c
void Quaternion_Scalar(float *r, float *q, float scalar) {
  r[0] = q[0] * scalar;
  r[1] = q[1] * scalar;
  r[2] = q[2] * scalar;
  r[3] = q[3] * scalar;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Quaternion_Scalar>:
  endbr64
  movss  (%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi)
  movss  0x4(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,0x4(%rdi)
  movss  0x8(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,0x8(%rdi)
  mulss  0xc(%rsi),%xmm0
  movss  %xmm0,0xc(%rdi)
  ret
```

assistant: ```c
void Quaternion_Scalar(float *r, float *q, float scalar) {
  r[0] = q[0] * scalar;
  r[1] = q[1] * scalar;
  r[2] = q[2] * scalar;
  r[3] = q[3] * scalar;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<str_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L1:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L1
  mov    -0xc(%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
  leave
  ret
```

assistant: ```c
int str_to_int(char *str) {
  int len = strlen(str);
  int result = 0;
  int i;
  for (i = 0; i < len; i++) {
    result += str[i] - '0';
    result *= 10;
  }
  return result / 10;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<str_to_int>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L10
  mov    %rbx,%rcx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rsi
  mov    $0x0,%edx
L23:
  movsbl (%rcx),%eax
  lea    -0x30(%rdx,%rax,1),%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L23
L3:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  pop    %rbx
  ret
L10:
  mov    $0x0,%edx
  jmp    L3
```

assistant: ```c
int str_to_int(char *str) {
  int len = strlen(str);
  int result = 0;
  int i;
  for (i = 0; i < len; i++) {
    result += str[i] - '0';
    result *= 10;
  }
  return result / 10;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<str_to_int>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L14
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rcx
  xor    %eax,%eax
  xchg   %ax,%ax
L8:
  movsbl (%rdi),%edx
  add    $0x1,%rdi
  lea    -0x30(%rax,%rdx,1),%r8d
  lea    (%r8,%r8,4),%eax
  add    %eax,%eax
  cmp    %rdi,%rcx
  jne    L8
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L14:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int str_to_int(char *str) {
  int len = strlen(str);
  int result = 0;
  int i;
  for (i = 0; i < len; i++) {
    result += str[i] - '0';
    result *= 10;
  }
  return result / 10;
}
```