ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sendHeaders>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x140,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x138(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L38
  mov    %rsp,%rbp
  mov    %rsi,%rcx
  lea    D54(%rip),%rdx
  xor    %eax,%eax
  mov    $0xff,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
L42:
  mov    $0x10,%edx
  test   %ebx,%ebx
  je     L30
  lea    D50(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
L29:
  mov    $0xb,%edx
  lea    D18(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  mov    $0xa,%edx
  lea    D29(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  mov    $0x4,%edx
  lea    D12(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <send@plt>
  mov    0x138(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x140,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  lea    D66(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  jmp    L29
  nopl   0x0(%rax)
L38:
  mov    %rsp,%rbp
  lea    D37(%rip),%rdx
  mov    $0xff,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
  jmp    L42
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D54
type: string
name: unpack_data
arguments: label: D50
type: string
name: unpack_data
arguments: label: D18
type: string
name: unpack_data
arguments: label: D29
type: string
name: unpack_data
arguments: label: D12
type: string
name: unpack_data
arguments: label: D66
type: string
name: unpack_data
arguments: label: D37
type: string

tool: D54: GET %s HTTP/1.0


tool: D50: Icy-MetaData:1


tool: D18: User-Agent:

tool: D29: cPige/1.5

tool: D12: 



tool: D66: Icy-MetaData:0


tool: D37: GET / HTTP/1.0


assistant: ```c
int sendHeaders(int serversocket, char *mountpoint, int metadata) {
  int ret = 0;
  char headers[256 + sizeof("cPige/1.5") + 16 + 11 + 4];
  if (mountpoint == NULL)
    snprintf(headers, 255, "GET / HTTP/1.0\r\n");
  else
    snprintf(headers, 255, "GET %s HTTP/1.0\r\n", mountpoint);
  if (metadata != 0)
    strncat(headers, "Icy-MetaData:1\r\n", 16);
  else
    strncat(headers, "Icy-MetaData:0\r\n", 16);
  strncat(headers, "User-Agent:", 11);
  strncat(headers, "cPige/1.5", sizeof("cPige/1.5"));
  strncat(headers, "\r\n\r\n", 4);
  ret = send(serversocket, headers, strlen(headers), 0);
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<clock_gettime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    %rax,%rsi
  mov    $0x2a,%edi
  call   <_syscall5@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L9
  mov    -0x4(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  mov    D3(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L10
L9:
  mov    -0x4(%rbp),%eax
L10:
  leave
  ret
```

assistant: ```c
int clock_gettime(clockid_t clock_id, struct timespec *tp) {
  int ret = (int)_syscall5(42, (void *)clock_id, (void *)tp, ((void *)0), ((void *)0), ((void *)0));
  if (ret < 0) {
    errno = -ret;
    return -1;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<clock_gettime>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    %rdi,%rsi
  mov    $0x2a,%edi
  call   <_syscall5@plt>
  mov    %eax,%edx
  test   %eax,%eax
  js     L2
L12:
  mov    %edx,%eax
  add    $0x8,%rsp
  ret
L2:
  neg    %eax
  mov    D8(%rip),%rdx
  mov    %eax,(%rdx)
  mov    $0xffffffff,%edx
  jmp    L12
```

assistant: ```c
int clock_gettime(clockid_t clock_id, struct timespec *tp) {
  int ret = (int)_syscall5(42, (void *)clock_id, (void *)tp, ((void *)0), ((void *)0), ((void *)0));
  if (ret < 0) {
    errno = -ret;
    return -1;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<clock_gettime>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    %rdi,%rsi
  mov    $0x2a,%edi
  call   <_syscall5@plt>
  test   %eax,%eax
  js     L4
  mov    %eax,%r8d
L7:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
L4:
  mov    D9(%rip),%rdx
  neg    %eax
  mov    $0xffffffff,%r8d
  mov    %eax,(%rdx)
  jmp    L7
```

assistant: ```c
int clock_gettime(clockid_t clock_id, struct timespec *tp) {
  int ret = (int)_syscall5(42, (void *)clock_id, (void *)tp, ((void *)0), ((void *)0), ((void *)0));
  if (ret < 0) {
    errno = -ret;
    return -1;
  }
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clock_gettime>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    %rdi,%rsi
  mov    $0x2a,%edi
  call   <_syscall5@plt>
  test   %eax,%eax
  js     L0
  mov    %eax,%r8d
L10:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
L0:
  mov    D4(%rip),%rdx
  neg    %eax
  mov    $0xffffffff,%r8d
  mov    %eax,(%rdx)
  jmp    L10
```

assistant: ```c
int clock_gettime(clockid_t clock_id, struct timespec *tp) {
  int ret = (int)_syscall5(42, (void *)clock_id, (void *)tp, ((void *)0), ((void *)0), ((void *)0));
  if (ret < 0) {
    errno = -ret;
    return -1;
  }
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<be32getv>:
  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
  shl    $0x18,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  shl    $0x10,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  or     %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t be32getv(void *p) {
  uint8_t *a;
  uint32_t v;
  a = p;
  v = a[0] << 24;
  v |= a[1] << 16;
  v |= a[2] << 8;
  v |= a[3];
  return v;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<be32getv>:
  endbr64
  mov    0x4(%rdi),%edx
  shl    $0x10,%edx
  mov    (%rdi),%eax
  shl    $0x18,%eax
  or     %eax,%edx
  or     0xc(%rdi),%edx
  mov    0x8(%rdi),%eax
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t be32getv(void *p) {
  uint8_t *a;
  uint32_t v;
  a = p;
  v = a[0] << 24;
  v |= a[1] << 16;
  v |= a[2] << 8;
  v |= a[3];
  return v;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<be32getv>:
  endbr64
  mov    0x4(%rdi),%edx
  mov    (%rdi),%eax
  shl    $0x10,%edx
  shl    $0x18,%eax
  or     %eax,%edx
  mov    0x8(%rdi),%eax
  or     0xc(%rdi),%edx
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t be32getv(void *p) {
  uint8_t *a;
  uint32_t v;
  a = p;
  v = a[0] << 24;
  v |= a[1] << 16;
  v |= a[2] << 8;
  v |= a[3];
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<be32getv>:
  endbr64
  mov    0x4(%rdi),%edx
  mov    (%rdi),%eax
  shl    $0x10,%edx
  shl    $0x18,%eax
  or     %eax,%edx
  mov    0x8(%rdi),%eax
  or     0xc(%rdi),%edx
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t be32getv(void *p) {
  uint8_t *a;
  uint32_t v;
  a = p;
  v = a[0] << 24;
  v |= a[1] << 16;
  v |= a[2] << 8;
  v |= a[3];
  return v;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strupr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L17
L16:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  addq   $0x1,-0x20(%rbp)
L17:
  mov    -0x20(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L16
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strupr(char *str) {
  size_t len;
  size_t i;
  assert(str != NULL);
  len = strlen(str);
  for (i = 0; i < len; ++i)
    str[i] = toupper(str[i]);
  return str;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strupr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  je     L12
  mov    %r12,%rbx
  lea    (%r12,%rax,1),%rbp
L16:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L16
L12:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strupr(char *str) {
  size_t len;
  size_t i;
  assert(str != NULL);
  len = strlen(str);
  for (i = 0; i < len; ++i)
    str[i] = toupper(str[i]);
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strupr>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  mov    %rdi,%r12
  push   %rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  je     L13
  mov    %r12,%rbx
  lea    (%r12,%rax,1),%rbp
  nop
L12:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L12
L13:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strupr(char *str) {
  size_t len;
  size_t i;
  assert(str != NULL);
  len = strlen(str);
  for (i = 0; i < len; ++i)
    str[i] = toupper(str[i]);
  return str;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strupr>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  mov    %rdi,%r12
  push   %rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  je     L1
  mov    %r12,%rbx
  lea    (%rax,%r12,1),%rbp
  nop
L0:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L0
L1:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strupr(char *str) {
  size_t len;
  size_t i;
  assert(str != NULL);
  len = strlen(str);
  for (i = 0; i < len; ++i)
    str[i] = toupper(str[i]);
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<gcd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L3
  mov    $0x1,%eax
  jmp    L11
L3:
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <gcd>
L11:
  leave
  ret
```

assistant: ```c
int gcd(int x, int y) {
  if (y == 0)
    return 1;
  else
    return gcd(y, x % y);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gcd>:
  endbr64
  mov    $0x1,%eax
  test   %esi,%esi
  jne    L4
  ret
L4:
  sub    $0x8,%rsp
  mov    %edi,%edx
  mov    %esi,%edi
  mov    %edx,%eax
  cltd
  idiv   %esi
  mov    %edx,%esi
  call   <gcd>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gcd(int x, int y) {
  if (y == 0)
    return 1;
  else
    return gcd(y, x % y);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gcd>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  jne    L10
  jmp    L18
  nopl   0x0(%rax)
L8:
  mov    %edx,%esi
L10:
  cltd
  idiv   %esi
  mov    %esi,%eax
  test   %edx,%edx
  jne    L8
L18:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int gcd(int x, int y) {
  if (y == 0)
    return 1;
  else
    return gcd(y, x % y);
}
```
user: What is the c source code of the assembly code below:

```asm
<gcd>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  jne    L27
  jmp    L25
  nopl   0x0(%rax)
L23:
  mov    %edx,%esi
L27:
  cltd
  idiv   %esi
  mov    %esi,%eax
  test   %edx,%edx
  jne    L23
L25:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int gcd(int x, int y) {
  if (y == 0)
    return 1;
  else
    return gcd(y, x % y);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<pclin_crosscorr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x28(%rbp)
  movss  %xmm0,-0x2c(%rbp)
  mov    %esi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x44(%rbp)
  mov    %r9,-0x50(%rbp)
  mov    -0x44(%rbp),%edx
  mov    -0x40(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <pclin_zerodc@plt>
  mov    0x18(%rbp),%edx
  mov    -0x50(%rbp),%rax
  mov    0x10(%rbp),%rsi
  mov    %rax,%rdi
  call   <pclin_zerodc@plt>
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L5
L17:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x14(%rbp),%edi
  mov    0x18(%rbp),%r8d
  mov    -0x50(%rbp),%rcx
  mov    -0x44(%rbp),%edx
  mov    -0x40(%rbp),%rsi
  mov    -0x38(%rbp),%rax
  movd   %edi,%xmm0
  mov    %r8d,%r9d
  mov    0x10(%rbp),%r8
  mov    %rax,%rdi
  call   <pclin_innerproduct@plt>
  movd   %xmm0,%eax
  mov    %eax,(%rbx)
  movss  -0x14(%rbp),%xmm0
  addss  -0x2c(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
  addl   $0x1,-0x18(%rbp)
L5:
  mov    -0x18(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L17
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void pclin_crosscorr(float *out, float step, int n, float *pdx1, float *pf1, int n1, float *pdx2, float *pf2, int n2) {
  int i;
  float x;
  pclin_zerodc(pdx1, pf1, n1);
  pclin_zerodc(pdx2, pf2, n2);
  x = 0.0;
  for (i = 0; i < n; i++) {
    out[i] = pclin_innerproduct(pdx1, pf1, n1, pdx2, pf2, n2, x);
    x += step;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<pclin_crosscorr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  movss  %xmm0,0xc(%rsp)
  mov    %esi,%ebx
  mov    %rdx,%rbp
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %r9,%r14
  mov    %r8d,%edx
  mov    %rcx,%rsi
  mov    %rbp,%rdi
  call   <pclin_zerodc@plt>
  mov    0x58(%rsp),%edx
  mov    0x50(%rsp),%rsi
  mov    %r14,%rdi
  call   <pclin_zerodc@plt>
  test   %ebx,%ebx
  jle    L4
  mov    %ebx,%eax
  mov    %r15,%rbx
  lea    -0x1(%rax),%eax
  lea    0x4(%r15,%rax,4),%r15
  pxor   %xmm3,%xmm3
  movss  %xmm3,0x8(%rsp)
L17:
  movss  0x8(%rsp),%xmm0
  mov    0x58(%rsp),%r9d
  mov    0x50(%rsp),%r8
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <pclin_innerproduct@plt>
  movss  %xmm0,(%rbx)
  movss  0x8(%rsp),%xmm1
  addss  0xc(%rsp),%xmm1
  movss  %xmm1,0x8(%rsp)
  add    $0x4,%rbx
  cmp    %r15,%rbx
  jne    L17
L4:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void pclin_crosscorr(float *out, float step, int n, float *pdx1, float *pf1, int n1, float *pdx2, float *pf2, int n2) {
  int i;
  float x;
  pclin_zerodc(pdx1, pf1, n1);
  pclin_zerodc(pdx2, pf2, n2);
  x = 0.0;
  for (i = 0; i < n; i++) {
    out[i] = pclin_innerproduct(pdx1, pf1, n1, pdx2, pf2, n2, x);
    x += step;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<pclin_crosscorr>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  mov    %r8d,%edx
  mov    %rbx,%rdi
  sub    $0x18,%rsp
  mov    %esi,(%rsp)
  mov    0x58(%rsp),%r15d
  mov    %rcx,%rsi
  movss  %xmm0,0x4(%rsp)
  call   <pclin_zerodc@plt>
  mov    0x50(%rsp),%rsi
  mov    %r15d,%edx
  mov    %r13,%rdi
  call   <pclin_zerodc@plt>
  mov    (%rsp),%r10d
  test   %r10d,%r10d
  jle    L8
  lea    -0x1(%r10),%edx
  pxor   %xmm1,%xmm1
  lea    0x4(%r14,%rdx,4),%rax
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L3:
  mov    0x50(%rsp),%r8
  movaps %xmm1,%xmm0
  mov    %r15d,%r9d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  movss  %xmm1,(%rsp)
  call   <pclin_innerproduct@plt>
  add    $0x4,%r14
  movss  (%rsp),%xmm1
  addss  0x4(%rsp),%xmm1
  movss  %xmm0,-0x4(%r14)
  cmp    0x8(%rsp),%r14
  jne    L3
L8:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void pclin_crosscorr(float *out, float step, int n, float *pdx1, float *pf1, int n1, float *pdx2, float *pf2, int n2) {
  int i;
  float x;
  pclin_zerodc(pdx1, pf1, n1);
  pclin_zerodc(pdx2, pf2, n2);
  x = 0.0;
  for (i = 0; i < n; i++) {
    out[i] = pclin_innerproduct(pdx1, pf1, n1, pdx2, pf2, n2, x);
    x += step;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<pclin_crosscorr>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  mov    %r8d,%edx
  mov    %rbx,%rdi
  sub    $0x18,%rsp
  mov    %esi,(%rsp)
  mov    0x58(%rsp),%r15d
  mov    %rcx,%rsi
  movss  %xmm0,0x4(%rsp)
  call   <pclin_zerodc@plt>
  mov    0x50(%rsp),%rsi
  mov    %r15d,%edx
  mov    %r13,%rdi
  call   <pclin_zerodc@plt>
  mov    (%rsp),%r10d
  test   %r10d,%r10d
  jle    L10
  lea    -0x1(%r10),%edx
  pxor   %xmm1,%xmm1
  lea    0x4(%r14,%rdx,4),%rax
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L14:
  mov    0x50(%rsp),%r8
  movaps %xmm1,%xmm0
  mov    %r15d,%r9d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  movss  %xmm1,(%rsp)
  call   <pclin_innerproduct@plt>
  add    $0x4,%r14
  movss  (%rsp),%xmm1
  addss  0x4(%rsp),%xmm1
  movss  %xmm0,-0x4(%r14)
  cmp    0x8(%rsp),%r14
  jne    L14
L10:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void pclin_crosscorr(float *out, float step, int n, float *pdx1, float *pf1, int n1, float *pdx2, float *pf2, int n2) {
  int i;
  float x;
  pclin_zerodc(pdx1, pf1, n1);
  pclin_zerodc(pdx2, pf2, n2);
  x = 0.0;
  for (i = 0; i < n; i++) {
    out[i] = pclin_innerproduct(pdx1, pf1, n1, pdx2, pf2, n2, x);
    x += step;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<eval_exponent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %al,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jns    L13
  mov    $0x0,%eax
  jmp    L57
L13:
  cmpq   $0x0,-0x18(%rbp)
  jne    L36
  mov    $0x1,%eax
  jmp    L57
L36:
  cmpq   $0x1,-0x18(%rbp)
  jne    L17
  mov    -0x10(%rbp),%rax
  jmp    L57
L17:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  movsbl -0x4(%rbp),%eax
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <eval_exponent>
  imul   -0x10(%rbp),%rax
L57:
  leave
  ret
```

assistant: ```c
long eval_exponent(char op, long a, long b) {
  if (b < 0)
    return 0;
  if (b == 0)
    return 1;
  if (b == 1)
    return a;
  return a * eval_exponent(op, a, b - 1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<eval_exponent>:
  endbr64
  mov    $0x0,%eax
  test   %rdx,%rdx
  js     L27
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x1,%eax
  je     L3
  mov    %rsi,%rax
  cmp    $0x1,%rdx
  je     L3
  sub    $0x1,%rdx
  movsbl %dil,%edi
  call   <eval_exponent>
  imul   %rbx,%rax
L3:
  pop    %rbx
  ret
L27:
  ret
```

assistant: ```c
long eval_exponent(char op, long a, long b) {
  if (b < 0)
    return 0;
  if (b == 0)
    return 1;
  if (b == 1)
    return a;
  return a * eval_exponent(op, a, b - 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<eval_exponent>:
  endbr64
  xor    %eax,%eax
  test   %rdx,%rdx
  js     L32
  mov    $0x1,%eax
  je     L32
  mov    %rsi,%rax
  cmp    $0x1,%rdx
  je     L32
  cmp    $0x2,%rdx
  je     L26
  sub    $0x18,%rsp
  sub    $0x2,%rdx
  movsbl %dil,%edi
  mov    %rsi,0x8(%rsp)
  call   <eval_exponent>
  mov    0x8(%rsp),%rsi
  add    $0x18,%rsp
  imul   %rsi,%rax
  imul   %rsi,%rax
  ret
  nopl   0x0(%rax)
L32:
  ret
  nopl   0x0(%rax)
L26:
  imul   %rsi,%rax
  ret
```

assistant: ```c
long eval_exponent(char op, long a, long b) {
  if (b < 0)
    return 0;
  if (b == 0)
    return 1;
  if (b == 1)
    return a;
  return a * eval_exponent(op, a, b - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<eval_exponent>:
  endbr64
  xor    %eax,%eax
  test   %rdx,%rdx
  js     L6
  mov    $0x1,%eax
  je     L6
  mov    %rsi,%rax
  cmp    $0x1,%rdx
  je     L6
  cmp    $0x2,%rdx
  je     L11
  sub    $0x18,%rsp
  sub    $0x2,%rdx
  movsbl %dil,%edi
  mov    %rsi,0x8(%rsp)
  call   <eval_exponent>
  mov    0x8(%rsp),%rsi
  add    $0x18,%rsp
  imul   %rsi,%rax
  imul   %rsi,%rax
  ret
  nopl   0x0(%rax)
L6:
  ret
  nopl   0x0(%rax)
L11:
  imul   %rsi,%rax
  ret
```

assistant: ```c
long eval_exponent(char op, long a, long b) {
  if (b < 0)
    return 0;
  if (b == 0)
    return 1;
  if (b == 1)
    return a;
  return a * eval_exponent(op, a, b - 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<match_regex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  movl   $0x1,-0x38(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  sub    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  mov    %rax,%r10
  mov    $0x0,%r11d
  mov    -0x38(%rbp),%eax
  cltq
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L42:
  cmp    %rdx,%rsp
  je     L33
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L42
L33:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L55
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L55:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%edx
  mov    -0x30(%rbp),%rsi
  mov    -0x48(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rdi
  call   <regexec@plt>
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  je     L13
  mov    -0x34(%rbp),%eax
  jmp    L23
L13:
  mov    $0x0,%eax
L23:
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int match_regex(regex_t *r, const char *to_match) {
  const char *p = to_match;
  const int n_matches = 1;
  regmatch_t m[n_matches];
  int nomatch = regexec(r, p, n_matches, m, 0);
  if (nomatch) {
    return nomatch;
  } else {
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<match_regex>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  mov    $0x0,%r8d
  mov    $0x1,%edx
  call   <regexec@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_regex(regex_t *r, const char *to_match) {
  const char *p = to_match;
  const int n_matches = 1;
  regmatch_t m[n_matches];
  int nomatch = regexec(r, p, n_matches, m, 0);
  if (nomatch) {
    return nomatch;
  } else {
    return 0;
  }
}
```