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

```asm
<write_be32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <write_be16@plt>
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    $0x4,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <write_at_be16@plt>
  nop
  leave
  ret
```

assistant: ```c
void write_be32(void *dest, uint32_t val) {
  write_be16(dest, val >> 16);
  write_at_be16(dest, val >> 0, sizeof(uint16_t));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<write_be32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  sar    $0x10,%esi
  call   <write_be16@plt>
  mov    $0x4,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <write_at_be16@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void write_be32(void *dest, uint32_t val) {
  write_be16(dest, val >> 16);
  write_at_be16(dest, val >> 0, sizeof(uint16_t));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<write_be32>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  sar    $0x10,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <write_be16@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    $0x4,%edx
  pop    %rbp
  pop    %r12
  jmp    1060 <write_at_be16@plt>
```

assistant: ```c
void write_be32(void *dest, uint32_t val) {
  write_be16(dest, val >> 16);
  write_at_be16(dest, val >> 0, sizeof(uint16_t));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_be32>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  sar    $0x10,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <write_be16@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    $0x4,%edx
  pop    %rbp
  pop    %r12
  jmp    1060 <write_at_be16@plt>
```

assistant: ```c
void write_be32(void *dest, uint32_t val) {
  write_be16(dest, val >> 16);
  write_at_be16(dest, val >> 0, sizeof(uint16_t));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<oct_to_dec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L7
L2:
  mov    -0x14(%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
  mov    %edx,%eax
  imul   -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  shll   $0x3,-0x4(%rbp)
  mov    -0x14(%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,-0x14(%rbp)
L7:
  cmpl   $0x0,-0x14(%rbp)
  jne    L2
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int oct_to_dec(int n) {
  int res = 0;
  int i = 1;
  while (n) {
    res = res + ((n % 10) * i);
    i = i * 8;
    n = n / 10;
  }
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<oct_to_dec>:
  endbr64
  test   %edi,%edi
  je     L10
  mov    $0x1,%ecx
  mov    $0x0,%esi
L27:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  imul   %ecx,%edi
  add    %edi,%esi
  shl    $0x3,%ecx
  mov    %eax,%edi
  test   %eax,%eax
  jne    L27
L17:
  mov    %esi,%eax
  ret
L10:
  mov    %edi,%esi
  jmp    L17
```

assistant: ```c
int oct_to_dec(int n) {
  int res = 0;
  int i = 1;
  while (n) {
    res = res + ((n % 10) * i);
    i = i * 8;
    n = n / 10;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<oct_to_dec>:
  endbr64
  test   %edi,%edi
  je     L16
  mov    $0x1,%ecx
  xor    %r8d,%r8d
L1:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  imul   %ecx,%edi
  shl    $0x3,%ecx
  add    %edi,%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L1
  mov    %r8d,%eax
  ret
  nop
L16:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int oct_to_dec(int n) {
  int res = 0;
  int i = 1;
  while (n) {
    res = res + ((n % 10) * i);
    i = i * 8;
    n = n / 10;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<oct_to_dec>:
  endbr64
  test   %edi,%edi
  je     L13
  mov    $0x1,%ecx
  xor    %r8d,%r8d
L19:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  imul   %ecx,%edi
  shl    $0x3,%ecx
  add    %edi,%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L19
  mov    %r8d,%eax
  ret
  nop
L13:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int oct_to_dec(int n) {
  int res = 0;
  int i = 1;
  while (n) {
    res = res + ((n % 10) * i);
    i = i * 8;
    n = n / 10;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<psi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  movss  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtss2sd -0x20(%rbp),%xmm0
  mulsd  %xmm0,%xmm2
  movss  -0x18(%rbp),%xmm0
  mulss  %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm2,%xmm0
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float psi(x, sigma)
float x, sigma;
{
  float result;
  result = (2.0 * x) / ((2.0 * sigma * sigma) + (x * x));
  return (result);
}
```
user: What is the c source code of the assembly code below:

```asm
<psi>:
  endbr64
  pxor   %xmm2,%xmm2
  cvtsd2ss %xmm0,%xmm2
  cvtsd2ss %xmm1,%xmm1
  cvtss2sd %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  addsd  %xmm0,%xmm0
  movapd %xmm1,%xmm3
  addsd  %xmm1,%xmm3
  mulsd  %xmm3,%xmm1
  mulss  %xmm2,%xmm2
  cvtss2sd %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float psi(x, sigma)
float x, sigma;
{
  float result;
  result = (2.0 * x) / ((2.0 * sigma * sigma) + (x * x));
  return (result);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<psi>:
  endbr64
  cvtsd2ss %xmm1,%xmm1
  cvtss2sd %xmm1,%xmm1
  movapd %xmm1,%xmm3
  pxor   %xmm2,%xmm2
  addsd  %xmm1,%xmm3
  cvtsd2ss %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  mulss  %xmm2,%xmm2
  addsd  %xmm0,%xmm0
  mulsd  %xmm3,%xmm1
  cvtss2sd %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float psi(x, sigma)
float x, sigma;
{
  float result;
  result = (2.0 * x) / ((2.0 * sigma * sigma) + (x * x));
  return (result);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<psi>:
  endbr64
  cvtsd2ss %xmm1,%xmm1
  cvtss2sd %xmm1,%xmm1
  movapd %xmm1,%xmm3
  pxor   %xmm2,%xmm2
  addsd  %xmm1,%xmm3
  cvtsd2ss %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  mulss  %xmm2,%xmm2
  addsd  %xmm0,%xmm0
  mulsd  %xmm3,%xmm1
  cvtss2sd %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

assistant: ```c
float psi(x, sigma)
float x, sigma;
{
  float result;
  result = (2.0 * x) / ((2.0 * sigma * sigma) + (x * x));
  return (result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bstrcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L3
L15:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L3:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *bstrcpy(char *dst, const char *src) {
  while (*src) {
    *dst++ = *src++;
  }
  *dst = 0;
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bstrcpy>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L6
L1:
  add    $0x1,%rsi
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L1
L6:
  movb   $0x0,(%rax)
  ret
```

assistant: ```c
char *bstrcpy(char *dst, const char *src) {
  while (*src) {
    *dst++ = *src++;
  }
  *dst = 0;
  return dst;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bstrcpy>:
  endbr64
  movzbl (%rsi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L8
  xchg   %ax,%ax
L1:
  add    $0x1,%rsi
  mov    %dl,(%rax)
  add    $0x1,%rax
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L1
L8:
  movb   $0x0,(%rax)
  ret
```

assistant: ```c
char *bstrcpy(char *dst, const char *src) {
  while (*src) {
    *dst++ = *src++;
  }
  *dst = 0;
  return dst;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bstrcpy>:
  endbr64
  movzbl (%rsi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L13
  xchg   %ax,%ax
L17:
  add    $0x1,%rsi
  mov    %dl,(%rax)
  add    $0x1,%rax
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L17
L13:
  movb   $0x0,(%rax)
  ret
```

assistant: ```c
char *bstrcpy(char *dst, const char *src) {
  while (*src) {
    *dst++ = *src++;
  }
  *dst = 0;
  return dst;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<stage_cu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0x81010401,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x4,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0xbf810000,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0xb,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x22,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x17,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x10,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x25,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x1f,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x29,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x16,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x20,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x1d,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x24,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void stage_cu() {
  Xil_Out32(0x50001004, 0);
  Xil_Out32(0x50001000, 0x81010401);
  Xil_Out32(0x50001004, 4);
  Xil_Out32(0x50001000, 0xBF810000);
  Xil_Out32(0x50002004, 0);
  Xil_Out32(0x50002008, 0xB);
  Xil_Out32(0x5000200C, 0x22);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x17);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 16);
  Xil_Out32(0x50002008, 0x25);
  Xil_Out32(0x5000200C, 0x1F);
  Xil_Out32(0x50002010, 0x29);
  Xil_Out32(0x50002014, 0x16);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 32);
  Xil_Out32(0x50002008, 0x1D);
  Xil_Out32(0x5000200C, 0x24);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x0);
  Xil_Out32(0x50002000, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<stage_cu>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0x81010401,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x4,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0xbf810000,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0xb,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x22,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x17,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x10,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x25,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x1f,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x29,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x16,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x20,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x1d,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x24,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x0,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void stage_cu() {
  Xil_Out32(0x50001004, 0);
  Xil_Out32(0x50001000, 0x81010401);
  Xil_Out32(0x50001004, 4);
  Xil_Out32(0x50001000, 0xBF810000);
  Xil_Out32(0x50002004, 0);
  Xil_Out32(0x50002008, 0xB);
  Xil_Out32(0x5000200C, 0x22);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x17);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 16);
  Xil_Out32(0x50002008, 0x25);
  Xil_Out32(0x5000200C, 0x1F);
  Xil_Out32(0x50002010, 0x29);
  Xil_Out32(0x50002014, 0x16);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 32);
  Xil_Out32(0x50002008, 0x1D);
  Xil_Out32(0x5000200C, 0x24);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x0);
  Xil_Out32(0x50002000, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<stage_cu>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0x81010401,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x4,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0xbf810000,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0xb,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x22,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x17,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x10,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x25,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x1f,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x29,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x16,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x20,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x1d,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x24,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  add    $0x8,%rsp
  jmp    1050 <Xil_Out32@plt>
```

assistant: ```c
void stage_cu() {
  Xil_Out32(0x50001004, 0);
  Xil_Out32(0x50001000, 0x81010401);
  Xil_Out32(0x50001004, 4);
  Xil_Out32(0x50001000, 0xBF810000);
  Xil_Out32(0x50002004, 0);
  Xil_Out32(0x50002008, 0xB);
  Xil_Out32(0x5000200C, 0x22);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x17);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 16);
  Xil_Out32(0x50002008, 0x25);
  Xil_Out32(0x5000200C, 0x1F);
  Xil_Out32(0x50002010, 0x29);
  Xil_Out32(0x50002014, 0x16);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 32);
  Xil_Out32(0x50002008, 0x1D);
  Xil_Out32(0x5000200C, 0x24);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x0);
  Xil_Out32(0x50002000, 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<stage_cu>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0x81010401,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  mov    $0x4,%esi
  mov    $0x50001004,%edi
  call   <Xil_Out32@plt>
  mov    $0xbf810000,%esi
  mov    $0x50001000,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0xb,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x22,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x17,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x10,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x25,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x1f,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  mov    $0x29,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  mov    $0x16,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  call   <Xil_Out32@plt>
  mov    $0x20,%esi
  mov    $0x50002004,%edi
  call   <Xil_Out32@plt>
  mov    $0x1d,%esi
  mov    $0x50002008,%edi
  call   <Xil_Out32@plt>
  mov    $0x24,%esi
  mov    $0x5000200c,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002010,%edi
  call   <Xil_Out32@plt>
  xor    %esi,%esi
  mov    $0x50002014,%edi
  call   <Xil_Out32@plt>
  mov    $0x1,%esi
  mov    $0x50002000,%edi
  add    $0x8,%rsp
  jmp    1050 <Xil_Out32@plt>
```

assistant: ```c
void stage_cu() {
  Xil_Out32(0x50001004, 0);
  Xil_Out32(0x50001000, 0x81010401);
  Xil_Out32(0x50001004, 4);
  Xil_Out32(0x50001000, 0xBF810000);
  Xil_Out32(0x50002004, 0);
  Xil_Out32(0x50002008, 0xB);
  Xil_Out32(0x5000200C, 0x22);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x17);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 16);
  Xil_Out32(0x50002008, 0x25);
  Xil_Out32(0x5000200C, 0x1F);
  Xil_Out32(0x50002010, 0x29);
  Xil_Out32(0x50002014, 0x16);
  Xil_Out32(0x50002000, 1);
  Xil_Out32(0x50002004, 32);
  Xil_Out32(0x50002008, 0x1D);
  Xil_Out32(0x5000200C, 0x24);
  Xil_Out32(0x50002010, 0x0);
  Xil_Out32(0x50002014, 0x0);
  Xil_Out32(0x50002000, 1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L127
L39:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L10
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L127
L10:
  mov    $0xffffffff,%eax
  jmp    L138
L127:
  cmpl   $0x0,-0xc(%rbp)
  jle    L39
  movl   $0x0,-0xc(%rbp)
  jmp    L54
L75:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  je     L54
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x8(%rbp)
L54:
  cmpl   $0x0,-0xc(%rbp)
  jle    L75
  cmpl   $0x0,-0x8(%rbp)
  je     L98
  cmpl   $0x1,-0x8(%rbp)
  jne    L23
L98:
  mov    $0x0,%eax
  jmp    L138
L23:
  movl   $0x2,-0xc(%rbp)
  jmp    L22
L4:
  addl   $0x1,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L22
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L22
  mov    $0x0,%eax
  jmp    L138
L22:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L4
  mov    $0x1,%eax
L138:
  leave
  ret
```

assistant: ```c
int is_prime(char *a) {
  int i = 0, l, n = 0;
  l = strlen(a) - 1;
  while (i < 1) {
    i = i + 1;
    if (a[i] < '0' || a[i] > '9') {
      return -1;
    }
  }
  i = 0;
  while (i < 1) {
    i = i + 1;
    n += a[i] - '0';
    if (i != l - 1) {
      n *= 10;
    }
  }
  if (n == 0 || n == 1) {
    return 0;
  }
  i = 2;
  while (i < n) {
    i = i + 1;
    if (n % i == 0 && i != n) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movzbl 0x1(%rbx),%esi
  lea    -0x30(%rsi),%edx
  cmp    $0x9,%dl
  ja     L36
  movsbl %sil,%esi
  sub    $0x30,%esi
  cmp    $0x3,%eax
  je     L70
  lea    (%rsi,%rsi,4),%esi
  add    %esi,%esi
L70:
  mov    $0x0,%eax
  mov    $0x2,%ecx
  cmp    $0x1,%esi
  jbe    L51
L17:
  cmp    %esi,%ecx
  jge    L67
  add    $0x1,%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L17
  cmp    %esi,%ecx
  je     L17
  mov    $0x0,%eax
  jmp    L51
L67:
  mov    $0x1,%eax
L51:
  pop    %rbx
  ret
L36:
  mov    $0xffffffff,%eax
  jmp    L51
```

assistant: ```c
int is_prime(char *a) {
  int i = 0, l, n = 0;
  l = strlen(a) - 1;
  while (i < 1) {
    i = i + 1;
    if (a[i] < '0' || a[i] > '9') {
      return -1;
    }
  }
  i = 0;
  while (i < 1) {
    i = i + 1;
    n += a[i] - '0';
    if (i != l - 1) {
      n *= 10;
    }
  }
  if (n == 0 || n == 1) {
    return 0;
  }
  i = 2;
  while (i < n) {
    i = i + 1;
    if (n % i == 0 && i != n) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movzbl 0x1(%rbx),%ecx
  mov    $0xffffffff,%r8d
  sub    $0x30,%ecx
  cmp    $0x9,%cl
  jbe    L12
L8:
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L12:
  movsbl %cl,%ecx
  cmp    $0x3,%eax
  je     L52
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
L52:
  xor    %r8d,%r8d
  mov    $0x2,%esi
  cmp    $0x1,%ecx
  jle    L8
  nopw   0x0(%rax,%rax,1)
L13:
  cmp    %ecx,%esi
  jge    L6
  mov    %ecx,%eax
  add    $0x1,%esi
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L13
  cmp    %ecx,%esi
  je     L13
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L6:
  mov    $0x1,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_prime(char *a) {
  int i = 0, l, n = 0;
  l = strlen(a) - 1;
  while (i < 1) {
    i = i + 1;
    if (a[i] < '0' || a[i] > '9') {
      return -1;
    }
  }
  i = 0;
  while (i < 1) {
    i = i + 1;
    n += a[i] - '0';
    if (i != l - 1) {
      n *= 10;
    }
  }
  if (n == 0 || n == 1) {
    return 0;
  }
  i = 2;
  while (i < n) {
    i = i + 1;
    if (n % i == 0 && i != n) {
      return 0;
    }
  }
  return 1;
}
```