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
<fls64>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L20
  movslq %edi,%rdi
  mov    $0x1,%eax
  movabs $0xffffffff00000000,%rdx
  test   %rdx,%rdi
  je     L20
  movabs $0xffff000000000000,%rax
  test   %rax,%rdi
  je     L2
  movabs $0xff00000000000000,%rax
  test   %rax,%rdi
  je     L41
  movabs $0xf000000000000000,%rax
  test   %rax,%rdi
  je     L47
  movabs $0xc000000000000000,%rax
  and    %rax,%rdi
  cmp    $0x1,%rdi
  sbb    %eax,%eax
  and    $0xfffffffd,%eax
  add    $0x40,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  ret
L2:
  mov    $0x21,%eax
  ret
L41:
  mov    $0x31,%eax
  ret
L47:
  mov    $0x39,%eax
  ret
```

assistant: ```c
int fls64(uint64_t n) {
  int v;
  if (!n)
    return 0;
  v = 64;
  if ((n & 0xFFFFFFFF00000000ULL) == 0) {
    n <<= 32;
    v -= 32;
  }
  if ((n & 0xFFFF000000000000ULL) == 0) {
    n <<= 16;
    v -= 16;
  }
  if ((n & 0xFF00000000000000ULL) == 0) {
    n <<= 8;
    v -= 8;
  }
  if ((n & 0xF000000000000000ULL) == 0) {
    n <<= 4;
    v -= 4;
  }
  if ((n & 0xC000000000000000ULL) == 0) {
    n <<= 2;
    v -= 2;
  }
  if ((n & 0x8000000000000000ULL) == 0) {
    n <<= 1;
    v -= 1;
  }
  return v;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fls64>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L15
  movslq %edi,%rdi
  mov    $0x1,%eax
  movabs $0xffffffff00000000,%rdx
  test   %rdx,%rdi
  je     L15
  movabs $0xffff000000000000,%rax
  test   %rax,%rdi
  je     L3
  movabs $0xff00000000000000,%rax
  test   %rax,%rdi
  je     L36
  movabs $0xf000000000000000,%rax
  test   %rax,%rdi
  je     L21
  movabs $0xc000000000000000,%rax
  and    %rax,%rdi
  cmp    $0x1,%rdi
  sbb    %eax,%eax
  and    $0xfffffffd,%eax
  add    $0x40,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  ret
L3:
  mov    $0x21,%eax
  ret
L36:
  mov    $0x31,%eax
  ret
L21:
  mov    $0x39,%eax
  ret
```

assistant: ```c
int fls64(uint64_t n) {
  int v;
  if (!n)
    return 0;
  v = 64;
  if ((n & 0xFFFFFFFF00000000ULL) == 0) {
    n <<= 32;
    v -= 32;
  }
  if ((n & 0xFFFF000000000000ULL) == 0) {
    n <<= 16;
    v -= 16;
  }
  if ((n & 0xFF00000000000000ULL) == 0) {
    n <<= 8;
    v -= 8;
  }
  if ((n & 0xF000000000000000ULL) == 0) {
    n <<= 4;
    v -= 4;
  }
  if ((n & 0xC000000000000000ULL) == 0) {
    n <<= 2;
    v -= 2;
  }
  if ((n & 0x8000000000000000ULL) == 0) {
    n <<= 1;
    v -= 1;
  }
  return v;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<br_str_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L6
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L6:
  nop
  leave
  ret
```

assistant: ```c
void br_str_free(void *str) {
  if (str)
    free(str);
}
```
user: What is the c source code of the assembly code below?

```asm
<br_str_free>:
  endbr64
  test   %rdi,%rdi
  je     L6
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L6:
  ret
```

assistant: ```c
void br_str_free(void *str) {
  if (str)
    free(str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<br_str_free>:
  endbr64
  test   %rdi,%rdi
  je     L8
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L8:
  ret
```

assistant: ```c
void br_str_free(void *str) {
  if (str)
    free(str);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<br_str_free>:
  endbr64
  test   %rdi,%rdi
  je     L0
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L0:
  ret
```

assistant: ```c
void br_str_free(void *str) {
  if (str)
    free(str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<debug>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L30:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jne    L6
  mov    $0x5b,%edi
  call   <putchar@plt>
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jne    L4
  mov    $0x5d,%edi
  call   <putchar@plt>
L4:
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L30
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D9: %ld 

assistant: ```c
void debug(long *a, int n, int idx) {
  int i = 0;
  for (i = 0; i < n; i++) {
    if (i == idx) {
      printf("[");
    }
    printf("%ld ", a[i]);
    if (i == idx) {
      printf("]");
    }
  }
  printf("\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<debug>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L23
  mov    %rdi,%r13
  mov    %edx,%r12d
  mov    %esi,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r14
  jmp    L26
L33:
  mov    $0x5b,%edi
  call   <putchar@plt>
  mov    0x0(%r13,%rbx,8),%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5d,%edi
  call   <putchar@plt>
L31:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L23
L26:
  cmp    %ebx,%r12d
  je     L33
  mov    0x0(%r13,%rbx,8),%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L31
L23:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_infered: string

tool: D2: %ld 

assistant: ```c
void debug(long *a, int n, int idx) {
  int i = 0;
  for (i = 0; i < n; i++) {
    if (i == idx) {
      printf("[");
    }
    printf("%ld ", a[i]);
    if (i == idx) {
      printf("]");
    }
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<debug>:
  endbr64
  test   %esi,%esi
  jle    L47
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L0:
  mov    (%r12,%rbx,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r13
  je     L14
L6:
  cmp    %ebx,%ebp
  jne    L0
  mov    $0x5b,%edi
  call   <putchar@plt>
  mov    (%r12,%rbx,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    $0x5d,%edi
  call   <putchar@plt>
  cmp    %rbx,%r13
  jne    L6
L14:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <putchar@plt>
L47:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: label: D1
type_infered: string

tool: D1: %ld 

assistant: ```c
void debug(long *a, int n, int idx) {
  int i = 0;
  for (i = 0; i < n; i++) {
    if (i == idx) {
      printf("[");
    }
    printf("%ld ", a[i]);
    if (i == idx) {
      printf("]");
    }
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<debug>:
  endbr64
  test   %esi,%esi
  jle    L44
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L4:
  mov    (%r12,%rbx,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r13
  je     L5
L26:
  cmp    %ebx,%ebp
  jne    L4
  mov    $0x5b,%edi
  call   <putchar@plt>
  mov    (%r12,%rbx,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    $0x5d,%edi
  call   <putchar@plt>
  cmp    %rbx,%r13
  jne    L26
L5:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <putchar@plt>
L44:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: label: D2
data_type: string

tool: D2: %ld 

assistant: ```c
void debug(long *a, int n, int idx) {
  int i = 0;
  for (i = 0; i < n; i++) {
    if (i == idx) {
      printf("[");
    }
    printf("%ld ", a[i]);
    if (i == idx) {
      printf("]");
    }
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<skiplab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L53
L51:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L13
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L17
  addq   $0x1,-0x8(%rbp)
  jmp    L16
L17:
  mov    $0x0,%eax
  jmp    L4
L13:
  nop
  addq   $0x1,-0x8(%rbp)
L53:
  mov    -0x18(%rbp),%rax
  add    $0x6,%rax
  cmp    %rax,-0x8(%rbp)
  jb     L51
L16:
  mov    -0x8(%rbp),%rax
L4:
  pop    %rbp
  ret
```

assistant: ```c
char *skiplab(char *p) {
  char *ptr;
  for (ptr = p; ptr < &p[6]; ptr++) {
    if (*ptr == ' ')
      continue;
    if (*ptr == '\t') {
      ptr++;
      break;
    }
    return (0);
  }
  return (ptr);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<skiplab>:
  endbr64
  lea    0x6(%rdi),%rax
L10:
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L14
  add    $0x1,%rdi
  cmp    $0x9,%dl
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
L14:
  add    $0x1,%rdi
  cmp    %rax,%rdi
  jne    L10
  ret
```

assistant: ```c
char *skiplab(char *p) {
  char *ptr;
  for (ptr = p; ptr < &p[6]; ptr++) {
    if (*ptr == ' ')
      continue;
    if (*ptr == '\t') {
      ptr++;
      break;
    }
    return (0);
  }
  return (ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<skiplab>:
  endbr64
  lea    0x6(%rdi),%rdx
L19:
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  cmp    $0x20,%al
  je     L5
  cmp    $0x9,%al
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
  nop
L5:
  cmp    %rdi,%rdx
  jne    L19
  mov    %rdx,%rax
  ret
```

assistant: ```c
char *skiplab(char *p) {
  char *ptr;
  for (ptr = p; ptr < &p[6]; ptr++) {
    if (*ptr == ' ')
      continue;
    if (*ptr == '\t') {
      ptr++;
      break;
    }
    return (0);
  }
  return (ptr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<skiplab>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x20,%al
  je     L21
  add    $0x1,%rdi
  cmp    $0x9,%al
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L21:
  movzbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rdx
  cmp    $0x20,%al
  je     L7
L47:
  mov    %rdx,%rdi
  add    $0x1,%rdi
  cmp    $0x9,%al
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
  nop
L7:
  movzbl 0x2(%rdi),%eax
  lea    0x2(%rdi),%rdx
  cmp    $0x20,%al
  jne    L47
  movzbl 0x3(%rdi),%eax
  lea    0x3(%rdi),%rdx
  cmp    $0x20,%al
  jne    L47
  movzbl 0x4(%rdi),%eax
  lea    0x4(%rdi),%rdx
  cmp    $0x20,%al
  jne    L47
  movzbl 0x5(%rdi),%eax
  lea    0x5(%rdi),%rdx
  cmp    $0x20,%al
  jne    L47
  lea    0x6(%rdi),%rax
  ret
```

assistant: ```c
char *skiplab(char *p) {
  char *ptr;
  for (ptr = p; ptr < &p[6]; ptr++) {
    if (*ptr == ' ')
      continue;
    if (*ptr == '\t') {
      ptr++;
      break;
    }
    return (0);
  }
  return (ptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nmult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,-0x30(%rbp)
  mov    %rsi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    %r8d,-0x48(%rbp)
  mov    %r9d,-0x4c(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    $0x0,%ebx
  jmp    L48
L59:
  mov    $0x0,%r13d
  jmp    L52
L13:
  mov    %ebx,%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    %ebx,%eax
  imul   -0x4c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  add    %r13d,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  mov    $0x0,%r12d
  jmp    L20
L30:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x24(%rbp),%eax
  add    %r12d,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm2
  mov    %r12d,%eax
  imul   -0x4c(%rbp),%eax
  add    %r13d,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm2,%xmm0
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  add    $0x1,%r12d
L20:
  cmp    -0x28(%rbp),%r12d
  jl     L30
  add    $0x1,%r13d
L52:
  cmp    -0x4c(%rbp),%r13d
  jl     L13
  add    $0x1,%ebx
L48:
  cmp    -0x44(%rbp),%ebx
  jl     L59
  nop
  nop
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
void nmult(float *A, float *B, float *C, int ra, int ca, int cb) {
  register int i, j, k;
  int u, v, t, rb;
  rb = ca;
#pragma omp parallel for private(i, j, k, v, u, t)
  for (i = 0; i < ra; ++i) {
    for (j = 0; j < cb; ++j) {
      v = i * rb;
      u = i * cb;
      t = j + u;
      C[t] = 0.;
      for (k = 0; k < rb; ++k) {
        C[t] += A[k + v] * B[j + k * cb];
      }
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nmult>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,-0x20(%rsp)
  mov    %rdx,-0x18(%rsp)
  mov    %ecx,-0x24(%rsp)
  test   %ecx,%ecx
  jle    L11
  mov    %rdi,%rbx
  mov    %r8d,%ebp
  mov    %r9d,%r11d
  movslq %r9d,%rdi
  shl    $0x2,%rdi
  mov    $0x0,%r15d
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  lea    -0x1(%r8),%eax
  mov    %rax,-0x10(%rsp)
  lea    0x4(%rbx),%rax
  mov    %rax,-0x8(%rsp)
  jmp    L63
L72:
  movslq %r14d,%rax
  mov    -0x18(%rsp),%rsi
  lea    (%rsi,%rax,4),%r8
  mov    -0x20(%rsp),%r10
  movslq %r15d,%rax
  lea    (%rbx,%rax,4),%r12
  add    -0x10(%rsp),%rax
  mov    -0x8(%rsp),%rsi
  lea    (%rsi,%rax,4),%rsi
  mov    $0x0,%r9d
L17:
  mov    %r8,%rcx
  movl   $0x0,(%r8)
  test   %ebp,%ebp
  jle    L50
  mov    %r10,%rdx
  mov    %r12,%rax
L49:
  movss  (%rax),%xmm0
  mulss  (%rdx),%xmm0
  addss  (%rcx),%xmm0
  movss  %xmm0,(%rcx)
  add    $0x4,%rax
  add    %rdi,%rdx
  cmp    %rsi,%rax
  jne    L49
L50:
  add    $0x1,%r9d
  add    $0x4,%r8
  add    $0x4,%r10
  cmp    %r9d,%r11d
  jne    L17
L21:
  add    $0x1,%r13d
  add    %r11d,%r14d
  add    %ebp,%r15d
  cmp    %r13d,-0x24(%rsp)
  je     L11
L63:
  test   %r11d,%r11d
  jg     L72
  jmp    L21
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void nmult(float *A, float *B, float *C, int ra, int ca, int cb) {
  register int i, j, k;
  int u, v, t, rb;
  rb = ca;
#pragma omp parallel for private(i, j, k, v, u, t)
  for (i = 0; i < ra; ++i) {
    for (j = 0; j < cb; ++j) {
      v = i * rb;
      u = i * cb;
      t = j + u;
      C[t] = 0.;
      for (k = 0; k < rb; ++k) {
        C[t] += A[k + v] * B[j + k * cb];
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<nmult>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,-0x18(%rsp)
  mov    %ecx,-0x1c(%rsp)
  test   %ecx,%ecx
  jle    L42
  mov    %rdi,%rbx
  lea    -0x1(%r8),%eax
  movslq %r9d,%rdi
  mov    %rdx,%r12
  mov    %rax,-0x10(%rsp)
  lea    0x4(%rbx),%rax
  mov    %r8d,%r11d
  mov    %r9d,%r10d
  mov    %rax,-0x8(%rsp)
  shl    $0x2,%rdi
  xor    %r15d,%r15d
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  pxor   %xmm2,%xmm2
  nopl   0x0(%rax)
L21:
  test   %r10d,%r10d
  jle    L2
  movslq %r15d,%rax
  mov    -0x8(%rsp),%rsi
  mov    -0x18(%rsp),%r9
  xor    %r8d,%r8d
  lea    (%r12,%rax,4),%rcx
  movslq %r14d,%rax
  lea    (%rbx,%rax,4),%rbp
  add    -0x10(%rsp),%rax
  lea    (%rsi,%rax,4),%rsi
  nopl   0x0(%rax)
L44:
  movl   $0x0,(%rcx)
  mov    %r9,%rdx
  mov    %rbp,%rax
  movaps %xmm2,%xmm1
  test   %r11d,%r11d
  jle    L4
  nopl   0x0(%rax)
L11:
  movss  (%rax),%xmm0
  mulss  (%rdx),%xmm0
  add    $0x4,%rax
  add    %rdi,%rdx
  addss  %xmm0,%xmm1
  movss  %xmm1,(%rcx)
  cmp    %rax,%rsi
  jne    L11
L4:
  add    $0x1,%r8d
  add    $0x4,%rcx
  add    $0x4,%r9
  cmp    %r8d,%r10d
  jne    L44
L2:
  add    $0x1,%r13d
  add    %r11d,%r14d
  add    %r10d,%r15d
  cmp    %r13d,-0x1c(%rsp)
  jne    L21
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void nmult(float *A, float *B, float *C, int ra, int ca, int cb) {
  register int i, j, k;
  int u, v, t, rb;
  rb = ca;
#pragma omp parallel for private(i, j, k, v, u, t)
  for (i = 0; i < ra; ++i) {
    for (j = 0; j < cb; ++j) {
      v = i * rb;
      u = i * cb;
      t = j + u;
      C[t] = 0.;
      for (k = 0; k < rb; ++k) {
        C[t] += A[k + v] * B[j + k * cb];
      }
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<nmult>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,-0x10(%rsp)
  mov    %ecx,-0x4(%rsp)
  test   %ecx,%ecx
  jle    L24
  mov    %r9d,%r11d
  test   %r9d,%r9d
  jle    L24
  movslq %r8d,%r14
  lea    -0x1(%r8),%eax
  movslq %r9d,%r15
  mov    %rdi,%r10
  lea    0x4(%rdi,%rax,4),%rsi
  mov    %rdx,%r12
  mov    %r8d,%ebx
  shl    $0x2,%r14
  lea    0x0(,%r15,4),%rdi
  xor    %r13d,%r13d
  pxor   %xmm2,%xmm2
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L10:
  mov    -0x10(%rsp),%r9
  lea    (%r12,%r13,4),%rcx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L59:
  movl   $0x0,(%rcx)
  mov    %r9,%rdx
  mov    %r10,%rax
  movaps %xmm2,%xmm1
  test   %ebx,%ebx
  jle    L33
  nopl   0x0(%rax,%rax,1)
L43:
  movss  (%rax),%xmm0
  mulss  (%rdx),%xmm0
  add    $0x4,%rax
  add    %rdi,%rdx
  addss  %xmm0,%xmm1
  movss  %xmm1,(%rcx)
  cmp    %rsi,%rax
  jne    L43
L33:
  add    $0x1,%r8d
  add    $0x4,%rcx
  add    $0x4,%r9
  cmp    %r8d,%r11d
  jne    L59
  add    $0x1,%ebp
  add    %r14,%r10
  add    %r14,%rsi
  add    %r15,%r13
  cmp    %ebp,-0x4(%rsp)
  jne    L10
L24:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void nmult(float *A, float *B, float *C, int ra, int ca, int cb) {
  register int i, j, k;
  int u, v, t, rb;
  rb = ca;
#pragma omp parallel for private(i, j, k, v, u, t)
  for (i = 0; i < ra; ++i) {
    for (j = 0; j < cb; ++j) {
      v = i * rb;
      u = i * cb;
      t = j + u;
      C[t] = 0.;
      for (k = 0; k < rb; ++k) {
        C[t] += A[k + v] * B[j + k * cb];
      }
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sum_omp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <sum_omp_tasks@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double sum_omp(double *v, int low, int high) {
  double s = 0;
#pragma omp parallel
  {
#pragma omp single nowait
    s = sum_omp_tasks(v, low, high);
  }
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sum_omp>:
  endbr64
  sub    $0x8,%rsp
  call   <sum_omp_tasks@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
double sum_omp(double *v, int low, int high) {
  double s = 0;
#pragma omp parallel
  {
#pragma omp single nowait
    s = sum_omp_tasks(v, low, high);
  }
  return s;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum_omp>:
  endbr64
  jmp    1050 <sum_omp_tasks@plt>
```

assistant: ```c
double sum_omp(double *v, int low, int high) {
  double s = 0;
#pragma omp parallel
  {
#pragma omp single nowait
    s = sum_omp_tasks(v, low, high);
  }
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sum_omp>:
  endbr64
  jmp    1050 <sum_omp_tasks@plt>
```

assistant: ```c
double sum_omp(double *v, int low, int high) {
  double s = 0;
#pragma omp parallel
  {
#pragma omp single nowait
    s = sum_omp_tasks(v, low, high);
  }
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<platform_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <serial_putc@plt>
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
int platform_puts(const char *buf, int n) {
  int i;
  for (i = 0; i < n; i++)
    serial_putc(buf[i]);
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<platform_puts>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r12d
  test   %esi,%esi
  jle    L7
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rbp
L4:
  movsbl (%rbx),%edi
  call   <serial_putc@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L4
L7:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int platform_puts(const char *buf, int n) {
  int i;
  for (i = 0; i < n; i++)
    serial_putc(buf[i]);
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<platform_puts>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x1(%rdi,%rax,1),%rbp
  nopw   0x0(%rax,%rax,1)
L17:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <serial_putc@plt>
  cmp    %rbp,%rbx
  jne    L17
L18:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int platform_puts(const char *buf, int n) {
  int i;
  for (i = 0; i < n; i++)
    serial_putc(buf[i]);
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<platform_puts>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x1(%rdi,%rax,1),%rbp
  nopw   0x0(%rax,%rax,1)
L10:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <serial_putc@plt>
  cmp    %rbp,%rbx
  jne    L10
L11:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int platform_puts(const char *buf, int n) {
  int i;
  for (i = 0; i < n; i++)
    serial_putc(buf[i]);
  return n;
}
```