ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<return_output>:
  endbr64
  push   %r13
  lea    D21(%rip),%r13
  xor    %eax,%eax
  push   %r12
  mov    %r13,%rdx
  mov    %rdi,%r12
  lea    D27(%rip),%rdi
  push   %rbp
  mov    %esi,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x1,%edx
  xor    %eax,%eax
  call   <outputchildren@plt>
  mov    %r13,%rdx
  mov    %ebp,%esi
  xor    %eax,%eax
  pop    %rbp
  lea    D23(%rip),%rdi
  pop    %r12
  pop    %r13
  jmp    1060 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D21
type: string
name: disassemble_data
arguments: data_label: D27
type: string
name: disassemble_data
arguments: data_label: D23
type: string

tool: 

tool: D27: %*s<description><p>


tool: D23: %*s</p></description>


assistant: ```c
void return_output(struct cnode *cnode, unsigned int indent) {
  printf("%*s<description><p>\n", indent, "");
  outputchildren(cnode, indent, 1);
  printf("%*s</p></description>\n", indent, "");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<return_output>:
  endbr64
  push   %r13
  lea    D15(%rip),%r13
  xor    %eax,%eax
  push   %r12
  mov    %r13,%rdx
  mov    %rdi,%r12
  lea    D11(%rip),%rdi
  push   %rbp
  mov    %esi,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x1,%edx
  xor    %eax,%eax
  call   <outputchildren@plt>
  mov    %r13,%rdx
  mov    %ebp,%esi
  xor    %eax,%eax
  pop    %rbp
  lea    D17(%rip),%rdi
  pop    %r12
  pop    %r13
  jmp    1060 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type_infered: string
name: extract_symbolic_data
arguments: tag: D11
type_infered: string
name: extract_symbolic_data
arguments: tag: D17
type_infered: string

tool: 

tool: D11: %*s<description><p>


tool: D17: %*s</p></description>


assistant: ```c
void return_output(struct cnode *cnode, unsigned int indent) {
  printf("%*s<description><p>\n", indent, "");
  outputchildren(cnode, indent, 1);
  printf("%*s</p></description>\n", indent, "");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_be32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  shl    $0x10,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t read_be32(const void *src) {
  const uint8_t *s = src;
  return (((uint32_t)s[0]) << 24) | (((uint32_t)s[1]) << 16) | (((uint32_t)s[2]) << 8) | (((uint32_t)s[3]) << 0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<read_be32>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x18,%eax
  mov    0x8(%rdi),%edx
  shl    $0x10,%edx
  or     %edx,%eax
  or     0x18(%rdi),%eax
  mov    0x10(%rdi),%rdx
  shl    $0x8,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t read_be32(const void *src) {
  const uint8_t *s = src;
  return (((uint32_t)s[0]) << 24) | (((uint32_t)s[1]) << 16) | (((uint32_t)s[2]) << 8) | (((uint32_t)s[3]) << 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_be32>:
  endbr64
  mov    (%rdi),%eax
  mov    0x8(%rdi),%edx
  shl    $0x18,%eax
  shl    $0x10,%edx
  or     %edx,%eax
  mov    0x10(%rdi),%rdx
  or     0x18(%rdi),%eax
  shl    $0x8,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t read_be32(const void *src) {
  const uint8_t *s = src;
  return (((uint32_t)s[0]) << 24) | (((uint32_t)s[1]) << 16) | (((uint32_t)s[2]) << 8) | (((uint32_t)s[3]) << 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_be32>:
  endbr64
  mov    (%rdi),%eax
  mov    0x8(%rdi),%edx
  shl    $0x18,%eax
  shl    $0x10,%edx
  or     %edx,%eax
  mov    0x10(%rdi),%rdx
  or     0x18(%rdi),%eax
  shl    $0x8,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
uint32_t read_be32(const void *src) {
  const uint8_t *s = src;
  return (((uint32_t)s[0]) << 24) | (((uint32_t)s[1]) << 16) | (((uint32_t)s[2]) << 8) | (((uint32_t)s[3]) << 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<generate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x48(%rbp),%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0x24(%rbp)
  jmp    L37
L12:
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x48(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movl   $0x1,(%rax)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  movl   $0x1,-0x20(%rbp)
  jmp    L23
L0:
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rsi
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rsi,%rax
  add    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x20(%rbp)
L23:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jl     L0
  addl   $0x1,-0x24(%rbp)
L37:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L12
  mov    -0x40(%rbp),%rax
  mov    -0x34(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int **generate(int numRows, int *returnSize, int **returnColumnSizes) {
  int i, j;
  int **triangle = malloc(numRows * sizeof(int *));
  *returnColumnSizes = malloc(numRows * sizeof(int *));
  for (i = 0; i < numRows; i++) {
    int num = i + 1;
    (*returnColumnSizes)[i] = num;
    triangle[i] = malloc(num * sizeof(int));
    triangle[i][0] = 1;
    triangle[i][num - 1] = 1;
    for (j = 1; j <= num - 2; j++) {
      triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
    }
  }
  *returnSize = numRows;
  return triangle;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<generate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r14d
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%r15
  lea    0x0(,%rdi,8),%ebx
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,(%rsp)
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  test   %r14d,%r14d
  jle    L0
  mov    $0x0,%r12d
  mov    $0x0,%ebx
L22:
  mov    %ebx,%r13d
  add    $0x1,%ebx
  mov    (%r15),%rax
  mov    %ebx,(%rax,%r12,1)
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,0x0(%rbp)
  movl   $0x1,(%rax)
  mov    0x0(%rbp),%rax
  movl   $0x1,(%rax,%r12,1)
  cmp    $0x1,%r13d
  jle    L37
  lea    -0x2(%r13),%eax
  lea    0x4(,%rax,4),%r9
  mov    $0x0,%eax
L18:
  mov    -0x8(%rsi),%rcx
  mov    %rax,%r8
  add    $0x4,%rax
  mov    (%rsi),%rdi
  mov    (%rcx,%rax,1),%edx
  add    (%rcx,%r8,1),%edx
  mov    %edx,(%rdi,%rax,1)
  cmp    %r9,%rax
  jne    L18
L37:
  add    $0x4,%r12
  add    $0x8,%rbp
  cmp    %ebx,%r14d
  jne    L22
L0:
  mov    0x8(%rsp),%rax
  mov    %r14d,(%rax)
  mov    (%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **generate(int numRows, int *returnSize, int **returnColumnSizes) {
  int i, j;
  int **triangle = malloc(numRows * sizeof(int *));
  *returnColumnSizes = malloc(numRows * sizeof(int *));
  for (i = 0; i < numRows; i++) {
    int num = i + 1;
    (*returnColumnSizes)[i] = num;
    triangle[i] = malloc(num * sizeof(int));
    triangle[i][0] = 1;
    triangle[i][num - 1] = 1;
    for (j = 1; j <= num - 2; j++) {
      triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
    }
  }
  *returnSize = numRows;
  return triangle;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<generate>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%r15d
  mov    %rdx,%rbx
  mov    %rbp,%r13
  mov    %r15d,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %ebp,%ebp
  jle    L49
  xor    %r15d,%r15d
  nopl   0x0(%rax,%rax,1)
L45:
  lea    0x1(%r15),%edi
  mov    %edi,(%rax,%r15,4)
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%r12,%r15,8)
  movl   $0x1,(%rax)
  movl   $0x1,(%rax,%r15,4)
  cmp    $0x1,%r15
  jbe    L2
  lea    -0x2(%r15),%edi
  mov    -0x8(%r12,%r15,8),%rsi
  mov    $0x1,%edx
  add    $0x2,%rdi
  nopw   0x0(%rax,%rax,1)
L7:
  mov    (%rsi,%rdx,4),%ecx
  add    -0x4(%rsi,%rdx,4),%ecx
  mov    %ecx,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L7
L2:
  add    $0x1,%r15
  cmp    %rbp,%r15
  je     L49
  mov    (%rbx),%rax
  jmp    L45
  nopl   0x0(%rax)
L49:
  mov    %r13d,(%r14)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **generate(int numRows, int *returnSize, int **returnColumnSizes) {
  int i, j;
  int **triangle = malloc(numRows * sizeof(int *));
  *returnColumnSizes = malloc(numRows * sizeof(int *));
  for (i = 0; i < numRows; i++) {
    int num = i + 1;
    (*returnColumnSizes)[i] = num;
    triangle[i] = malloc(num * sizeof(int));
    triangle[i][0] = 1;
    triangle[i][num - 1] = 1;
    for (j = 1; j <= num - 2; j++) {
      triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
    }
  }
  *returnSize = numRows;
  return triangle;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<generate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%r15d
  mov    %rdx,%rbx
  mov    %rbp,%r13
  mov    %r15d,%edi
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %ebp,%ebp
  jle    L48
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L72:
  lea    0x1(%r15),%edi
  mov    %r15d,%r14d
  mov    %edi,(%rax,%r15,4)
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,(%r12,%r15,8)
  mov    %rax,%rcx
  movl   $0x1,(%rax)
  movl   $0x1,(%rax,%r15,4)
  cmp    $0x1,%r15
  jbe    L77
  mov    -0x8(%r12,%r15,8),%rdx
  sub    %rdx,%rax
  cmp    $0xc,%rax
  jbe    L53
  cmp    $0x2,%r15d
  je     L53
  lea    -0x2(%r15),%eax
  lea    -0x1(%r15),%edi
  cmp    $0x2,%eax
  jbe    L51
  mov    %edi,%eax
  shr    $0x2,%eax
  lea    -0x1(%rax),%esi
  mov    $0x4,%eax
  shl    $0x4,%rsi
  add    $0x14,%rsi
  nopl   0x0(%rax)
L39:
  movdqu -0x4(%rdx,%rax,1),%xmm0
  movdqu (%rdx,%rax,1),%xmm2
  paddd  %xmm2,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L39
  mov    %edi,%esi
  and    $0xfffffffc,%esi
  lea    0x1(%rsi),%eax
  cmp    %esi,%edi
  je     L77
L73:
  sub    %esi,%r14d
  lea    -0x1(%r14),%edi
  cmp    $0x2,%r14d
  je     L38
  lea    0x0(,%rsi,4),%r8
  movq   (%rdx,%rsi,4),%xmm0
  mov    %edi,%esi
  movq   0x4(%rdx,%r8,1),%xmm1
  and    $0xfffffffe,%esi
  add    %esi,%eax
  paddd  %xmm1,%xmm0
  movq   %xmm0,0x4(%rcx,%r8,1)
  cmp    %esi,%edi
  je     L77
L38:
  cltq
  mov    (%rdx,%rax,4),%esi
  add    -0x4(%rdx,%rax,4),%esi
  mov    %esi,(%rcx,%rax,4)
L77:
  add    $0x1,%r15
  cmp    %r15,%rbp
  je     L48
L99:
  mov    (%rbx),%rax
  jmp    L72
  nopl   0x0(%rax)
L53:
  lea    -0x2(%r14),%edi
  mov    $0x1,%eax
  add    $0x2,%rdi
  nopl   (%rax)
L58:
  mov    -0x4(%rdx,%rax,4),%esi
  add    (%rdx,%rax,4),%esi
  mov    %esi,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L58
  add    $0x1,%r15
  cmp    %r15,%rbp
  jne    L99
L48:
  mov    0x8(%rsp),%rax
  mov    %r13d,(%rax)
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  xor    %esi,%esi
  mov    $0x1,%eax
  jmp    L73
```

assistant: ```c
int **generate(int numRows, int *returnSize, int **returnColumnSizes) {
  int i, j;
  int **triangle = malloc(numRows * sizeof(int *));
  *returnColumnSizes = malloc(numRows * sizeof(int *));
  for (i = 0; i < numRows; i++) {
    int num = i + 1;
    (*returnColumnSizes)[i] = num;
    triangle[i] = malloc(num * sizeof(int));
    triangle[i][0] = 1;
    triangle[i][num - 1] = 1;
    for (j = 1; j <= num - 2; j++) {
      triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
    }
  }
  *returnSize = numRows;
  return triangle;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skip_white>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L11
L29:
  addq   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L29
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L29
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *skip_white(char *ptr) {
  while (*ptr == ' ' || *ptr == '\t')
    ptr++;
  return ptr;
}
```
user: What is the c source code of the assembly code below:

```asm
<skip_white>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L2
  cmp    $0x9,%dl
  jne    L39
L2:
  add    $0x1,%rax
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  je     L2
  cmp    $0x9,%dl
  je     L2
L39:
  ret
```

assistant: ```c
char *skip_white(char *ptr) {
  while (*ptr == ' ' || *ptr == '\t')
    ptr++;
  return ptr;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<skip_white>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  cmp    $0x20,%dl
  jne    L18
  nop
L4:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L4
L18:
  cmp    $0x9,%dl
  je     L4
  ret
```

assistant: ```c
char *skip_white(char *ptr) {
  while (*ptr == ' ' || *ptr == '\t')
    ptr++;
  return ptr;
}
```
user: What is the c source code of the assembly code below?

```asm
<skip_white>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  cmp    $0x20,%dl
  jne    L23
  nop
L5:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L5
L23:
  cmp    $0x9,%dl
  je     L5
  ret
```

assistant: ```c
char *skip_white(char *ptr) {
  while (*ptr == ' ' || *ptr == '\t')
    ptr++;
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uwsgi_str2_num>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x30(%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int uwsgi_str2_num(char *str) {
  int num = 0;
  num = 10 * (str[0] - 48);
  num += str[1] - 48;
  return num;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<uwsgi_str2_num>:
  endbr64
  movsbl (%rdi),%eax
  lea    -0xf0(%rax,%rax,4),%edx
  movsbl 0x1(%rdi),%eax
  lea    -0x30(%rax,%rdx,2),%eax
  ret
```

assistant: ```c
int uwsgi_str2_num(char *str) {
  int num = 0;
  num = 10 * (str[0] - 48);
  num += str[1] - 48;
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<uwsgi_str2_num>:
  endbr64
  movsbl (%rdi),%eax
  lea    -0xf0(%rax,%rax,4),%edx
  movsbl 0x1(%rdi),%eax
  lea    -0x30(%rax,%rdx,2),%eax
  ret
```

assistant: ```c
int uwsgi_str2_num(char *str) {
  int num = 0;
  num = 10 * (str[0] - 48);
  num += str[1] - 48;
  return num;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uwsgi_str2_num>:
  endbr64
  movsbl (%rdi),%eax
  lea    -0xf0(%rax,%rax,4),%edx
  movsbl 0x1(%rdi),%eax
  lea    -0x30(%rax,%rdx,2),%eax
  ret
```

assistant: ```c
int uwsgi_str2_num(char *str) {
  int num = 0;
  num = 10 * (str[0] - 48);
  num += str[1] - 48;
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1020304,%esi
  lea    D24(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D33(%rip),%rdi
  call   <puts@plt>
  movabs $0x102030405060708,%rsi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  mov    $0x8,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D24
datatype: string
name: parse_data
arguments: data_tag: D33
datatype: string
name: parse_data
arguments: data_tag: D6
datatype: string
name: parse_data
arguments: data_tag: D16
datatype: string

tool: D24: %x


tool: D33: Printing z ...

tool: D6: %llx


tool: D16: Printing z done

assistant: ```c
void main() {
  int x;
  char *y;
  long long z;
  x = 0x01020304;
  z = 0x0102030405060708LL;
  printf("%x\n", x);
  y = (char *)&x;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("Printing z ...\n");
  printf("%llx\n", z);
  printf("Printing z done\n");
  y = (char *)&z;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("%x\n", y[4]);
  printf("%x\n", y[5]);
  printf("%x\n", y[6]);
  printf("%x\n", y[7]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D25(%rip),%rbp
  mov    $0x1020304,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  movabs $0x102030405060708,%rsi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0x8,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x6,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x5,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D25
type_guessed: string
name: read_data
arguments: label: D8
type_guessed: string
name: read_data
arguments: label: D10
type_guessed: string
name: read_data
arguments: label: D32
type_guessed: string

tool: D25: %x


tool: D8: Printing z ...

tool: D10: %llx


tool: D32: Printing z done

assistant: ```c
void main() {
  int x;
  char *y;
  long long z;
  x = 0x01020304;
  z = 0x0102030405060708LL;
  printf("%x\n", x);
  y = (char *)&x;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("Printing z ...\n");
  printf("%llx\n", z);
  printf("Printing z done\n");
  y = (char *)&z;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("%x\n", y[4]);
  printf("%x\n", y[5]);
  printf("%x\n", y[6]);
  printf("%x\n", y[7]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rbp
  mov    $0x1020304,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  movabs $0x102030405060708,%rsi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0x8,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x6,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x5,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_infered: string
name: unpack_data
arguments: data_tag: D15
type_infered: string
name: unpack_data
arguments: data_tag: D22
type_infered: string
name: unpack_data
arguments: data_tag: D10
type_infered: string

tool: D3: %x


tool: D15: Printing z ...

tool: D22: %llx


tool: D10: Printing z done

assistant: ```c
void main() {
  int x;
  char *y;
  long long z;
  x = 0x01020304;
  z = 0x0102030405060708LL;
  printf("%x\n", x);
  y = (char *)&x;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("Printing z ...\n");
  printf("%llx\n", z);
  printf("Printing z done\n");
  y = (char *)&z;
  printf("%x\n", y[0]);
  printf("%x\n", y[1]);
  printf("%x\n", y[2]);
  printf("%x\n", y[3]);
  printf("%x\n", y[4]);
  printf("%x\n", y[5]);
  printf("%x\n", y[6]);
  printf("%x\n", y[7]);
}
```