使用 x86 mov 指令实现 C 转换
Achieve C casting using x86 mov instructions
我正在看书 "Computer Systems: A Programmer's Perspective"。现在我不确定我是否理解何时使用不同的 mov 指令。这是练习：
Practice Problem 3.4
Assume variables v and p declared with types
src_t v;
dest_t *p;
where src_t and dest_t are data types declared with typedef. We wish
to use the appropriate data movement instruction to implement the
operation
*p = (dest_t) v;
where v is stored in the appropriately named portion of register %eax
(i.e., %eax, %ax, or %al), while pointer p is stored in register %edx.
For the following combinations of src_t and dest_t, write a line of
assembly code that does the appropriate transfer. Recall that when
performing a cast that involves both a size change and a change of
“signedness” in C, the operation should change the signedness first
(Section 2.2.6).
我正在对照 this blog post 检查我的解决方案，但我不确定我是否理解问题 3：
src_t dest_t My Solution Blog's Solution
char unsigned movzbl %al, (%edx) movsbl %al, (%edx)
第 3 条：我使用 movzbl，而博客作者使用 movsbl。在这种情况下，我不明白 movzbl 与 movsbl 的推理......如果 char 是负数，你最终会得到一个错误的数字，任何人都可以澄清为什么 movsbl 是正确的选择吗？
Recall that when performing a cast that involves both a size change
and a change of “signedness” in C, the operation should change the
signedness first (Section 2.2.6).
除非我看错了，否则这句话似乎与他给出的正确答案相冲突。
该规则意味着，从有符号字符开始：
- 符号字符
- -> 无符号字符
- -> 无符号整数
所以大小转换是从 unsigned char 到 unsigned int，所以 movz 是合适的。大概规则是错误的（与 C 实际指定的不匹配），movs 确实是 C 编译器会使用的规则。
Jester的评论说-1 char值的unsigned int值必须是0xffffffff，所以看起来像练习题最后一句关于转换signedness和size的顺序是问题所在。
我正在看书 "Computer Systems: A Programmer's Perspective"。现在我不确定我是否理解何时使用不同的 mov 指令。这是练习：
Practice Problem 3.4
Assume variables v and p declared with types
src_t v;
dest_t *p;
where src_t and dest_t are data types declared with typedef. We wish to use the appropriate data movement instruction to implement the operation
*p = (dest_t) v;
where v is stored in the appropriately named portion of register %eax (i.e., %eax, %ax, or %al), while pointer p is stored in register %edx.
For the following combinations of src_t and dest_t, write a line of assembly code that does the appropriate transfer. Recall that when performing a cast that involves both a size change and a change of “signedness” in C, the operation should change the signedness first (Section 2.2.6).
我正在对照 this blog post 检查我的解决方案，但我不确定我是否理解问题 3：
src_t dest_t My Solution Blog's Solution
char unsigned movzbl %al, (%edx) movsbl %al, (%edx)
第 3 条：我使用 movzbl，而博客作者使用 movsbl。在这种情况下，我不明白 movzbl 与 movsbl 的推理......如果 char 是负数，你最终会得到一个错误的数字，任何人都可以澄清为什么 movsbl 是正确的选择吗？
Recall that when performing a cast that involves both a size change and a change of “signedness” in C, the operation should change the signedness first (Section 2.2.6).
除非我看错了，否则这句话似乎与他给出的正确答案相冲突。
该规则意味着，从有符号字符开始：
- 符号字符
- -> 无符号字符
- -> 无符号整数
所以大小转换是从 unsigned char 到 unsigned int，所以 movz 是合适的。大概规则是错误的（与 C 实际指定的不匹配），movs 确实是 C 编译器会使用的规则。
Jester的评论说-1 char值的unsigned int值必须是0xffffffff，所以看起来像练习题最后一句关于转换signedness和size的顺序是问题所在。