因此，如果希望只存储一个字节，且试图在未对齐的地址直接存储，那么它就会进入到不正确的位置并会截断四字内剩余的字节。
Therefore, if you want to store only one byte, if you tried to do it directly on an unaligned address, it would both go into the wrong location and clobber the remaining bytes in the quadword.
如果每个地址引用一个存储字节，那么一个16位物理地址将允许处理器寻址64KB内存。
If each address references a byte of storage, a 16-bit physical address would allow a processor to address 64kb of memory.
返回的是什么？，What, does,malloc, return ？，它返回的是第一个字符，第一个字节的地址，现在在s2中存储的是什么？
Malloc It returns the address of this first byte so really the address of the first char here and so what gets stored in s2 now?
我在s1中存储的是用户输入的字符串的首地址，这样第一个字节存储在这块，新的堆的地方，这个是唯一的修正。
I store in s1 the address of the first byte that the user typed in and by the way that first bite happens to live in this new place called the heap and that's the only update to the story thus far.
big -endian意味着最高位字节存储在最低地址(按高位优先的顺序存储字)。
Big-endian means that the most significant byte has the lowest address (the word is stored big-end-first).
而little -endian意味着最低位字节存储在最低地址(按低位优先的顺序存储字)。
Little-endian means that the least significant byte has the lowest address (the word is stored little-end-first).
Big -endian是将高位字节存储在内存的低地址中，将低位字节存储在内存的高地址中。
Big-endian means that the most significant byte is stored at the lowest memory address and the least significant byte is stored at the highest memory address.
Little - endian是将低位字节存储在内存的低地址中，将高位字节存储在内存的高地址中。
Little-endian means that the least significant byte is stored at the lowest memory address and the most significant byte is stored at the highest memory address.
要让地址为736425的字节具有意义，必须要知道存储在该地址的值的类型。
To give meaning to the byte at address 736425, we must know the type of the value stored there.
要让地址为736425的字节具有意义，必须要知道存储在该地址的值的类型。
To give meaning to the byte at address 736425, we must know the type of the value stored there.
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