Question

我正在C读到一个方案(一个服务器/客户通信的实施)。

for (i = 0; i < len; i++)
 sprintf(nickmsg+i*2, "%02X", buf[i] & 0xFF);

What does this line do? I don t understand this especially: nickmsg+i*2. nickmsg is a char table and i is an integer. If it was just nickmsg, ok I ll understand but there what s the aim of this line ?

感谢。

Answer 1

Start at the address pointed to by nickmsg and then go an additional i * 2 * CHAR_BIT / 8 bytes in memory. From there, write the hex representation of buf[i] & 0xFF, which will occupy 2 * CHAR_BIT / 8 bytes. Repeat for each i.

Assuming buf looks like

buf[0] = 20
buf[1] = 12

Then the memory pointed to by nickmsg will look like:

nickmsg
|
|
|
+ + + + +
0 2 4 6 8
140C

在<代码><>>/代码>为“无效者”的名词时,在末尾填写“代码>sprintf。

Answer 2

It s converting the values in the buf array to their hexadecimal representation and storing them in the nickmsg array.

由于在<代码>buf <>>/code>中的每一数值上有所变动,因此,通过执行“双向”<代码>AND,以<0xFF的形式提取了最正确的8个轨道。页: 1

然后,它使用显示<代码>%02X”的格式,将每个数值印成2赫克斯数字。

它使用索引<代码>i* 2,储存每台海数。

Answer 3

nickmsg+i*2>正在将镍变数作为加固表的点,然后通过该表逐个条目2。

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