如何转换一个4字节的“字符串"?到uint32_t? [英] How to convert a 4-byte "string" to an uint32_t?

问题描述

基本上，我有一个字节串的数据，例如:\x00\x00\x00\x00 \x08\x00\x00\x00 \x05\x00\x00\x00(空格仅用于可见性，实际字节串中没有空格字节).数据是小端的.

Basically, I have a byte-string of data like: \x00\x00\x00\x00 \x08\x00\x00\x00 \x05\x00\x00\x00 (spaces are used only for visibility, there are no space bytes in the actual byte-string). The data is little-endian.

现在，我需要提取第二个4字节组(\x08\x00\x00\x00，它是128)并将其设置为无符号长.因此，uint32_t类型.

Now, I need to extract the second 4-byte group (\x08\x00\x00\x00, which is 128) and turn them it an unsigned long. So, uint32_t type.

基本上，我正在做的是:moveBlock(&gdata->str[4], &second_group, 4);

Basically, what I'm doing is: moveBlock(&gdata->str[4], &second_group, 4);

moveBlock是宏的位置:#define moveBlock(src,dest,size) memmove(dest,src,size). 我使用该宏是因为如果有人想知道的话，我个人更喜欢参数的顺序.

Where moveBlock is a macro: #define moveBlock(src,dest,size) memmove(dest,src,size). I use the macro because I personally prefer that order of parameters, if someone's wondering.

gdata->str是指向gchar *(的指针ref.) 而gdata是GString *(参考.). second_group被定义为uint32_t.

gdata->str is a pointer to a gchar *(ref. here) and gdata is a GString *(ref. here). second_group is defined as an uint32_t.

因此，这有时 有效，但并非总是如此.老实说，我不知道我在做什么错！

So, this works sometimes, but not always. I honestly don't know what I'm doing wrong!

谢谢！

P.S:代码有点冗长和怪异，我认为遍历所有代码都不重要.除非有人要求，否则我不会不必要地使问题混乱

P.S: The code it a bit lengthy and weird, and I don't think that going through it all would be relevant. Unless someone asks for it, I won't unnecessarily clutter the question

推荐答案

这是干净的便携式版本:

Here's the clean portable version:

unsigned char *p = (void *)data_source;
uint32_t x = p[0] + 256U*p[1] + 65536U*p[2] + 16777216U*p[3];

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