由4个字符组成的十六进制数 [英] Number made of 4 chars with hex

问题描述

IAM编程在C + +和IAM与另一个愚蠢的问题。如果我有4个字符，像这样：

Iam programming in C++ and Iam comming with another "stupid" problem. If I have 4 chars like these:

char a = 0x90
char b = 0x01
char c = 0x00
char d = 0x00

当这一切表示十六进制数0x00000190，数。
如何将这些字符转换为一个int？我知道我可以做

when that all means hexadecimal number 0x00000190 which is 400 decimal number. How do I convert these chars to one int? I know i can do

int number = a;

但这只会将一个字符转换为int。

but this will convert only one char to int. Could anybody help please?

推荐答案

您可以使用：

int number = (a & 0xFF)
           | ((b & 0xFF) << 8) 
           | ((c & 0xFF) << 16)
           | ((d & 0xFF) << 24);

使用无符号值会更简单。

It would be simpler with unsigned values.

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