如何将十六进制数转换为整数？ [英] How to convert a hexadecimal number to integer?

问题描述

考虑到我在uint8_t变量中有一个hexa值。现在我该如何将其转换为十进制值？

Consider that I'm having a hexa value in the uint8_t variable. Now how can i convert this to decimal value?

uint8_t one[]={0x00,0x00,0xea,0x60};
int a=*one;
NSLog(@"%d",a);

输出为：0



我试过这样的这样的事情

Output is : 0

I have tried some this like this

uint32_t one={0x0000EA60};
int a=one;
NSLog(@"%d",a);

现在我输出60000.





为什么会这样？

在我的项目中，我喜欢uint8_t变量（这是其他原因的bcz）或者引导我将uint8_t转换为uint32_t变量？

请帮我完成这个通过在第一种情况下获得60000的输出。

提前感谢。

Now i got the output 60000.


Why this happens ?
In my project i'm having like uint8_t variables (this is bcz of other reason) or guide me to convert uint8_t to uint32_t variables?
Please help me to complete this by getting output as 60000 on the first case.
Thanks in advance.

推荐答案

您必须通过添加以创建32位值字节元素移动到正确的位置：

You must create the 32-bit value by adding the byte elements shifted to the correct position:

uint32_t a = (one[3] << 24) | (one[2] << 16) | (one[1] << 8) | one[0];

当然必须是：

It must be of course:

uint32_t a = (one[0] << 24) | (one[1] << 16) | (one[2] << 8) | one[3];

[/ EDIT]



您也可以使用C铸造：

[/EDIT]

You may also use C casting:

uint32_t a = *((uint32_t*)one);

但是演员需要 endianess [ ^ ]（字节顺序）与 one 使用的相同array。

But casting requires that the endianess[^] (byte order) of your system is the same as used by your one array.

int a = *one;

与

is the same as

int a = one[0];

因此结果为零。

hence the zero result.

unsigned char *bytes = {0x00,0x00,0xea,0x60};
NSData *data = [NSData dataWithBytes:bytes length:4];
NSData *timeInt = [data subdataWithRange:NSMakeRange(0,4)];//Ex: 0000ea60
interval = CFSwapInt16BigToHost(*(int*)([timeInt bytes]));
NSLog(@"%d",interval);

输出为：60000： - ）

The Output is : 60000 :-)

这篇关于如何将十六进制数转换为整数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！