将unsigned char []转换为int的问题

问题描述

大家好：


我有一个unsigned char数组（大小为4）：


unsigned char array [4];


array [0] = 0x00;

array [1] = 0x00;

array [2] = 0x02;

array [3] = 0xe7;


我想将array []转换为整数。


0x00 0x00 0x02 0xe7 = 00000000 00000000 00000010 11100111 = 743

感谢任何帮助。

解决方案

C没有'' t保证UCHAR_MAX< = UINT_MAX也不保证UINT_MAX是> = 0xFFFFFFFF

。但是如果

你可以保证数组中的值是< = 0xFF：


#define lengthof（a）（sizeof） （a）/ sizeof（a [0]））

const unsigned char array [4] = {0,0,0x02,0xe7};

unsigned long ui = 0;

unsigned long mult = 1;


/ *

看起来你想要大端转换。对于小的

endian，只需按正常顺序索引数组

* /

for（unsigned i = 0; i< 4; ++ i）{

ui + = mult * array [lengthof（array）-1-i];

mult<< = 8;

}


断言（ui == 743）;


否则用uint32_t和
uint8_t类型在stdint.h中，这可能就是你想要的。

C并不保证UCHAR_MAX< = UINT_MAX没有r是否保证UINT_MAX为> = 0xFFFFFFFF。但是，如果你可以保证数组中的值是< = 0xFF：
#define lengthof（a）（sizeof（a）/ sizeof（a [ 0]））

const unsigned char array [4] = {0,0,0x02,0xe7};
unsigned long ui = 0;
unsigned long mult = 1;

/ *
看起来你想要大端转换。对于小的
endian，只需按正常顺序索引数组
* /

< snip>


你不需要为小端做那个。无论是小b $ b还是大端，数组[0]总是0，

数组[1]总是0，

array [2]将始终为0x02，并且

array [3]将始终为0xe7。

Hi all:

I have an unsigned char array (size 4):

unsigned char array[4];

array[0] = 0x00;
array[1] = 0x00;
array[2] = 0x02;
array[3] = 0xe7;

I would like to convert array[] to an integer.

0x00 0x00 0x02 0xe7 = 00000000 00000000 00000010 11100111 = 743

Any help is appreciated.

解决方案

C doesn''t guarantee that UCHAR_MAX <= UINT_MAX nor does it guarantee
that UINT_MAX is >= 0xFFFFFFFF. But here is something that will work if
you can guarantee the values in the array are <= 0xFF:

#define lengthof(a) (sizeof(a)/sizeof(a[0]))

const unsigned char array[4] = { 0, 0, 0x02, 0xe7 };
unsigned long ui = 0;
unsigned long mult = 1;

/*
it looks like you wanted big endian conversion. For little
endian, just index the array in the regular order
*/
for (unsigned i = 0; i < 4; ++i) {
ui += mult * array[lengthof(array)-1-i];
mult <<= 8;
}

assert(ui == 743);

Otherwise replace unsigned long and unsigned char with the uint32_t and
uint8_t types in stdint.h which is probably what you intended anyway.

C doesn''t guarantee that UCHAR_MAX <= UINT_MAX nor does it guarantee
that UINT_MAX is >= 0xFFFFFFFF. But here is something that will work if
you can guarantee the values in the array are <= 0xFF:

#define lengthof(a) (sizeof(a)/sizeof(a[0]))

const unsigned char array[4] = { 0, 0, 0x02, 0xe7 };
unsigned long ui = 0;
unsigned long mult = 1;

/*
it looks like you wanted big endian conversion. For little
endian, just index the array in the regular order
*/

<snip>

You don''t need to do that for little endian. Whether it is little
or big endian, array[0] will always be 0,
array[1] will always be 0,
array[2] will always be 0x02, and
array[3] will always be 0xe7.

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