将unsigned char []转换为int的问题 [英] Question converting unsigned char [] to int
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。
嗨
首先你需要了解BCD表示之间的关系。
如3表示为0111而不是111. 4将被视为0100
而不是100.所以如果给你一个数字，比如0xe7，这意味着
11100111，将它移到右边4.
int temp = array [3]>> 4;
这意味着
temp = 00001110（= 0x0e）
int temp2 = array [3]<< 4;
temp2 = 01110000（0x70）
类似地你可以通过移位分割所有元素并得到4比特
当量。转换它们以得到确切的BCD号。
希望有所帮助。
badri
>我有一个unsigned char数组（大小为4）：
unsigned char数组[4];
数组[0] = 0x00;
数组[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
任何帮助表示赞赏。
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。
我写道：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
任何帮助表示赞赏。
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.
hi
first u need to understand the relationship between BCD representation.
like 3 is represented as 0111 and not 111. 4 will be considered as 0100
and not 100.so if u r given a number , say 0xe7, which means
11100111,shift it to right by 4.
int temp = array[3] >> 4;
which means
temp = 00001110 (= 0x0e)
int temp2 = array[3] << 4;
temp2 = 01110000 (0x70)
similarly u can split up all the elements by shifting and get the 4 bit
equivalents. convert them to get the exact BCD no.
hope that was of some help.
badri
> 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.
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.
Me wrote: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.
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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